Shell to be modified#include stdlib.h #include unistd.h .pdf
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Shell to be modified #include #include #include #include #include #include #include #include // Simplifed xv6 shell. #define MAXARGS 10 // All commands have at least a type. Have looked at the type, the code // typically casts the *cmd to some specific cmd type. struct cmd { int type; // ' ' (exec), | (pipe), '<' or '>' for redirection }; struct execcmd { int type; // ' ' char *argv[MAXARGS]; // arguments to the command to be exec-ed }; struct redircmd { int type; // < or > struct cmd *cmd; // the command to be run (e.g., an execcmd) char *file; // the input/output file int mode; // the mode to open the file with int fd; // the file descriptor number to use for the file }; struct pipecmd { int type; // | struct cmd *left; // left side of pipe struct cmd *right; // right side of pipe }; int fork1(void); // Fork but exits on failure. struct cmd *parsecmd(char*); // Execute cmd. Never returns. void runcmd(struct cmd *cmd) { int p[2], r; struct execcmd *ecmd; struct pipecmd *pcmd; struct redircmd *rcmd; if(cmd == 0) exit(0); switch(cmd->type){ default: fprintf(stderr, "unknown runcmd\n"); exit(-1); case ' ': ecmd = (struct execcmd*)cmd; if(ecmd->argv[0] == 0) exit(0); fprintf(stderr, "exec not implemented\n"); // Your code here ... break; case '>': case '<': rcmd = (struct redircmd*)cmd; fprintf(stderr, "redir not implemented\n"); // Your code here ... runcmd(rcmd->cmd); break; case '|': pcmd = (struct pipecmd*)cmd; fprintf(stderr, "pipe not implemented\n"); // Your code here ... break; } exit(0); } int getcmd(char *buf, int nbuf) { if (isatty(fileno(stdin))) fprintf(stdout, "$ "); memset(buf, 0, nbuf); fgets(buf, nbuf, stdin); if(buf[0] == 0) // EOF return -1; return 0; } int main(void) { static char buf[100]; int fd, r; // Read and run input commands. while(getcmd(buf, sizeof(buf)) >= 0){ if(buf[0] == 'c' && buf[1] == 'd' && buf[2] == ' '){ // Clumsy but will have to do for now. // Chdir has no effect on the parent if run in the child. buf[strlen(buf)-1] = 0; // chop \n if(chdir(buf+3) < 0) fprintf(stderr, "cannot cd %s\n", buf+3); continue; } if(fork1() == 0) runcmd(parsecmd(buf)); wait(&r); } exit(0); } int fork1(void) { int pid; pid = fork(); if(pid == -1) perror("fork"); return pid; } struct cmd* execcmd(void) { struct execcmd *cmd; cmd = malloc(sizeof(*cmd)); memset(cmd, 0, sizeof(*cmd)); cmd->type = ' '; return (struct cmd*)cmd; } struct cmd* redircmd(struct cmd *subcmd, char *file, int type) { struct redircmd *cmd; cmd = malloc(sizeof(*cmd)); memset(cmd, 0, sizeof(*cmd)); cmd->type = type; cmd->cmd = subcmd; cmd->file = file; cmd->mode = (type == '<') ? O_RDONLY : O_WRONLY|O_CREAT|O_TRUNC; cmd->fd = (type == '<') ? 0 : 1; return (struct cmd*)cmd; } struct cmd* pipecmd(struct cmd *left, struct cmd *right) { struct pipecmd *cmd; cmd = malloc(sizeof(*cmd)); memset(cmd, 0, sizeof(*cmd)); cmd->type = '|'; cmd->left = left; cmd->right = right; return (struct cmd*)cmd; } // Parsing char whitespace[] = " \t\r\n\v"; char symbols[] = "<|>"; int ge.
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![Shell to be modified
#include
#include
#include
#include
#include
#include
#include
#include
// Simplifed xv6 shell.
#define MAXARGS 10
// All commands have at least a type. Have looked at the type, the code
// typically casts the *cmd to some specific cmd type.
struct cmd {
int type; // ' ' (exec), | (pipe), '<' or '>' for redirection
};
struct execcmd {
int type; // ' '
char *argv[MAXARGS]; // arguments to the command to be exec-ed
};
struct redircmd {
int type; // < or >
struct cmd *cmd; // the command to be run (e.g., an execcmd)
char *file; // the input/output file
int mode; // the mode to open the file with
int fd; // the file descriptor number to use for the file
};
struct pipecmd {](https://image.slidesharecdn.com/shelltobemodifiedincludestdlib-231007064549-dcc48cdb/85/Shell-to-be-modified-include-stdlib-h-include-unistd-h-pdf-1-320.jpg)
![int type; // |
struct cmd *left; // left side of pipe
struct cmd *right; // right side of pipe
};
int fork1(void); // Fork but exits on failure.
struct cmd *parsecmd(char*);
// Execute cmd. Never returns.
void
runcmd(struct cmd *cmd)
{
int p[2], r;
struct execcmd *ecmd;
struct pipecmd *pcmd;
struct redircmd *rcmd;
if(cmd == 0)
exit(0);
switch(cmd->type){
default:
fprintf(stderr, "unknown runcmdn");
exit(-1);
case ' ':
ecmd = (struct execcmd*)cmd;
if(ecmd->argv[0] == 0)
exit(0);
fprintf(stderr, "exec not implementedn");
// Your code here ...
break;
case '>':
case '<':
rcmd = (struct redircmd*)cmd;](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
httplinux.die.netman3execfork() creates a new process by.docx
http://linux.die.net/man/3/exec
fork() creates a new process by duplicating the calling process. The new process, referred to as the child, is an exact duplicate of the calling process, referred to as the parent#include <unistd.h>pid_t fork(void);
The exec() family of functions replaces the current process image with a new process image. The functions described in this manual page are front-ends for execve(2). (See the manual page for execve(2) for further details about the replacement of the current process image.)
The exec() family of functions include execl, execlp, execle, execv, execvp, and execvpe to execute a file.
The ANSI prototype for execl() is:
int execl(const char *path, const char *arg0,..., const char *argn, 0)
http://www.cems.uwe.ac.uk/~irjohnso/coursenotes/lrc/system/pc/pc4.htm #inciude <stdio.h> #inciude <unistd.h> main() { execl("/bin/ls", "ls", "-l", 0); printf("Can only get here on error\n"); }
The first parameter to execl() in this example is the full pathname to the ls command. This is the file whose contents will be run, provided the process has execute permission on the file. The rest of the execl() parameters provide the strings to which the argv array elements in the new program will point. In this example, it means that the ls program will see the string ls pointed to by its argv[0], and the string -l pointed to by itsargv[1]. In addition to making all these parameters available to the new program, the exec() calls also pass a value for the variable: extern char **environ;
This variable has the same format as the argv variable except that the items passed via environ are the values in the environment of the process (like any exported shell variables), rather than the command line parameters. In the case of execl(), the value of the environ variable in the new program will be a copy of the value of this variable in the calling process.
The execl() version of exec() is fine in the circumstances where you can ex-plicitly list all of the parameters, as in the previous example. Now suppose you want to write a program that doesn't just run ls, but will run any program you wish, and pass it any number of appropriate command line parameters. Obviously, execl() won't do the job.
The example program below, which implements this requirement, shows, however, that the system call execv() will perform as required: #inciude <stdio.h> main(int argc, char **argv) { if (argc==1) { printf("Usage: run <command> [<paraneters>]\n"); exit(1) } execv(argv[l], &argv[1)); printf("Sorry... couldn't run that!\n"); }
The prototype for execv() shows that it only takes two parameters, the first is the full pathname to the command to execute and the second is the argv value you want to pass into the new program. In the previous example this value was derived from the argv value passed into the run command, so that the run command can take the command line parameter values you pass it and just pass them on. int execl(.
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txt
sleep 5
ls -latr
sleep 3
pwd
sleep 1
wc /etc/passwd
CommandNode.c
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "CommandNode.h"
/**
*CommandNode.c
*The function bodies for a linked list of commands
**/
/**
Example of linked list initialization in code:
//First command node (head of list)
headCommand = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(headCommand, command, index, NULL);
index = 1;
.....
//Later command nodes
while ((linesize = getline(&line, &len, fp)) >= 0) {
...
index++;
nextCommand1 = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(nextCommand1, command, index, NULL);
InsertCommandAfter(headCommand, nextCommand1);
}
**/
//create a new command node. usually nextCmd can be NULL and function InsertCommandAfter can be called to insert after head node.
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd) {
//this is useful if you store a string (char *): strcpy(thisNode->command, cmd);
for (int i = 0; i < 20; i++)
for (int j = 0; j < 20; j++)
thisNode->command[i][j] = cmd[i][j];
thisNode->index = ind;
thisNode->nextCommandPtr = nextCmd;
return;
}
//insert node newNode after thisNode
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode) {
CommandNode* tmpNext = NULL;
tmpNext = thisNode->nextCommandPtr;
thisNode->nextCommandPtr = newNode;
newNode->nextCommandPtr = tmpNext;
return;
}
//get next command node in linked list
CommandNode* GetNextCommand(CommandNode* thisNode) {
return thisNode->nextCommandPtr;
}
//find a command based on the pid
CommandNode* FindCommand(CommandNode* cmd, int pid) {
CommandNode* tmpNext = cmd;
while (tmpNext != NULL) {
if (tmpNext->PID == pid) { return tmpNext; }
tmpNext = tmpNext->nextCommandPtr;
}
return NULL;
}
CommandNode.h
/**
*CommandNode.h
*The struct for a Command Node and function prototypes for a linked list of commands
**/
typedef struct command_struct {
char command[20][20];
int index;
int PID;
int starttime;
struct command_struct* nextCommandPtr;
} CommandNode;
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd);
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode);
CommandNode* GetNextCommand(CommandNode* thisNode);
CommandNode* FindCommand(CommandNode* cmd, int pid);
HWdetails&helperFile/HW_details1.jpg
HWdetails&helperFile/HW_details1_1.jpg
HWdetails&helperFile/read_parse_file.c
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
FILE * fp;
char * line = NULL;
size_t len = 0;
ssize_t read;
char newString[20][20];
int i,j,ctr;
fp = fopen("commands.txt", "r");
if (fp == NULL)
exit.
Help Needed!UNIX Shell and History Feature This project consists.pdf
Help Needed!
UNIX Shell and History Feature
This project consists of designing a C program to serve as a shell interface
that accepts user commands and then executes each command in a separate
process. This project can be completed on any Linux,
UNIX,orMacOS X system.
A shell interface gives the user a prompt, after which the next command
is entered. The example below illustrates the prompt
osh> and the user’s
next command:
cat prog.c. (This command displays the le prog.c on the
terminal using the
UNIX cat command.)
osh> cat prog.c
One technique for implementing a shell interface is to have the parent process
rst read what the user enters on the command line (in this case,
cat
prog.c), and then create a separate child process that performs the command.
Unless otherwise specied, the parent process waits for the child to exit
before continuing. This is similar in functionality to the new process creation
illustrated in Figure 3.10. However,
UNIX shells typically also allow the child
process to run in the background, or concurrently. To accomplish this, we add
an ampersand (&) at the end of the command. Thus, if we rewrite the above
command as
osh> cat prog.c &
the parent and child processes will run concurrently.
The separate child process is created using the
fork() system call, and the
user’s command is executed using one of the system calls in the
exec() family
A C program that provides the general operations of a command-line shell
is supplied in Figure 3.36. The
main() function presents the prompt osh->
and outlines the steps to be taken after input from the user has been read. The
main() function continually loops as long as should run equals 1; when the
user enters
exit at the prompt, your program will set should run to 0 and
terminate.
This project is organized into two parts: (1) creating the child process and
executing the command in the child, and (2) modifying the shell to allow a
history feature.
#include
#include
#define MAXLINE 80 /* The maximum length command */
int main(void)
{
char *args[MAXLINE/2 + 1]; /* command line arguments */
int should
run = 1; /* flag to determine when to exit program */
while (should run) {
printf(\"osh>\");
}
fflush(stdout);
/**
* After reading user input, the steps are:
* (1) fork a child process using fork()
* (2) the child process will invoke execvp()
* (3) if command included &, parent will invoke wait()
*/
return 0;
}
Part I — Creating a Child Process
The rst task is to modify the
main() function in the above program so that a child
process is forked and executes the command specied by the user. This will
require parsing what the user has entered into separate tokens and storing the
tokens in an array of character strings (
args in the above program. For example, if the
user enters the command
ps -ael at the osh> prompt, the values stored in the
args array are:
args[0] = \"ps\"
args[1] = \"-ael\"
args[2] = NULL
This args array will be passed to the execvp() function, which has the
following prot.
Степан Кольцов — Rust — лучше, чем C++
Rust — это современный, практический, быстрый и безопасный язык программирования. Некоторые говорят, что Rust — это как C++, если бы его писал человек, знающий Haskell.
Система типов Rust решает главную проблему C++ — небезопасность. C++ очень легко сделать ошибки, которые приведут к поломкам (например, use after free). Rust позволяет писать безопасный код, сохраняя при этом выразительность и околонулевые накладные расходы C++. В докладе будут подробно описаны механизмы языка, которые контролируют безопасность программы.
Хотя в данный момент Rust ещё не подходит для использования в продакшне, его всё равно стоит изучать. Во-первых, потому что это очень интересный подход к программированию, а во-вторых, потому что через несколько лет для разработки требовательных к ресурсам программ будет необходим именно Rust или другой похожий инструмент.
Program Assignment Process ManagementObjective This program a.docx
Program Assignment : Process Management
Objective: This program assignment is given to the Operating Systems course to allow the students to figure out how a single process (parent process) creates a child process and how they work on Unix/Linux(/Mac OS X/Windows) environment. Additionally, student should combine the code for describing inter-process communication into this assignment. Both parent and child processes interact with each other through shared memory-based communication scheme or message passing scheme.
Environment: Unix/Linux environment (VM Linux or Triton Server, or Mac OS X), Windows platform
Language: C or C++, Java
Requirements:
i. You have wide range of choices for this assignment. First, design your program to explain the basic concept of the process management in Unix Kernel. This main idea will be evolved to show your understanding on inter-process communication, file processing, etc.
ii. Refer to the following system calls:
- fork(), getpid(), family of exec(), wait(), sleep() system calls for process management
- shmget(), shmat(), shmdt(), shmctl() for shared memory support or
- msgget(), msgsnd(), msgrcv(), msgctl(), etc. for message passing support
iii. The program should present that two different processes, both parent and child, execute as they are supposed to.
iv. The output should contain the screen capture of the execution procedure of the program.
v. Interaction between parent and child processes can be provided through inter-process communication schemes, such as shared-memory or message passing schemes.
vi. Result should be organized as a document which explains the overview of your program, code, execution results, and the conclusion including justification of your program, lessons you've learned, comments, etc.
Note:
i. In addition, please try to understand how the local and global variables work across the processes
ii. read() or write () functions are used to understand how they work on the different processes.
iii. For extra credit, you can also incorporate advanced features, like socket or thread functions, into your code.
Examples:
1. Process Creation and IPC with Shared Memory Scheme
=============================================================
#include <stdio.h>
#include <sys/shm.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>
int main(){
pid_t pid;
int segment_id; //allocate the memory
char *shared_memory; //pointer to memory
const int size = 4096;
segment_id = shmget(IPC_PRIVATE, size, S_IRUSR | S_IWUSR);
shared_memory = (char *) shmat(segment_id, NULL, 0);
pid = fork();
if(pid < 0) { //error
fprintf(stderr, "Fork failed");
return 1;
}
else if(pid == 0){ //child process
char *child_shared_memory;
child_shared_memory = (char *) shmat(segment_id,NULL,0); //attach mem
sprintf(child_shared_memory, "Hello parent process!"); //write to the shared mem
shmdt(child_shared_memory);
}
else ...
-- This is the shell-c Test- --shell -test sub #include -ctype-h- -- C.pdf
// This is the shell.c
Test: ./shell -test sub
#include <ctype.h> /* Character types */
#include <stdio.h> /* Standard buffered input/output */
#include <stdlib.h> /* Standard library functions */
#include <string.h> /* String operations */
#include <sys/types.h> /* Data types */
#include <sys/wait.h> /* Declarations for waiting */
#include <unistd.h> /* Standard symbolic constants and types */
#include "smp1_tests.h" /* Built-in test system */
/* DEFINE SECTION */
#define SHELL_BUFFER_SIZE 256 /* Size of the Shell input buffer */
#define SHELL_MAX_ARGS 8 /* Maximum number of arguments parsed */
#define SHELL_HISTORY_SIZE 10
/* VARIABLE SECTION */
enum { STATE_SPACE, STATE_NON_SPACE }; /* Parser states */
char shell_history[SHELL_HISTORY_SIZE][SHELL_BUFFER_SIZE];
int shell_history_next_index = 0;
int imthechild(const char *path_to_exec, char *const args[])
{
// TO-DO P5.1
return execvp(path_to_exec, args) ? -1 : 0;
}
void imtheparent(pid_t child_pid, int run_in_background)
{
int child_return_val, child_error_code;
/* fork returned a positive pid so we are the parent */
fprintf(stderr,
" Parent says 'child process has been forked with pid=%d'\n",
child_pid);
if (run_in_background) {
fprintf(stderr,
" Parent says 'run_in_background=1 ... so we're not waiting for the child'\n");
return;
}
// TO-DO P5.4
wait(&child_return_val);
/* Use the WEXITSTATUS to extract the status code from the return value */
child_error_code = WEXITSTATUS(child_return_val);
fprintf(stderr,
" Parent says 'wait() returned so the child with pid=%d is finished.'\n",
child_pid);
if (child_error_code != 0) {
fprintf(stderr,
" Parent says 'Child process %d failed with code %d'\n",
child_pid, child_error_code);
}
}
int get_shell_command(char *buffer, int size)
{
int i = 0;
char c;
while (i < size - 1 && (c = getchar()) != EOF && c != '\n') {
buffer[i++] = c;
}
buffer[i] = '\0';
return i;
}
void parse_command(char *command, char **exec_args, int *exec_bg)
{
int argc = 0;
*exec_bg = 0;
while (*command != '\0') {
/* Strip whitespace. */
while (isspace(*command)) {
++command;
}
if (*command == '\0') {
break;
}
/* Save the argument. */
if (argc < SHELL_MAX_ARGS) {
exec_args[argc++] = command;
}
while (*command != '\0' && !isspace(*command)) {
++command;
}
if (*command != '\0') {
*command++ = '\0';
}
}
exec_args[argc] = NULL;
/* Check for background execution request. */
if (argc > 0 && !strcmp(exec_args[argc - 1], "&")) {
*exec_bg = 1;
exec_args[--argc] = NULL;
}
}
void run_shell_command;
int main(int argc, char **argv)
{
pid_t shell_pid, pid_from_fork;
int n_read, i, exec_argc, parser_state, run_in_background;
char buffer[SHELL_BUFFER_SIZE];
char *exec_argv[SHELL_MAX_ARGS + 1];
// TO-DO new variables for P5.2, P5.3, P5.6
if (argc > 1 && !strcmp(argv[1], "-test"))
{
return run_smp1_tests(argc - 1, argv + 1);
}
shell_pid = getpid();
int command_counter = 1;
while (1)
{
/* The Shell runs in an infinite loop, processing input. */
if (exec_argc > 0)
{
command_counter++;
}
// TO-DO P5.2
.
IO redirection in C shellPlease implement input output redirect.pdf
I/O redirection in C shell
Please implement input / output redirection in the code posted below. Here are the specs:
Your shell will need to support file redirection. Use the same syntax as defined in the Bash shell:
a single \'>\' implies that one needs to redirect the standard output of the program being started to
the referenced file while a single \'<\' implies the same with standard input. The double \'>>\'
implies that standard output will append to an existing file rather than create a new file (similar
behavior is required for the \'<<\' operator and standard input). You do not need to implement
support for the Bash pipeline operator \'|\'.
Before calling exec to begin execution, the child process may have to close stdin (file desriptor
0) and/or stdout (file descriptor 0), open the corresponding file and use the dup2 system call to
make it the appropriate file descriptor. Don\'t forget to use the close system call to close the old
file descriptor.
#include
#include
#include
#include
#include
#include
using namespace std;
int change_dir(char* args[])
{
if (args[1] == NULL)
{
chdir(getenv(\"HOME\"));
return 1;
}
else
{
if (chdir(args[1]) == -1)
{
cout << \"No such directory\" << args[1];
return -1;
}
}
return 0;
}
int helper_func(char* args[])
{
cout << \"Builtin commands so far are; cd, and help.\ \";
return 0;
}
int main()
{
while(true)
{
cout << \"$ \";
char cmd[128];
cin.getline(cmd,128);
vector args;
char *line = strtok(cmd, \" \");
char *tmp = line;
while (tmp != NULL)
{
args.push_back(tmp);
tmp = strtok(NULL, \" \");
}
char** argv = new char*[args.size() + 1];
for (int i = 0; i < args.size(); i++)
argv[i] = args[i];
argv[args.size()] = NULL;
if (strcmp(cmd, \"exit\") == 0)
{
{
exit(0);
}
}
//else if (strcmp(args[0], \"cd\") == 0) change_dir(args);
if (strcmp(cmd, \"cd\") == 0)
{
change_dir(argv);
}
if (strcmp(cmd, \"help\") == 0)
{
helper_func(argv);
}
else
{
pid_t pid;
pid_t wpid;
int status;
pid = fork();
if (pid == 0)
{
if (execvp(args[0], argv) == -1)
{
perror(\"cmd\");
}
exit(EXIT_FAILURE);
}
else if (pid < 0)
{
perror(\"cmd\");
}
{
do {
wpid = waitpid(pid, &status, WUNTRACED);
} while (!WIFEXITED(status) && !WIFSIGNALED(status));
}
}
}
signal(SIGINT, SIG_IGN);
signal(SIGTERM, SIG_DFL);
return 0;
}
Solution
#include
#include
#include
#include
#include
#include
using namespace std;
int change_dir(char* args[])
{
if (args[1] == NULL)
{
chdir(getenv(\"HOME\"));
return 1;
}
else
{
if (chdir(args[1]) == -1)
{
cout << \"No such directory\" << args[1];
return -1;
}
}
return 0;
}
int helper_func(char* args[])
{
cout << \"Builtin commands so far are; cd, and help.\ \";
return 0;
}
int main()
{
while(true)
{
cout << \"$ \";
char cmd[128];
cin.getline(cmd,128);
vector args;
char *line = strtok(cmd, \" \");
char *tmp = line;
while (tmp != NULL)
{
args.push_back(tmp);
tmp = strtok(NULL, \" \");
}
char** argv = new char*[args.size() + 1];
for (int i = 0; i < args.size(); i++)
argv[i] = args[i];
argv[args.size()] = NULL;
if (strcmp(cmd, \.
Add a 3rd field help that contains a short help string for each of t.pdf
Add a 3rd field help that contains a short help string for each of the commands you were to
implement in assignment #3. Make sure that your array(s) are big enough to handle 5 extra items
beyond your initialization. To save time only include help for exercises 4, 5, 6 and 8 in this
assignment, and use No help for the other entries.
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
void ctrlCHandler(int signum)
{
fprintf(stderr,"Command server terminated using C\n");
exit(1);
}
char * fExport(char *cmd, char *tokensleft[])
{
setenv(tokensleft[0],tokensleft[1],1);
return "Command 'export' was received";
}
char * fChdir(char *cmd,char *tokensleft[])
{
int ch=chdir(tokensleft[0]);
if(ch<0)
perror("chdir change of directory not successful\n");
else
printf("chdir change of directory successful\n");
return "Command 'chdir' was received\n";
}
char * fAccess(char *cmd,char *tokensleft[])
{
int exists =0;
for(int i=0;tokensleft[i]; i++) {
exists =0;
if(access(tokensleft[i],F_OK)==0){
exists = 1;
printf("file %s exists\n",tokensleft[i]);
}else{
printf("file %s does not exists\n",tokensleft[i]);
}
if (exists == 1){
if(access(tokensleft[i],R_OK)==0) {
printf("file %s is readable\n",tokensleft[i]);
}else{printf("file %s is not readable\n",tokensleft[i]);}
if(access(tokensleft[i],W_OK)==0) {
printf("file %s is writeable\n",tokensleft[i]);
}else{
printf("file %s is not writeable\n",tokensleft[i]);
}
if(access(tokensleft[i],X_OK)==0) {
printf("file %s is executeable\n",tokensleft[i]);
}else{
printf("file %s is not executeable\n",tokensleft[i]);
}
}// exists if
} //for
return "Command 'acsess' was received\n";
}
char * fChmod(char *cmd,char *tokensleft[])
{
unsigned int octalPerm;
sscanf(tokensleft[0],"%o",&octalPerm);
for(int i=1;tokensleft[i]; i++) {
if(chmod(tokensleft[i],octalPerm)==0 ){
chmod(tokensleft[i],octalPerm);
}else{
printf("Error: %s \n",strerror(errno));
}
}
return "Command 'chmod' was received";
}
char * fPath(char *cmd,char *tokensleft[])
{
char *pathLink;
char actualPath[PATH_MAX+1];
char *pointer;
char *bName;
char *dName;
for(int i=0;tokensleft[i]; i++) {
pathLink = tokensleft[i];
pointer =realpath(pathLink,actualPath);
bName = basename(actualPath);
dName = dirname(tokensleft[i]);
if (pointer){
printf("The Real path of %s is: %s\n",tokensleft[i],actualPath);
printf("The Dir name path of %s is: %s\n",tokensleft[i],dName);
printf("The Base name of %s is: %s\n",tokensleft[i],bName);
}else{
printf("Error: %s \n",strerror(errno));
}
}
return "Command 'path' was received";
}
char * fTouch(char *cmd,char *tokensleft[])
{
extern int optind,optopt,opterr;
struct FLAG{
bool aFlag;
bool mFlag;
} flags = { false, false };
int t1 = time(NULL), t2 = time(NULL);
int argc = 0;
int flag;
for (int i = 0; tokensleft[i]; i++) {
argc++;
}
while ((flag = getopt(argc, tokensleft, "m.
Recommended
httplinux.die.netman3execfork() creates a new process by.docx
http://linux.die.net/man/3/exec
fork() creates a new process by duplicating the calling process. The new process, referred to as the child, is an exact duplicate of the calling process, referred to as the parent#include <unistd.h>pid_t fork(void);
The exec() family of functions replaces the current process image with a new process image. The functions described in this manual page are front-ends for execve(2). (See the manual page for execve(2) for further details about the replacement of the current process image.)
The exec() family of functions include execl, execlp, execle, execv, execvp, and execvpe to execute a file.
The ANSI prototype for execl() is:
int execl(const char *path, const char *arg0,..., const char *argn, 0)
http://www.cems.uwe.ac.uk/~irjohnso/coursenotes/lrc/system/pc/pc4.htm #inciude <stdio.h> #inciude <unistd.h> main() { execl("/bin/ls", "ls", "-l", 0); printf("Can only get here on error\n"); }
The first parameter to execl() in this example is the full pathname to the ls command. This is the file whose contents will be run, provided the process has execute permission on the file. The rest of the execl() parameters provide the strings to which the argv array elements in the new program will point. In this example, it means that the ls program will see the string ls pointed to by its argv[0], and the string -l pointed to by itsargv[1]. In addition to making all these parameters available to the new program, the exec() calls also pass a value for the variable: extern char **environ;
This variable has the same format as the argv variable except that the items passed via environ are the values in the environment of the process (like any exported shell variables), rather than the command line parameters. In the case of execl(), the value of the environ variable in the new program will be a copy of the value of this variable in the calling process.
The execl() version of exec() is fine in the circumstances where you can ex-plicitly list all of the parameters, as in the previous example. Now suppose you want to write a program that doesn't just run ls, but will run any program you wish, and pass it any number of appropriate command line parameters. Obviously, execl() won't do the job.
The example program below, which implements this requirement, shows, however, that the system call execv() will perform as required: #inciude <stdio.h> main(int argc, char **argv) { if (argc==1) { printf("Usage: run <command> [<paraneters>]\n"); exit(1) } execv(argv[l], &argv[1)); printf("Sorry... couldn't run that!\n"); }
The prototype for execv() shows that it only takes two parameters, the first is the full pathname to the command to execute and the second is the argv value you want to pass into the new program. In the previous example this value was derived from the argv value passed into the run command, so that the run command can take the command line parameter values you pass it and just pass them on. int execl(.
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txt
sleep 5
ls -latr
sleep 3
pwd
sleep 1
wc /etc/passwd
CommandNode.c
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "CommandNode.h"
/**
*CommandNode.c
*The function bodies for a linked list of commands
**/
/**
Example of linked list initialization in code:
//First command node (head of list)
headCommand = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(headCommand, command, index, NULL);
index = 1;
.....
//Later command nodes
while ((linesize = getline(&line, &len, fp)) >= 0) {
...
index++;
nextCommand1 = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(nextCommand1, command, index, NULL);
InsertCommandAfter(headCommand, nextCommand1);
}
**/
//create a new command node. usually nextCmd can be NULL and function InsertCommandAfter can be called to insert after head node.
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd) {
//this is useful if you store a string (char *): strcpy(thisNode->command, cmd);
for (int i = 0; i < 20; i++)
for (int j = 0; j < 20; j++)
thisNode->command[i][j] = cmd[i][j];
thisNode->index = ind;
thisNode->nextCommandPtr = nextCmd;
return;
}
//insert node newNode after thisNode
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode) {
CommandNode* tmpNext = NULL;
tmpNext = thisNode->nextCommandPtr;
thisNode->nextCommandPtr = newNode;
newNode->nextCommandPtr = tmpNext;
return;
}
//get next command node in linked list
CommandNode* GetNextCommand(CommandNode* thisNode) {
return thisNode->nextCommandPtr;
}
//find a command based on the pid
CommandNode* FindCommand(CommandNode* cmd, int pid) {
CommandNode* tmpNext = cmd;
while (tmpNext != NULL) {
if (tmpNext->PID == pid) { return tmpNext; }
tmpNext = tmpNext->nextCommandPtr;
}
return NULL;
}
CommandNode.h
/**
*CommandNode.h
*The struct for a Command Node and function prototypes for a linked list of commands
**/
typedef struct command_struct {
char command[20][20];
int index;
int PID;
int starttime;
struct command_struct* nextCommandPtr;
} CommandNode;
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd);
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode);
CommandNode* GetNextCommand(CommandNode* thisNode);
CommandNode* FindCommand(CommandNode* cmd, int pid);
HWdetails&helperFile/HW_details1.jpg
HWdetails&helperFile/HW_details1_1.jpg
HWdetails&helperFile/read_parse_file.c
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
FILE * fp;
char * line = NULL;
size_t len = 0;
ssize_t read;
char newString[20][20];
int i,j,ctr;
fp = fopen("commands.txt", "r");
if (fp == NULL)
exit.
Help Needed!UNIX Shell and History Feature This project consists.pdf
Help Needed!
UNIX Shell and History Feature
This project consists of designing a C program to serve as a shell interface
that accepts user commands and then executes each command in a separate
process. This project can be completed on any Linux,
UNIX,orMacOS X system.
A shell interface gives the user a prompt, after which the next command
is entered. The example below illustrates the prompt
osh> and the user’s
next command:
cat prog.c. (This command displays the le prog.c on the
terminal using the
UNIX cat command.)
osh> cat prog.c
One technique for implementing a shell interface is to have the parent process
rst read what the user enters on the command line (in this case,
cat
prog.c), and then create a separate child process that performs the command.
Unless otherwise specied, the parent process waits for the child to exit
before continuing. This is similar in functionality to the new process creation
illustrated in Figure 3.10. However,
UNIX shells typically also allow the child
process to run in the background, or concurrently. To accomplish this, we add
an ampersand (&) at the end of the command. Thus, if we rewrite the above
command as
osh> cat prog.c &
the parent and child processes will run concurrently.
The separate child process is created using the
fork() system call, and the
user’s command is executed using one of the system calls in the
exec() family
A C program that provides the general operations of a command-line shell
is supplied in Figure 3.36. The
main() function presents the prompt osh->
and outlines the steps to be taken after input from the user has been read. The
main() function continually loops as long as should run equals 1; when the
user enters
exit at the prompt, your program will set should run to 0 and
terminate.
This project is organized into two parts: (1) creating the child process and
executing the command in the child, and (2) modifying the shell to allow a
history feature.
#include
#include
#define MAXLINE 80 /* The maximum length command */
int main(void)
{
char *args[MAXLINE/2 + 1]; /* command line arguments */
int should
run = 1; /* flag to determine when to exit program */
while (should run) {
printf(\"osh>\");
}
fflush(stdout);
/**
* After reading user input, the steps are:
* (1) fork a child process using fork()
* (2) the child process will invoke execvp()
* (3) if command included &, parent will invoke wait()
*/
return 0;
}
Part I — Creating a Child Process
The rst task is to modify the
main() function in the above program so that a child
process is forked and executes the command specied by the user. This will
require parsing what the user has entered into separate tokens and storing the
tokens in an array of character strings (
args in the above program. For example, if the
user enters the command
ps -ael at the osh> prompt, the values stored in the
args array are:
args[0] = \"ps\"
args[1] = \"-ael\"
args[2] = NULL
This args array will be passed to the execvp() function, which has the
following prot.
Степан Кольцов — Rust — лучше, чем C++
Rust — это современный, практический, быстрый и безопасный язык программирования. Некоторые говорят, что Rust — это как C++, если бы его писал человек, знающий Haskell.
Система типов Rust решает главную проблему C++ — небезопасность. C++ очень легко сделать ошибки, которые приведут к поломкам (например, use after free). Rust позволяет писать безопасный код, сохраняя при этом выразительность и околонулевые накладные расходы C++. В докладе будут подробно описаны механизмы языка, которые контролируют безопасность программы.
Хотя в данный момент Rust ещё не подходит для использования в продакшне, его всё равно стоит изучать. Во-первых, потому что это очень интересный подход к программированию, а во-вторых, потому что через несколько лет для разработки требовательных к ресурсам программ будет необходим именно Rust или другой похожий инструмент.
Program Assignment Process ManagementObjective This program a.docx
Program Assignment : Process Management
Objective: This program assignment is given to the Operating Systems course to allow the students to figure out how a single process (parent process) creates a child process and how they work on Unix/Linux(/Mac OS X/Windows) environment. Additionally, student should combine the code for describing inter-process communication into this assignment. Both parent and child processes interact with each other through shared memory-based communication scheme or message passing scheme.
Environment: Unix/Linux environment (VM Linux or Triton Server, or Mac OS X), Windows platform
Language: C or C++, Java
Requirements:
i. You have wide range of choices for this assignment. First, design your program to explain the basic concept of the process management in Unix Kernel. This main idea will be evolved to show your understanding on inter-process communication, file processing, etc.
ii. Refer to the following system calls:
- fork(), getpid(), family of exec(), wait(), sleep() system calls for process management
- shmget(), shmat(), shmdt(), shmctl() for shared memory support or
- msgget(), msgsnd(), msgrcv(), msgctl(), etc. for message passing support
iii. The program should present that two different processes, both parent and child, execute as they are supposed to.
iv. The output should contain the screen capture of the execution procedure of the program.
v. Interaction between parent and child processes can be provided through inter-process communication schemes, such as shared-memory or message passing schemes.
vi. Result should be organized as a document which explains the overview of your program, code, execution results, and the conclusion including justification of your program, lessons you've learned, comments, etc.
Note:
i. In addition, please try to understand how the local and global variables work across the processes
ii. read() or write () functions are used to understand how they work on the different processes.
iii. For extra credit, you can also incorporate advanced features, like socket or thread functions, into your code.
Examples:
1. Process Creation and IPC with Shared Memory Scheme
=============================================================
#include <stdio.h>
#include <sys/shm.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>
int main(){
pid_t pid;
int segment_id; //allocate the memory
char *shared_memory; //pointer to memory
const int size = 4096;
segment_id = shmget(IPC_PRIVATE, size, S_IRUSR | S_IWUSR);
shared_memory = (char *) shmat(segment_id, NULL, 0);
pid = fork();
if(pid < 0) { //error
fprintf(stderr, "Fork failed");
return 1;
}
else if(pid == 0){ //child process
char *child_shared_memory;
child_shared_memory = (char *) shmat(segment_id,NULL,0); //attach mem
sprintf(child_shared_memory, "Hello parent process!"); //write to the shared mem
shmdt(child_shared_memory);
}
else ...
-- This is the shell-c Test- --shell -test sub #include -ctype-h- -- C.pdf
// This is the shell.c
Test: ./shell -test sub
#include <ctype.h> /* Character types */
#include <stdio.h> /* Standard buffered input/output */
#include <stdlib.h> /* Standard library functions */
#include <string.h> /* String operations */
#include <sys/types.h> /* Data types */
#include <sys/wait.h> /* Declarations for waiting */
#include <unistd.h> /* Standard symbolic constants and types */
#include "smp1_tests.h" /* Built-in test system */
/* DEFINE SECTION */
#define SHELL_BUFFER_SIZE 256 /* Size of the Shell input buffer */
#define SHELL_MAX_ARGS 8 /* Maximum number of arguments parsed */
#define SHELL_HISTORY_SIZE 10
/* VARIABLE SECTION */
enum { STATE_SPACE, STATE_NON_SPACE }; /* Parser states */
char shell_history[SHELL_HISTORY_SIZE][SHELL_BUFFER_SIZE];
int shell_history_next_index = 0;
int imthechild(const char *path_to_exec, char *const args[])
{
// TO-DO P5.1
return execvp(path_to_exec, args) ? -1 : 0;
}
void imtheparent(pid_t child_pid, int run_in_background)
{
int child_return_val, child_error_code;
/* fork returned a positive pid so we are the parent */
fprintf(stderr,
" Parent says 'child process has been forked with pid=%d'\n",
child_pid);
if (run_in_background) {
fprintf(stderr,
" Parent says 'run_in_background=1 ... so we're not waiting for the child'\n");
return;
}
// TO-DO P5.4
wait(&child_return_val);
/* Use the WEXITSTATUS to extract the status code from the return value */
child_error_code = WEXITSTATUS(child_return_val);
fprintf(stderr,
" Parent says 'wait() returned so the child with pid=%d is finished.'\n",
child_pid);
if (child_error_code != 0) {
fprintf(stderr,
" Parent says 'Child process %d failed with code %d'\n",
child_pid, child_error_code);
}
}
int get_shell_command(char *buffer, int size)
{
int i = 0;
char c;
while (i < size - 1 && (c = getchar()) != EOF && c != '\n') {
buffer[i++] = c;
}
buffer[i] = '\0';
return i;
}
void parse_command(char *command, char **exec_args, int *exec_bg)
{
int argc = 0;
*exec_bg = 0;
while (*command != '\0') {
/* Strip whitespace. */
while (isspace(*command)) {
++command;
}
if (*command == '\0') {
break;
}
/* Save the argument. */
if (argc < SHELL_MAX_ARGS) {
exec_args[argc++] = command;
}
while (*command != '\0' && !isspace(*command)) {
++command;
}
if (*command != '\0') {
*command++ = '\0';
}
}
exec_args[argc] = NULL;
/* Check for background execution request. */
if (argc > 0 && !strcmp(exec_args[argc - 1], "&")) {
*exec_bg = 1;
exec_args[--argc] = NULL;
}
}
void run_shell_command;
int main(int argc, char **argv)
{
pid_t shell_pid, pid_from_fork;
int n_read, i, exec_argc, parser_state, run_in_background;
char buffer[SHELL_BUFFER_SIZE];
char *exec_argv[SHELL_MAX_ARGS + 1];
// TO-DO new variables for P5.2, P5.3, P5.6
if (argc > 1 && !strcmp(argv[1], "-test"))
{
return run_smp1_tests(argc - 1, argv + 1);
}
shell_pid = getpid();
int command_counter = 1;
while (1)
{
/* The Shell runs in an infinite loop, processing input. */
if (exec_argc > 0)
{
command_counter++;
}
// TO-DO P5.2
.
IO redirection in C shellPlease implement input output redirect.pdf
I/O redirection in C shell
Please implement input / output redirection in the code posted below. Here are the specs:
Your shell will need to support file redirection. Use the same syntax as defined in the Bash shell:
a single \'>\' implies that one needs to redirect the standard output of the program being started to
the referenced file while a single \'<\' implies the same with standard input. The double \'>>\'
implies that standard output will append to an existing file rather than create a new file (similar
behavior is required for the \'<<\' operator and standard input). You do not need to implement
support for the Bash pipeline operator \'|\'.
Before calling exec to begin execution, the child process may have to close stdin (file desriptor
0) and/or stdout (file descriptor 0), open the corresponding file and use the dup2 system call to
make it the appropriate file descriptor. Don\'t forget to use the close system call to close the old
file descriptor.
#include
#include
#include
#include
#include
#include
using namespace std;
int change_dir(char* args[])
{
if (args[1] == NULL)
{
chdir(getenv(\"HOME\"));
return 1;
}
else
{
if (chdir(args[1]) == -1)
{
cout << \"No such directory\" << args[1];
return -1;
}
}
return 0;
}
int helper_func(char* args[])
{
cout << \"Builtin commands so far are; cd, and help.\ \";
return 0;
}
int main()
{
while(true)
{
cout << \"$ \";
char cmd[128];
cin.getline(cmd,128);
vector args;
char *line = strtok(cmd, \" \");
char *tmp = line;
while (tmp != NULL)
{
args.push_back(tmp);
tmp = strtok(NULL, \" \");
}
char** argv = new char*[args.size() + 1];
for (int i = 0; i < args.size(); i++)
argv[i] = args[i];
argv[args.size()] = NULL;
if (strcmp(cmd, \"exit\") == 0)
{
{
exit(0);
}
}
//else if (strcmp(args[0], \"cd\") == 0) change_dir(args);
if (strcmp(cmd, \"cd\") == 0)
{
change_dir(argv);
}
if (strcmp(cmd, \"help\") == 0)
{
helper_func(argv);
}
else
{
pid_t pid;
pid_t wpid;
int status;
pid = fork();
if (pid == 0)
{
if (execvp(args[0], argv) == -1)
{
perror(\"cmd\");
}
exit(EXIT_FAILURE);
}
else if (pid < 0)
{
perror(\"cmd\");
}
{
do {
wpid = waitpid(pid, &status, WUNTRACED);
} while (!WIFEXITED(status) && !WIFSIGNALED(status));
}
}
}
signal(SIGINT, SIG_IGN);
signal(SIGTERM, SIG_DFL);
return 0;
}
Solution
#include
#include
#include
#include
#include
#include
using namespace std;
int change_dir(char* args[])
{
if (args[1] == NULL)
{
chdir(getenv(\"HOME\"));
return 1;
}
else
{
if (chdir(args[1]) == -1)
{
cout << \"No such directory\" << args[1];
return -1;
}
}
return 0;
}
int helper_func(char* args[])
{
cout << \"Builtin commands so far are; cd, and help.\ \";
return 0;
}
int main()
{
while(true)
{
cout << \"$ \";
char cmd[128];
cin.getline(cmd,128);
vector args;
char *line = strtok(cmd, \" \");
char *tmp = line;
while (tmp != NULL)
{
args.push_back(tmp);
tmp = strtok(NULL, \" \");
}
char** argv = new char*[args.size() + 1];
for (int i = 0; i < args.size(); i++)
argv[i] = args[i];
argv[args.size()] = NULL;
if (strcmp(cmd, \.
Add a 3rd field help that contains a short help string for each of t.pdf
Add a 3rd field help that contains a short help string for each of the commands you were to
implement in assignment #3. Make sure that your array(s) are big enough to handle 5 extra items
beyond your initialization. To save time only include help for exercises 4, 5, 6 and 8 in this
assignment, and use No help for the other entries.
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
void ctrlCHandler(int signum)
{
fprintf(stderr,"Command server terminated using C\n");
exit(1);
}
char * fExport(char *cmd, char *tokensleft[])
{
setenv(tokensleft[0],tokensleft[1],1);
return "Command 'export' was received";
}
char * fChdir(char *cmd,char *tokensleft[])
{
int ch=chdir(tokensleft[0]);
if(ch<0)
perror("chdir change of directory not successful\n");
else
printf("chdir change of directory successful\n");
return "Command 'chdir' was received\n";
}
char * fAccess(char *cmd,char *tokensleft[])
{
int exists =0;
for(int i=0;tokensleft[i]; i++) {
exists =0;
if(access(tokensleft[i],F_OK)==0){
exists = 1;
printf("file %s exists\n",tokensleft[i]);
}else{
printf("file %s does not exists\n",tokensleft[i]);
}
if (exists == 1){
if(access(tokensleft[i],R_OK)==0) {
printf("file %s is readable\n",tokensleft[i]);
}else{printf("file %s is not readable\n",tokensleft[i]);}
if(access(tokensleft[i],W_OK)==0) {
printf("file %s is writeable\n",tokensleft[i]);
}else{
printf("file %s is not writeable\n",tokensleft[i]);
}
if(access(tokensleft[i],X_OK)==0) {
printf("file %s is executeable\n",tokensleft[i]);
}else{
printf("file %s is not executeable\n",tokensleft[i]);
}
}// exists if
} //for
return "Command 'acsess' was received\n";
}
char * fChmod(char *cmd,char *tokensleft[])
{
unsigned int octalPerm;
sscanf(tokensleft[0],"%o",&octalPerm);
for(int i=1;tokensleft[i]; i++) {
if(chmod(tokensleft[i],octalPerm)==0 ){
chmod(tokensleft[i],octalPerm);
}else{
printf("Error: %s \n",strerror(errno));
}
}
return "Command 'chmod' was received";
}
char * fPath(char *cmd,char *tokensleft[])
{
char *pathLink;
char actualPath[PATH_MAX+1];
char *pointer;
char *bName;
char *dName;
for(int i=0;tokensleft[i]; i++) {
pathLink = tokensleft[i];
pointer =realpath(pathLink,actualPath);
bName = basename(actualPath);
dName = dirname(tokensleft[i]);
if (pointer){
printf("The Real path of %s is: %s\n",tokensleft[i],actualPath);
printf("The Dir name path of %s is: %s\n",tokensleft[i],dName);
printf("The Base name of %s is: %s\n",tokensleft[i],bName);
}else{
printf("Error: %s \n",strerror(errno));
}
}
return "Command 'path' was received";
}
char * fTouch(char *cmd,char *tokensleft[])
{
extern int optind,optopt,opterr;
struct FLAG{
bool aFlag;
bool mFlag;
} flags = { false, false };
int t1 = time(NULL), t2 = time(NULL);
int argc = 0;
int flag;
for (int i = 0; tokensleft[i]; i++) {
argc++;
}
while ((flag = getopt(argc, tokensleft, "m.
C Exam Help
I am Marianna P. I am a C Exam Expert at programmingexamhelp.com. I hold a Bachelor of Information Technology from, the California Institute of Technology, United States. I have been helping students with their exams for the past 7 years. You can hire me to take your exam in C.
Visit programmingexamhelp.com or email support@programmingexamhelp.com. You can also call on +1 678 648 4277 for any assistance with the C Exam.
Write a C++ program 1. Study the function process_text() in file.pdf
Write a C++ program
1. Study the function process_text() in file \"text_processing.cpp\". What do the string member
functions find(), substr(), c_str() do?
//text_processing.cpp
#include \"util.h\"
#include \"text_processing.h\"
#include
using namespace std;
int process_text( fstream &ifs, fstream &ofs, char target[] )
{
string replace_str( \"XXX\" ); //hard-coded string for replacement
int tlen = strlen( target ); //string length of target
int max_len = 200; //maximum length of a line
char s[max_len+1];
clear_screen(); //clear the screen
while ( !ifs.eof() ) {
int i2, i1 = 0, len=0;
ifs.getline( s, max_len ); //get one line from file
string str( s ); //construct a string object
i2 = str.find ( target, i1 ); //find target
len = i2 - i1;
if ( i2 > -1 && len > 0 ){ //if target found
print_text( str.substr( i1, i2 ).c_str() ); //print up to target
ofs << str.substr( i1, i2 );
}
while ( i2 > -1 ) {
print_text_inverse ( target ); //highlight target
ofs << replace_str;
i1 = i2 + tlen; //new search position
i2 = str.find ( target, i1 ); //find next target
len = i2 - i1;
if ( i2 > -1 && len > 0 ){ //found target
print_text( str.substr( i1, len ).c_str() ); //print up to target
ofs << str.substr( i1, len );
}
}
len = str.length() - i1;
if ( len > 0 ) { //print the remainder
print_text( str.substr( i1, len ).c_str() );
ofs << str.substr( i1, len );
}
ofs << endl;
getnstr( s, 1 ); //prompt for
} //while ( !ifs.eof() )
restore_screen(); //restore the old screen
return 1;
}
//text_processing.h
#ifndef TEXT_PROCESSING_H
#define TEXT_PROCESSING_H
#include
#include
using namespace std;
int process_text( fstream &ifs, fstream &ofs, char target[] );
#endif
//util.h
#ifndef UTIL_H
#define UTIL_H
#include
#include
#include
#include
#include
using namespace std;
void usage ( char program_name[] );
void clear_screen();
void print_text( const char *s );
void print_text_inverse( const char *s );
void get_input_text( char *s, int max_len );
void restore_screen();
#endif
//util.cpp
#include \"util.h\"
void usage ( char program_name[] )
{
cout << \"Usage: \" << program_name << \" infile outfile search_string\" << endl;
return;
}
/*
The following are some curses functions. You can find
the details by the command \"man ncurses\". But its use
can be transparent to you.
*/
void clear_screen()
{
initscr();
scrollok ( stdscr, true ); //allow window to scroll
}
void print_text ( const char *s )
{
printw(\"%s\", s );
refresh();
}
void print_text_inverse( const char *s )
{
attron( A_REVERSE );
printw(\"%s\", s );
attroff( A_REVERSE );
refresh();
}
void get_input_text( char *s, int max_len )
{
getnstr( s, max_len );
}
void restore_screen()
{
endwin();
}
//str_main.cpp
#include \"util.h\"
#include \"text_processing.h\"
#include
using namespace std;
/*
Note that grace_open() must be defined in same file
as main(), otherwise the destructor of fstream will
automatically close the opened file as it finds that
its out of scope upon exiting the function.
*/
int grace_open( char *s, fst.
operating system ubuntu,linux,MacProgram will work only if you g.pdf
//operating system ubuntu,linux,Mac
Program will work only if you give command like
sum 2 3 or
sum 4 (sub 4 3)
#include
#include
#include
#include
/*Sum function implementation*/
int sum(int number1,int number2){
return number1+number2;
}
/*sub function implementation*/
int sub(int number1,int number2){
return number1-number2;
}
/*mul function implementation*/
int mul(int number1,int number2){
return number1*number2;
}
/*divide function implementation*/
float divide(int number1,int number2){
if(number2==0){
printf(\"We can not divide by 0 \ \");
return 0.0f;
}
return (float)(number1/number2);
}
char *substring(char *string, int position, int length)
{
char *pointer;
int c;
pointer = malloc(length+1);
if (pointer == NULL)
{
printf(\"Unable to allocate memory.\ \");
exit(1);
}
for (c = 0 ; c < length ; c++)
{
*(pointer+c) = *(string+position-1);
string++;
}
*(pointer+c) = \'\\0\';
return pointer;
}
/*Main Function start*/
int main(int argc, char *argv[]){
/*Variable declarations*/
int number1,number2;
char *operator,*str1,*str2;
char line[50];
char s[2] = \" \";
int errorFlag=0;// checking if user entered more then one operands
while(1){
/*User input*/
printf(\"Please Enter the command \");
gets(line);
int len=0;
for(int i=0; line[i]!=\'\\0\'; ++i){len++;}
if(len==0){
return 1;
}
/*Spliting into Tokens*/
operator = strtok(line, \" \");
if(strcmp(operator,\"bye\")==0){
break;
}
/*String to integer --- token1*/
number1=atoi(strtok(NULL, s));
/*Removing ( and ) from String*/
str2=strtok(NULL, s);
char *content;int length=0;
for(int i=0; str2[i]!=\'\\0\'; ++i){length++;}
if(str2[0]==\'(\'){
content=substring(str2,2,length-1);
char *opr=content;
int n1=atoi(strtok(NULL, s));
int n2=atoi(strtok(NULL, s));
/*If user entered more then two operands then it will return some token otherwise it will
return NULL pointer*/
if(strtok(NULL, s)!=NULL){
printf(\"You need to enter operator operand1 operand2 \ \");
errorFlag=1;
}else{
if(strcmp(opr,\"sum\")==0){
number2=sum(n1,n2);
}else if(strcmp(opr,\"sub\")==0){
number2=sub(n1,n2);
}else if(strcmp(opr,\"mul\")==0){
number2=mul(n1,n2);
} else if(strcmp(opr,\"div\")==0){
number2=divide(n1,n2);
}
}
}else{
number2=atoi(str2);
/*If user entered more then two operands then it will return some token otherwise it will return
NULL pointer*/
if(strtok(NULL, s)!=NULL){
printf(\"You need to enter operator operand1 operand2 \ \");
errorFlag=1;
}
}
if(errorFlag!=1){
/*Checking operator by strcmp function*/
if(strcmp(operator,\"sum\")==0){
printf(\"Result %d \ \",sum(number1,number2));
}else if(strcmp(operator,\"sub\")==0){
printf(\"Result %d \ \",sub(number1,number2));
}else if(strcmp(operator,\"mul\")==0){
printf(\"Result %d \ \",mul(number1,number2));
} else if(strcmp(operator,\"div\")==0){
printf(\"Result %.2f \ \",divide(number1,number2));
}
}
}
return 0;
}
/*******************Output***********************/
gopal@gopal:~/Desktop/chegg$ gcc Calculator.c
Calculator.c: In function ‘main’:
Calculator.c:67:3: warning: implici.
C programming language tutorial
Are you C programming beginners? Learn C programming language via javatpoint.com
-- C Program How might I modify this code to handle any number of comm.docx
( C2 , CO1 , PO3 ) (b) Table 1 shows a fragment of pseudo code. [Sedual l menuryukkan keratan dod sudo.] (i) Compute the frequency of each statement. [Kirakan Aekerapan setiap lequatoan.] (6 Marks/Markah) (ii) Determine the total frequency into time complexity by using Big-0 notation. [Tintukan jumiah knckerapan kepada mara karumitan mengezunakan nofasi Big-O. Table 1 (2 Marks/Markah) [Jachal t]
.
-- C Program How might I modify this code to handle any number of comm.docx
// C Program
How might I modify this code to handle any number of commands piped together? such as ls | wc or ls | wc -l | ps -elf etc..
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <sys/wait.h>
// keeps track of cmd parameters
typedef struct param {
char *param;
struct param *next;
} pt;
// parses individual commands
typedef struct cmd_list {
char *cmd;
int param_count;
pt *param_list;
char * input_file_name;
char *output_file_name;
struct cmd_list *next;
} cl;
// keeps track of main commands that are piped
typedef struct list {
cl *head;
cl *tail;
int count;
} pt;
void execommands(clt *cmds) {
clt *cmd = cmds;
for (int i = 1; i < cmds->count-1; i++) {
int pipes[2] = {-1, -1};
pid_t pid = -1;
clt *temp = NULL;
pipe(pipes);
pid = fork();
switch (pid) {
case -1:
perror("fork failed");
break;
case 0:
{
char **rhp_argv = NULL;
char *rhp = NULL;
pt *temp = NULL;
int i = 1;
int flag_count = 0;
char flag_string[100] = "";
temp = cmd->param_list;
if (dup2(pipes[STDIN_FILENO], STDIN_FILENO) < 0) {
perror("child process failed dup2");
_exit(EXIT_FAILURE);
}
close(pipes[STDIN_FILENO]);
close(pipes[STDOUT_FILENO]);
rhp = cmd->cmd;
if(0 != cmd->param_count) {
rhp_argv = (char **) calloc((cmd->param_count), sizeof(char *));
rhp_argv[0] = rhp;
while(temp != NULL) {
if (flag_count > 0 && temp->param[0] == '-')
memmove(temp->param, temp->param + 1, strlen(temp->param));
}
strcat(flag_string, temp->param)
flag_count++;
temp = temp->next
}
rhp_argv[i] = strdup(flag_string);
i++;
}
else {
rhp_argv = (char **) calloc(2, sizeof(char *));
rhp_argv[0] = rhp;
}
execvp(rhp_argv[0], rhp_argv);
perror("child cannot exec program");
_exit(EXIT_FAILURE);
if(cmd->param_count != 0) {
for(int celm = 0; celm < flag_count; i++) {
free(rhp_argv[i]);
rhp_argv[i] = NULL;
}
}
free(rhp_argv);
rhp_argv = NULL;
free(temp);
}
break;
default:
int status;
waitpid(pid, &status, 0);
break;
} // end switch
} // end for
}
.
Lab Assignment 4 CSE330 Spring 2014 Skeleton Code for ex.docx
/* Lab Assignment 4 CSE330 Spring 2014 */
/* Skeleton Code for ex0 of lab4 */
/* No code will be provided for ex1 of lab4 */
#include <stdio.h> /* for NULL */
#include <ctype.h> /* for atoi() */
#include <errno.h> /* for perror() */
#include <signal.h> /* for sigvec() etc. */
#include <sys/types.h> /* for <arpa/inet.h> */
#include <sys/socket.h> /* for PF_INET, etc. */
#include <netinet/in.h> /* for struct sockaddr_in */
#include <arpa/inet.h> /* for inet_addr() */
#include <sys/time.h>
#include "lab4.h"
/* this is in netinet/in.h; included here for reference only.
struct sockaddr_in {
short sin_family;
u_short sin_port;
struct in_addr sin_addr;
char sin_zero[8];
};
*/
/***************************************************************************/
#define LINESIZE 80
#define TYPE0TIMEOUT 3
#define TYPE1TIMEOUT 30
/* You should give a long timeout, like 30sec, otherwise */
/* duplicated responses from the server may cause some problem */
#define MAXRETRY 3
/***************************************************************************/
static char *RCSId="$Id: client2.c,v 4 2014/04/20 03:15:07 cse3300 Exp $";
/***************************************************************************/
/* some functions you may find useful if you implement them */
extern int stringToAddress(char *s, struct sockaddr_in *a);
extern int msgok(LABMSG *);
extern u_short msgchecksum(u_short *);
extern void alarmCatcher(void);
extern int tries; /* for retransmission routines */
int numtries;
/***** this will die ****/
FILE *logFile;
void die(char *s)
{
perror(s);
exit(2);
}
void startTimer(int secs)
{
alarm(secs);
}
void stopTimer(void)
{
alarm(0);
}
/* should change it by yourself if needed!!! */
void printResponse(LABMSG *mp, int ck)
{
int type;
mp->courseEtc = ntohs(mp->courseEtc);
type = ( (mp->courseEtc & MESSAGETYPE) !=0)?1:0;
printf("course=%d, Type=%d\n", mp->courseEtc&0x3fff,type);
if (ntohl(mp->cookie) != ck)
printf("Cookies don't match: sent %x received %x\n",mp->cookie,ck);
if (mp->courseEtc & REQRESP)
printf("response ");
else {
printf("request??\n");
return;
}
mp->result = ntohs(mp->result);
/*printf(" result = %x: ",mp->result);*/
if (mp->result&TRANSOUTCOME) { /* Check outcome */
printf("error: ");
switch (mp->result & 0x7fff) {
case ERROR_CHECKSUM:
printf("checksum failure\n");
break;
case ERROR_SYNTAX:
printf("syntax error\n");
break;
case ERROR_UNKSSN:
printf("unknown SSN %d\n", ntohl(mp->reqSSN) );
break;
case ERROR_SERVER:
printf("Unspecified Server Error\n");
default:
printf("Unknown Error.\n");
} /* case switch */
} else { /* successful transaction */
if(type)printf("Test succeeded.");
if (!type) /* Type 0 -- print SSN and Response */
printf(": %d -> %d\n", ntohl(mp->reqSSN), mp->result&0x7fff);
else
printf("\n"); /* XXX print number of responses */
}
}
/*-----------------------------------*/
/* added by ZXC:
this fun ...
C Homework Help
I am Simon M. I am a C Homework Expert at programminghomeworkhelp.com. I hold a Ph.D. in Programming, from Glasgow University, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C.
Visit programminghomeworkhelp.com or email support@programminghomeworkhelp.com.
You can also call on +1 678 648 4277 for any assistance with C Homework.
Linux seccomp(2) vs OpenBSD pledge(2)
seccomp is a computer security facility in the Linux kernel, pledge is a similar security facility in the OpenBSD kernel. In this presentation Giovanni Bechis will review the development story and progress of both kernel interfaces and will analyze the main differences. There will be some examples of implementations of security patches made for some important open source projects.
Sysprog17
The 17th session in eglug's system programming course, I only attend the session, the slides are not written by me
finalprojtemplatev5finalprojtemplate.gitignore# Ignore the b
finalprojtemplatev5/finalprojtemplate/.gitignore
# Ignore the build directory
build
# Ignore any executables
bin/*
# Ignore Mac specific files
.DS_Store
finalprojtemplatev5/finalprojtemplate/Makefile
# Script adapted from https://hiltmon.com/blog/2013/07/03/a-simple-c-plus-plus-project-structure/
CC := /usr/local/classes/eecs/spring2021/cs472/public/gcc/bin/g++
STRIPUTIL = strip
SRCDIR = src
BUILDDIR = build
TARGET = bin/cache_sim
# Handle debug case
DEBUG ?= 1
ifeq ($(DEBUG), 1)
CFLAGS += -g -Wall
else
CFLAGS += -DNDEBUG -O3
endif
SRCEXT = cpp
SOURCES = $(shell find $(SRCDIR) -type f -name *.$(SRCEXT))
OBJECTS = $(patsubst $(SRCDIR)/%,$(BUILDDIR)/%,$(SOURCES:.$(SRCEXT)=.o))
LDFLAGS += -Wl,-rpath,/usr/local/classes/eecs/spring2021/cs472/public/gcc/lib64
LIB += -pthread
INC += -I $(SRCDIR)
$(TARGET): $(OBJECTS)
@echo "Linking..."
@echo " $(CC) $^ -o $(TARGET) $(LIB) $(LDFLAGS)"; $(CC) $^ -o $(TARGET) $(LIB) $(LDFLAGS)
$(BUILDDIR)/%.o: $(SRCDIR)/%.$(SRCEXT)
@mkdir -p $(BUILDDIR)
@echo " $(CC) $(CFLAGS) $(INC) -c -o [email protected] $<"; $(CC) $(CFLAGS) $(INC) -c -o [email protected] $<
clean:
@echo "Cleaning...";
@echo " $(RM) -r $(BUILDDIR) $(TARGET)"; $(RM) -r $(BUILDDIR) $(TARGET)
.PHONY: clean
finalprojtemplatev5/finalprojtemplate/resources/simpletracefile==This is a simple tracefile for testing purposes L 08 L 808 L 1006 L 08 S 1805 L 1002 L 808 M 1806 L 1008 S 1005 S 1808 L 808 M 08 M 04 L 2804 M 808
finalprojtemplatev5/finalprojtemplate/resources/testconfig
1
230
8
16
3
1
0
13
finalprojtemplatev5/finalprojtemplate/src/CacheController.cppfinalprojtemplatev5/finalprojtemplate/src/CacheController.cpp/*
Cache Simulator (Starter Code) by Justin Goins
Oregon State University
Spring Term 2021
*/
#include"CacheController.h"
#include<iostream>
#include<fstream>
#include<regex>
#include<cmath>
usingnamespace std;
CacheController::CacheController(CacheInfo ci, string tracefile){
// store the configuration info
this->ci = ci;
this->inputFile = tracefile;
this->outputFile =this->inputFile +".out";
// compute the other cache parameters
this->ci.numByteOffsetBits = log2(ci.blockSize);
this->ci.numSetIndexBits = log2(ci.numberSets);
// initialize the counters
this->globalCycles =0;
this->globalHits =0;
this->globalMisses =0;
this->globalEvictions =0;
// create your cache structure
// ...
// manual test code to see if the cache is behaving properly
// will need to be changed slightly to match the function prototype
/*
cacheAccess(false, 0);
cacheAccess(false, 128);
cacheAccess(false, 256);
cacheAccess(false, 0);
cacheAccess(false, 128);
cacheAccess(false, 256);
*/
}
/*
Starts reading the tracefile and processing memory operations.
*/
voidCacheController::runTracefile(){
cout <<"Input tracefile: "<< inputFile << endl;
cout <<"Output file name: "<< outputFile << endl;
// process each input line
string line;
// define regular expressions that are used to locate ...
public class SLLT { protected SLLNodeT head, tail; pub.pdf
public class SLL { protected SLLNode head, tail; public SLL() { head = tail = null;
} public void addToTail(T e){ SLLNode newNode = new SLLNode(e); if(head ==
null) head = tail = newNode; else{ tail.next = newNode; tail =
tail.next; } } public void addToHead(T e) { SLLNode newNode = new
SLLNode(e, head); if(head == null) tail = newNode; head = newNode; }
public String toString(){ if (head == null) return "[]"; String out = "[";
SLLNode p = head; while (p.next != null){ out += p.info + " "; p = p.next;
} out += p.info + "]"; return out; } }
public class SLLNode { protected T info; protected SLLNode next; public SLLNode() {
this(null, null); } public SLLNode(T e) { this(e, null); } public SLLNode(T
e, SLLNode ptr) { info = e; next = ptr; } }
public class SortedSLL> extends SLL{ public void insert(T e){ } public void
merge(SortedSLL list){ } }
public class SortedSLL T extends Comparable T extends SLLT{ public void insert (T e) \{ \}
public void merge(SortedSLL T list) \{ \} public class SortedSLL T extends Comparable T
extends SLLT{ public void insert ( T e) \{ \} public void merge(SortedSLL T list) \{ \} \}
ublic class SLLT{ protected SLLNodeT head, tail; public SLL() \{ head = tail = null; \} public
void addToTail( T e) \{ SLLNode newNode = new SLLNodeT(e); if ( head == null) head = tail
= newNode; else\{ tail. next = newNode; tail = tail. next; \} \} public void addToHead( T e) \{
SLLNode T newNode = new SLLNodeT(e, head ); if (head == null) tail = newNode; head =
newNode; \} public string tostring()\{ if (head == null) return "[]"; String out = "["; SLLNode
Tp= head; while (p.next != null) \{ out +=p.info +" "; p=p next; \} out += p.info + "]"; return out;
\}
ublic class SLLNode T{ protected T info; protected SLLNode T next; public SLLNode() \{
this(null, null); \} public SLLNode( T e) \{ this (e, null); \} public SLLNode( T e, SLLNode T
ptr ){ info =e; next =ptr; \}
Write a subclass of SLL named SortedSLL that represents a sorted SinglyLinkedList (sorted in
nondecending order). The class should contain the following methods: 1- [40 Points] void insert
( T e) that inserts an SLLNode in its correct position so that the resulting list remains sorted.
Make sure to consider special cases such as: - The list is empty. - The list has one element. - The
inserted element is smaller than the head, or larger than the tail.. 2- [60 Points] void
merge(SortedSLL T list) that merges the calling list with the argument list. Note that this can be
done by calling the insert method for each element in list, but this is highly ineffiecient, answer
the following question: Q2.1: What is the complexity of this approach? Explain. [10 Points]
Instead, you should come up with a solution that traverse both lists just one time. Furthermore,
the argument list should be empty after the merge is completed. Q2.2: What is the complexity of
this approach? Explain. [10 Points] Hint: Make sure the class SortedSLL accepts types T that is
comparable. You can do this by writing .
Q1How would this be explained in the witness box, how would this lo.pdf
Q1:How would this be explained in the witness box, how would this look on the front page of
tomorrows newspaper, and how would I explain this to my family? This is the three quick tests
applies when an executive may not have time for a long ethical assessment process and he/she
needs to respond quickly to an ethical crisis.
A- True
B-False
Q8:A conflict is actual when a personal interest leads a person to act against the interests of an
employer or another person whose interests the person is obligated to serve. One the other hand,
a situation constitutes a potential conflict of interest when there is the possibility that a person
will fail to fulfill an obligation to act in the interests of another, even though the person has not
yet done so.
A-True
B-False
Q12:What are the kinds of conflict of interest?
A-Exercising biased judgment, engaging in direct competition, misusing a position, and violating
confidentiality.
B-Exercising biased judgment, engaging in direct competition, and stealing from the company.
C-Misusing a position, violating confidentiality and benefiting from the companys perks.
D-None of the above.
Q13 :Kohlberg's moral development stages in humans states that:
A-At the earliest phase the child defers to physical superiority, and does not understand that
others have consciousness and desires. The second phase has proper action serving to satisfy the
persons own needs. Subsequent phases progress through loyalty and affection within the sphere
of immediate family and friends. Then, there is a quantum lift in moral understanding that has
the person trying to develop impartial points of view, using self justified moral principles. At last
stage is that of seeing moral principles as being universal and consistent.
B-A child goes into two phases. The first phase, a child will learn right and wrong based on
rewards punishments. The second phase, a child acquired his/her moral understanding through
applying the morals of his family members.
C-A child has three stages of moral development. First, he/she accepts the moral understanding
of his family. Second, he/she acquires moral principles from school and friends. Third, he/she
develop his/her personal moral principles based on what he/she acquired from family and school.
D-None of the above.
Q15:What was/were the reasons of the collapse of Enron?
A-The conflict of interests between the two roles played by Arthur Anderson, as auditor but also
as consultant to Enron, and the lack of attention shown by members of the Enron board of
directors to the off-books financial entities with which Enron did business.
B-The bad economy at that time.
C-The arrogance of the top managers and their beliefs that they do no wrong..
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C Exam Help
I am Marianna P. I am a C Exam Expert at programmingexamhelp.com. I hold a Bachelor of Information Technology from, the California Institute of Technology, United States. I have been helping students with their exams for the past 7 years. You can hire me to take your exam in C.
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Write a C++ program 1. Study the function process_text() in file.pdf
Write a C++ program
1. Study the function process_text() in file \"text_processing.cpp\". What do the string member
functions find(), substr(), c_str() do?
//text_processing.cpp
#include \"util.h\"
#include \"text_processing.h\"
#include
using namespace std;
int process_text( fstream &ifs, fstream &ofs, char target[] )
{
string replace_str( \"XXX\" ); //hard-coded string for replacement
int tlen = strlen( target ); //string length of target
int max_len = 200; //maximum length of a line
char s[max_len+1];
clear_screen(); //clear the screen
while ( !ifs.eof() ) {
int i2, i1 = 0, len=0;
ifs.getline( s, max_len ); //get one line from file
string str( s ); //construct a string object
i2 = str.find ( target, i1 ); //find target
len = i2 - i1;
if ( i2 > -1 && len > 0 ){ //if target found
print_text( str.substr( i1, i2 ).c_str() ); //print up to target
ofs << str.substr( i1, i2 );
}
while ( i2 > -1 ) {
print_text_inverse ( target ); //highlight target
ofs << replace_str;
i1 = i2 + tlen; //new search position
i2 = str.find ( target, i1 ); //find next target
len = i2 - i1;
if ( i2 > -1 && len > 0 ){ //found target
print_text( str.substr( i1, len ).c_str() ); //print up to target
ofs << str.substr( i1, len );
}
}
len = str.length() - i1;
if ( len > 0 ) { //print the remainder
print_text( str.substr( i1, len ).c_str() );
ofs << str.substr( i1, len );
}
ofs << endl;
getnstr( s, 1 ); //prompt for
} //while ( !ifs.eof() )
restore_screen(); //restore the old screen
return 1;
}
//text_processing.h
#ifndef TEXT_PROCESSING_H
#define TEXT_PROCESSING_H
#include
#include
using namespace std;
int process_text( fstream &ifs, fstream &ofs, char target[] );
#endif
//util.h
#ifndef UTIL_H
#define UTIL_H
#include
#include
#include
#include
#include
using namespace std;
void usage ( char program_name[] );
void clear_screen();
void print_text( const char *s );
void print_text_inverse( const char *s );
void get_input_text( char *s, int max_len );
void restore_screen();
#endif
//util.cpp
#include \"util.h\"
void usage ( char program_name[] )
{
cout << \"Usage: \" << program_name << \" infile outfile search_string\" << endl;
return;
}
/*
The following are some curses functions. You can find
the details by the command \"man ncurses\". But its use
can be transparent to you.
*/
void clear_screen()
{
initscr();
scrollok ( stdscr, true ); //allow window to scroll
}
void print_text ( const char *s )
{
printw(\"%s\", s );
refresh();
}
void print_text_inverse( const char *s )
{
attron( A_REVERSE );
printw(\"%s\", s );
attroff( A_REVERSE );
refresh();
}
void get_input_text( char *s, int max_len )
{
getnstr( s, max_len );
}
void restore_screen()
{
endwin();
}
//str_main.cpp
#include \"util.h\"
#include \"text_processing.h\"
#include
using namespace std;
/*
Note that grace_open() must be defined in same file
as main(), otherwise the destructor of fstream will
automatically close the opened file as it finds that
its out of scope upon exiting the function.
*/
int grace_open( char *s, fst.
operating system ubuntu,linux,MacProgram will work only if you g.pdf
//operating system ubuntu,linux,Mac
Program will work only if you give command like
sum 2 3 or
sum 4 (sub 4 3)
#include
#include
#include
#include
/*Sum function implementation*/
int sum(int number1,int number2){
return number1+number2;
}
/*sub function implementation*/
int sub(int number1,int number2){
return number1-number2;
}
/*mul function implementation*/
int mul(int number1,int number2){
return number1*number2;
}
/*divide function implementation*/
float divide(int number1,int number2){
if(number2==0){
printf(\"We can not divide by 0 \ \");
return 0.0f;
}
return (float)(number1/number2);
}
char *substring(char *string, int position, int length)
{
char *pointer;
int c;
pointer = malloc(length+1);
if (pointer == NULL)
{
printf(\"Unable to allocate memory.\ \");
exit(1);
}
for (c = 0 ; c < length ; c++)
{
*(pointer+c) = *(string+position-1);
string++;
}
*(pointer+c) = \'\\0\';
return pointer;
}
/*Main Function start*/
int main(int argc, char *argv[]){
/*Variable declarations*/
int number1,number2;
char *operator,*str1,*str2;
char line[50];
char s[2] = \" \";
int errorFlag=0;// checking if user entered more then one operands
while(1){
/*User input*/
printf(\"Please Enter the command \");
gets(line);
int len=0;
for(int i=0; line[i]!=\'\\0\'; ++i){len++;}
if(len==0){
return 1;
}
/*Spliting into Tokens*/
operator = strtok(line, \" \");
if(strcmp(operator,\"bye\")==0){
break;
}
/*String to integer --- token1*/
number1=atoi(strtok(NULL, s));
/*Removing ( and ) from String*/
str2=strtok(NULL, s);
char *content;int length=0;
for(int i=0; str2[i]!=\'\\0\'; ++i){length++;}
if(str2[0]==\'(\'){
content=substring(str2,2,length-1);
char *opr=content;
int n1=atoi(strtok(NULL, s));
int n2=atoi(strtok(NULL, s));
/*If user entered more then two operands then it will return some token otherwise it will
return NULL pointer*/
if(strtok(NULL, s)!=NULL){
printf(\"You need to enter operator operand1 operand2 \ \");
errorFlag=1;
}else{
if(strcmp(opr,\"sum\")==0){
number2=sum(n1,n2);
}else if(strcmp(opr,\"sub\")==0){
number2=sub(n1,n2);
}else if(strcmp(opr,\"mul\")==0){
number2=mul(n1,n2);
} else if(strcmp(opr,\"div\")==0){
number2=divide(n1,n2);
}
}
}else{
number2=atoi(str2);
/*If user entered more then two operands then it will return some token otherwise it will return
NULL pointer*/
if(strtok(NULL, s)!=NULL){
printf(\"You need to enter operator operand1 operand2 \ \");
errorFlag=1;
}
}
if(errorFlag!=1){
/*Checking operator by strcmp function*/
if(strcmp(operator,\"sum\")==0){
printf(\"Result %d \ \",sum(number1,number2));
}else if(strcmp(operator,\"sub\")==0){
printf(\"Result %d \ \",sub(number1,number2));
}else if(strcmp(operator,\"mul\")==0){
printf(\"Result %d \ \",mul(number1,number2));
} else if(strcmp(operator,\"div\")==0){
printf(\"Result %.2f \ \",divide(number1,number2));
}
}
}
return 0;
}
/*******************Output***********************/
gopal@gopal:~/Desktop/chegg$ gcc Calculator.c
Calculator.c: In function ‘main’:
Calculator.c:67:3: warning: implici.
C programming language tutorial
Are you C programming beginners? Learn C programming language via javatpoint.com
-- C Program How might I modify this code to handle any number of comm.docx
( C2 , CO1 , PO3 ) (b) Table 1 shows a fragment of pseudo code. [Sedual l menuryukkan keratan dod sudo.] (i) Compute the frequency of each statement. [Kirakan Aekerapan setiap lequatoan.] (6 Marks/Markah) (ii) Determine the total frequency into time complexity by using Big-0 notation. [Tintukan jumiah knckerapan kepada mara karumitan mengezunakan nofasi Big-O. Table 1 (2 Marks/Markah) [Jachal t]
.
-- C Program How might I modify this code to handle any number of comm.docx
// C Program
How might I modify this code to handle any number of commands piped together? such as ls | wc or ls | wc -l | ps -elf etc..
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <sys/wait.h>
// keeps track of cmd parameters
typedef struct param {
char *param;
struct param *next;
} pt;
// parses individual commands
typedef struct cmd_list {
char *cmd;
int param_count;
pt *param_list;
char * input_file_name;
char *output_file_name;
struct cmd_list *next;
} cl;
// keeps track of main commands that are piped
typedef struct list {
cl *head;
cl *tail;
int count;
} pt;
void execommands(clt *cmds) {
clt *cmd = cmds;
for (int i = 1; i < cmds->count-1; i++) {
int pipes[2] = {-1, -1};
pid_t pid = -1;
clt *temp = NULL;
pipe(pipes);
pid = fork();
switch (pid) {
case -1:
perror("fork failed");
break;
case 0:
{
char **rhp_argv = NULL;
char *rhp = NULL;
pt *temp = NULL;
int i = 1;
int flag_count = 0;
char flag_string[100] = "";
temp = cmd->param_list;
if (dup2(pipes[STDIN_FILENO], STDIN_FILENO) < 0) {
perror("child process failed dup2");
_exit(EXIT_FAILURE);
}
close(pipes[STDIN_FILENO]);
close(pipes[STDOUT_FILENO]);
rhp = cmd->cmd;
if(0 != cmd->param_count) {
rhp_argv = (char **) calloc((cmd->param_count), sizeof(char *));
rhp_argv[0] = rhp;
while(temp != NULL) {
if (flag_count > 0 && temp->param[0] == '-')
memmove(temp->param, temp->param + 1, strlen(temp->param));
}
strcat(flag_string, temp->param)
flag_count++;
temp = temp->next
}
rhp_argv[i] = strdup(flag_string);
i++;
}
else {
rhp_argv = (char **) calloc(2, sizeof(char *));
rhp_argv[0] = rhp;
}
execvp(rhp_argv[0], rhp_argv);
perror("child cannot exec program");
_exit(EXIT_FAILURE);
if(cmd->param_count != 0) {
for(int celm = 0; celm < flag_count; i++) {
free(rhp_argv[i]);
rhp_argv[i] = NULL;
}
}
free(rhp_argv);
rhp_argv = NULL;
free(temp);
}
break;
default:
int status;
waitpid(pid, &status, 0);
break;
} // end switch
} // end for
}
.
Lab Assignment 4 CSE330 Spring 2014 Skeleton Code for ex.docx
/* Lab Assignment 4 CSE330 Spring 2014 */
/* Skeleton Code for ex0 of lab4 */
/* No code will be provided for ex1 of lab4 */
#include <stdio.h> /* for NULL */
#include <ctype.h> /* for atoi() */
#include <errno.h> /* for perror() */
#include <signal.h> /* for sigvec() etc. */
#include <sys/types.h> /* for <arpa/inet.h> */
#include <sys/socket.h> /* for PF_INET, etc. */
#include <netinet/in.h> /* for struct sockaddr_in */
#include <arpa/inet.h> /* for inet_addr() */
#include <sys/time.h>
#include "lab4.h"
/* this is in netinet/in.h; included here for reference only.
struct sockaddr_in {
short sin_family;
u_short sin_port;
struct in_addr sin_addr;
char sin_zero[8];
};
*/
/***************************************************************************/
#define LINESIZE 80
#define TYPE0TIMEOUT 3
#define TYPE1TIMEOUT 30
/* You should give a long timeout, like 30sec, otherwise */
/* duplicated responses from the server may cause some problem */
#define MAXRETRY 3
/***************************************************************************/
static char *RCSId="$Id: client2.c,v 4 2014/04/20 03:15:07 cse3300 Exp $";
/***************************************************************************/
/* some functions you may find useful if you implement them */
extern int stringToAddress(char *s, struct sockaddr_in *a);
extern int msgok(LABMSG *);
extern u_short msgchecksum(u_short *);
extern void alarmCatcher(void);
extern int tries; /* for retransmission routines */
int numtries;
/***** this will die ****/
FILE *logFile;
void die(char *s)
{
perror(s);
exit(2);
}
void startTimer(int secs)
{
alarm(secs);
}
void stopTimer(void)
{
alarm(0);
}
/* should change it by yourself if needed!!! */
void printResponse(LABMSG *mp, int ck)
{
int type;
mp->courseEtc = ntohs(mp->courseEtc);
type = ( (mp->courseEtc & MESSAGETYPE) !=0)?1:0;
printf("course=%d, Type=%d\n", mp->courseEtc&0x3fff,type);
if (ntohl(mp->cookie) != ck)
printf("Cookies don't match: sent %x received %x\n",mp->cookie,ck);
if (mp->courseEtc & REQRESP)
printf("response ");
else {
printf("request??\n");
return;
}
mp->result = ntohs(mp->result);
/*printf(" result = %x: ",mp->result);*/
if (mp->result&TRANSOUTCOME) { /* Check outcome */
printf("error: ");
switch (mp->result & 0x7fff) {
case ERROR_CHECKSUM:
printf("checksum failure\n");
break;
case ERROR_SYNTAX:
printf("syntax error\n");
break;
case ERROR_UNKSSN:
printf("unknown SSN %d\n", ntohl(mp->reqSSN) );
break;
case ERROR_SERVER:
printf("Unspecified Server Error\n");
default:
printf("Unknown Error.\n");
} /* case switch */
} else { /* successful transaction */
if(type)printf("Test succeeded.");
if (!type) /* Type 0 -- print SSN and Response */
printf(": %d -> %d\n", ntohl(mp->reqSSN), mp->result&0x7fff);
else
printf("\n"); /* XXX print number of responses */
}
}
/*-----------------------------------*/
/* added by ZXC:
this fun ...
C Homework Help
I am Simon M. I am a C Homework Expert at programminghomeworkhelp.com. I hold a Ph.D. in Programming, from Glasgow University, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C.
Visit programminghomeworkhelp.com or email support@programminghomeworkhelp.com.
You can also call on +1 678 648 4277 for any assistance with C Homework.
Linux seccomp(2) vs OpenBSD pledge(2)
seccomp is a computer security facility in the Linux kernel, pledge is a similar security facility in the OpenBSD kernel. In this presentation Giovanni Bechis will review the development story and progress of both kernel interfaces and will analyze the main differences. There will be some examples of implementations of security patches made for some important open source projects.
Sysprog17
The 17th session in eglug's system programming course, I only attend the session, the slides are not written by me
finalprojtemplatev5finalprojtemplate.gitignore# Ignore the b
finalprojtemplatev5/finalprojtemplate/.gitignore
# Ignore the build directory
build
# Ignore any executables
bin/*
# Ignore Mac specific files
.DS_Store
finalprojtemplatev5/finalprojtemplate/Makefile
# Script adapted from https://hiltmon.com/blog/2013/07/03/a-simple-c-plus-plus-project-structure/
CC := /usr/local/classes/eecs/spring2021/cs472/public/gcc/bin/g++
STRIPUTIL = strip
SRCDIR = src
BUILDDIR = build
TARGET = bin/cache_sim
# Handle debug case
DEBUG ?= 1
ifeq ($(DEBUG), 1)
CFLAGS += -g -Wall
else
CFLAGS += -DNDEBUG -O3
endif
SRCEXT = cpp
SOURCES = $(shell find $(SRCDIR) -type f -name *.$(SRCEXT))
OBJECTS = $(patsubst $(SRCDIR)/%,$(BUILDDIR)/%,$(SOURCES:.$(SRCEXT)=.o))
LDFLAGS += -Wl,-rpath,/usr/local/classes/eecs/spring2021/cs472/public/gcc/lib64
LIB += -pthread
INC += -I $(SRCDIR)
$(TARGET): $(OBJECTS)
@echo "Linking..."
@echo " $(CC) $^ -o $(TARGET) $(LIB) $(LDFLAGS)"; $(CC) $^ -o $(TARGET) $(LIB) $(LDFLAGS)
$(BUILDDIR)/%.o: $(SRCDIR)/%.$(SRCEXT)
@mkdir -p $(BUILDDIR)
@echo " $(CC) $(CFLAGS) $(INC) -c -o [email protected] $<"; $(CC) $(CFLAGS) $(INC) -c -o [email protected] $<
clean:
@echo "Cleaning...";
@echo " $(RM) -r $(BUILDDIR) $(TARGET)"; $(RM) -r $(BUILDDIR) $(TARGET)
.PHONY: clean
finalprojtemplatev5/finalprojtemplate/resources/simpletracefile==This is a simple tracefile for testing purposes L 08 L 808 L 1006 L 08 S 1805 L 1002 L 808 M 1806 L 1008 S 1005 S 1808 L 808 M 08 M 04 L 2804 M 808
finalprojtemplatev5/finalprojtemplate/resources/testconfig
1
230
8
16
3
1
0
13
finalprojtemplatev5/finalprojtemplate/src/CacheController.cppfinalprojtemplatev5/finalprojtemplate/src/CacheController.cpp/*
Cache Simulator (Starter Code) by Justin Goins
Oregon State University
Spring Term 2021
*/
#include"CacheController.h"
#include<iostream>
#include<fstream>
#include<regex>
#include<cmath>
usingnamespace std;
CacheController::CacheController(CacheInfo ci, string tracefile){
// store the configuration info
this->ci = ci;
this->inputFile = tracefile;
this->outputFile =this->inputFile +".out";
// compute the other cache parameters
this->ci.numByteOffsetBits = log2(ci.blockSize);
this->ci.numSetIndexBits = log2(ci.numberSets);
// initialize the counters
this->globalCycles =0;
this->globalHits =0;
this->globalMisses =0;
this->globalEvictions =0;
// create your cache structure
// ...
// manual test code to see if the cache is behaving properly
// will need to be changed slightly to match the function prototype
/*
cacheAccess(false, 0);
cacheAccess(false, 128);
cacheAccess(false, 256);
cacheAccess(false, 0);
cacheAccess(false, 128);
cacheAccess(false, 256);
*/
}
/*
Starts reading the tracefile and processing memory operations.
*/
voidCacheController::runTracefile(){
cout <<"Input tracefile: "<< inputFile << endl;
cout <<"Output file name: "<< outputFile << endl;
// process each input line
string line;
// define regular expressions that are used to locate ...
Similar to Shell to be modified#include stdlib.h #include unistd.h .pdf (20)
Write a C++ program 1. Study the function process_text() in file.pdf
Write a C++ program 1. Study the function process_text() in file.pdf
operating system ubuntu,linux,MacProgram will work only if you g.pdf
operating system ubuntu,linux,MacProgram will work only if you g.pdf
ExperiencesSharingOnEmbeddedSystemDevelopment_20160321
ExperiencesSharingOnEmbeddedSystemDevelopment_20160321
-- C Program How might I modify this code to handle any number of comm.docx
-- C Program How might I modify this code to handle any number of comm.docx
-- C Program How might I modify this code to handle any number of comm.docx
-- C Program How might I modify this code to handle any number of comm.docx
Lab Assignment 4 CSE330 Spring 2014 Skeleton Code for ex.docx
Lab Assignment 4 CSE330 Spring 2014 Skeleton Code for ex.docx
finalprojtemplatev5finalprojtemplate.gitignore# Ignore the b
finalprojtemplatev5finalprojtemplate.gitignore# Ignore the b
More from clarityvision
public class SLLT { protected SLLNodeT head, tail; pub.pdf
public class SLL { protected SLLNode head, tail; public SLL() { head = tail = null;
} public void addToTail(T e){ SLLNode newNode = new SLLNode(e); if(head ==
null) head = tail = newNode; else{ tail.next = newNode; tail =
tail.next; } } public void addToHead(T e) { SLLNode newNode = new
SLLNode(e, head); if(head == null) tail = newNode; head = newNode; }
public String toString(){ if (head == null) return "[]"; String out = "[";
SLLNode p = head; while (p.next != null){ out += p.info + " "; p = p.next;
} out += p.info + "]"; return out; } }
public class SLLNode { protected T info; protected SLLNode next; public SLLNode() {
this(null, null); } public SLLNode(T e) { this(e, null); } public SLLNode(T
e, SLLNode ptr) { info = e; next = ptr; } }
public class SortedSLL> extends SLL{ public void insert(T e){ } public void
merge(SortedSLL list){ } }
public class SortedSLL T extends Comparable T extends SLLT{ public void insert (T e) \{ \}
public void merge(SortedSLL T list) \{ \} public class SortedSLL T extends Comparable T
extends SLLT{ public void insert ( T e) \{ \} public void merge(SortedSLL T list) \{ \} \}
ublic class SLLT{ protected SLLNodeT head, tail; public SLL() \{ head = tail = null; \} public
void addToTail( T e) \{ SLLNode newNode = new SLLNodeT(e); if ( head == null) head = tail
= newNode; else\{ tail. next = newNode; tail = tail. next; \} \} public void addToHead( T e) \{
SLLNode T newNode = new SLLNodeT(e, head ); if (head == null) tail = newNode; head =
newNode; \} public string tostring()\{ if (head == null) return "[]"; String out = "["; SLLNode
Tp= head; while (p.next != null) \{ out +=p.info +" "; p=p next; \} out += p.info + "]"; return out;
\}
ublic class SLLNode T{ protected T info; protected SLLNode T next; public SLLNode() \{
this(null, null); \} public SLLNode( T e) \{ this (e, null); \} public SLLNode( T e, SLLNode T
ptr ){ info =e; next =ptr; \}
Write a subclass of SLL named SortedSLL that represents a sorted SinglyLinkedList (sorted in
nondecending order). The class should contain the following methods: 1- [40 Points] void insert
( T e) that inserts an SLLNode in its correct position so that the resulting list remains sorted.
Make sure to consider special cases such as: - The list is empty. - The list has one element. - The
inserted element is smaller than the head, or larger than the tail.. 2- [60 Points] void
merge(SortedSLL T list) that merges the calling list with the argument list. Note that this can be
done by calling the insert method for each element in list, but this is highly ineffiecient, answer
the following question: Q2.1: What is the complexity of this approach? Explain. [10 Points]
Instead, you should come up with a solution that traverse both lists just one time. Furthermore,
the argument list should be empty after the merge is completed. Q2.2: What is the complexity of
this approach? Explain. [10 Points] Hint: Make sure the class SortedSLL accepts types T that is
comparable. You can do this by writing .
Q1How would this be explained in the witness box, how would this lo.pdf
Q1:How would this be explained in the witness box, how would this look on the front page of
tomorrows newspaper, and how would I explain this to my family? This is the three quick tests
applies when an executive may not have time for a long ethical assessment process and he/she
needs to respond quickly to an ethical crisis.
A- True
B-False
Q8:A conflict is actual when a personal interest leads a person to act against the interests of an
employer or another person whose interests the person is obligated to serve. One the other hand,
a situation constitutes a potential conflict of interest when there is the possibility that a person
will fail to fulfill an obligation to act in the interests of another, even though the person has not
yet done so.
A-True
B-False
Q12:What are the kinds of conflict of interest?
A-Exercising biased judgment, engaging in direct competition, misusing a position, and violating
confidentiality.
B-Exercising biased judgment, engaging in direct competition, and stealing from the company.
C-Misusing a position, violating confidentiality and benefiting from the companys perks.
D-None of the above.
Q13 :Kohlberg's moral development stages in humans states that:
A-At the earliest phase the child defers to physical superiority, and does not understand that
others have consciousness and desires. The second phase has proper action serving to satisfy the
persons own needs. Subsequent phases progress through loyalty and affection within the sphere
of immediate family and friends. Then, there is a quantum lift in moral understanding that has
the person trying to develop impartial points of view, using self justified moral principles. At last
stage is that of seeing moral principles as being universal and consistent.
B-A child goes into two phases. The first phase, a child will learn right and wrong based on
rewards punishments. The second phase, a child acquired his/her moral understanding through
applying the morals of his family members.
C-A child has three stages of moral development. First, he/she accepts the moral understanding
of his family. Second, he/she acquires moral principles from school and friends. Third, he/she
develop his/her personal moral principles based on what he/she acquired from family and school.
D-None of the above.
Q15:What was/were the reasons of the collapse of Enron?
A-The conflict of interests between the two roles played by Arthur Anderson, as auditor but also
as consultant to Enron, and the lack of attention shown by members of the Enron board of
directors to the off-books financial entities with which Enron did business.
B-The bad economy at that time.
C-The arrogance of the top managers and their beliefs that they do no wrong..
Professor Lessiarty directs an underground network of n criminal maste.pdf
Professor Lessiarty directs an underground network of n criminal masterminds called Lessiartys
Web. He has heard that famed detective Sherlatch Houses is hot on his trail and needs to decide
whether to issue an attack on detective Houses. Professor Lessiarty decides to consult with the
Web and compiles a list of n polls, where each poll is a tuple matching a different Web members
name with their integer opinion on the topic. Opinion +x means they are for attacking Houses
with strength x, while opinion x means they are against attacking Houses with strength x.
Lessiarty wants to generate a list containing the names of the lg n Web members that have the
strongest opinions (breaking ties arbitrarily), so he can arrange a secret meeting to discuss with
them further. Describe an O(n)-time algorithm to generate Lessiartys list..
Project Advanced Inventory Management System Upgrade for ABCBio.pdf
Project: Advanced Inventory Management System Upgrade for ABC
Biopharmaceuticals Ltd.
Project Duration: 12-18 months
Budget: $1 million (fixed)
Project Description:
ABC Biopharmaceuticals Ltd. specializes in manufacturing in vitro diagnostic rapid test kits for
various diseases, including COVID-19, dengue, influenza, syphilis, and hepatitis A, B, and C.
The company has recognized the need to modernize its inventory management system to
improve operational efficiency, reduce costs, and ensure regulatory compliance. This project
aims to design and implement a customized inventory management solution tailored to the
specific needs of the biopharmaceutical industry.
Project Objectives:
Enhance Inventory Control: Implement a robust inventory management system to gain real-time
visibility into inventory levels, track product expiration dates, and reduce inventory holding
costs.
Streamline Procurement Processes: Optimize procurement processes to ensure timely
replenishment of stock, reduce lead times, and minimize excess inventory.
Minimize Stockouts: Implement demand forecasting and replenishment algorithms to minimize
the occurrence of stockouts and ensure product availability.
Regulatory Compliance: Ensure the new system complies with relevant regulatory requirements
for the biopharmaceutical industry, including traceability and reporting.
Cost Reduction: Identify cost-saving opportunities by reducing overstocking and implementing
efficient inventory practices.
Technical Requirements:
Inventory Management Software:
Customized inventory tracking and management software
Real-time inventory visibility
Barcode scanning capability for product tracking
Integration with accounting and procurement systems
User-friendly interface for staff
Demand Forecasting:
Advanced demand forecasting algorithms
Seasonal demand prediction
Historical sales data analysis
Forecast accuracy monitoring
Procurement Optimization:
Automated purchase order generation based on reorder points
Supplier performance tracking
Vendor management tools
Supplier communication and collaboration features
Regulatory Compliance:
Serialization and traceability features
Audit trail for compliance reporting
Integration with relevant regulatory databases and reporting systems
Data Security:
Robust data encryption
User authentication and access controls
Regular security audits and updates
Integration Capabilities
Seamless integration with existing accounting, procurement, and ERP systems.
API support for third-party integrations with regulatory databases and reporting tools.
Reporting and Analytics
Create customizable reports for inventory turnover, stockout analysis, supplier performance, and
compliance reporting.
Real-time analytics dashboard for insights.
Scalability:
Ability to scale the system as the company grows
Support for adding new product lines or categories
Technical Specification
Programming Languages and Frameworks:
Web-based application development using modern languages and framewo.
Scenario 3 Capacity and Legality Micah was a talented tennis player a.pdf
Scenario 3 Capacity and Legality Micah was a talented tennis player and started playing
competitively in 2019 when he was sixteen. Many of the tournaments required participants to
sign an exculpatory clause in order to participate. One of Micah's parents signed the agreements.
In 2021, Micah participated in the regional tennis championships in Columbus Georgia. During
the event, several players crashed into each other and Micah sustained an injury to his back,
leaving him partially paralyzed. Micah filed a negligence lawsuit against the tournament
organizers. The organizers could not find the exculpatory clause that Micah signed for the event.
The organizers argued that Micah must have signed the agreement to enter the tournament, but
even if he had not signed one, his participation in the event demonstrated his intent to be bound
by the terms of the agreement. Did Micah have contractual capacity to enter the contract in
2019? Why or why not? If Micah did not actually sign the exculpatory clause, could a court
conclude that he impliedly accepted its terms by participating in the tournament? Why or why
not? How would you rule in both of these situations? Support your answer with scholarly
material and case law..
Scenario 2 Types of Contracts Dr. Brown was an adjunct instructor at.pdf
Scenario 2: Types of Contracts Dr. Brown was an adjunct instructor at North State University
(NSU) in Michigan. Each semester, near the beginning of each term, the parties executed a
written contract that always included the following provision: The parties acknowledge that this
agreement may be cancelled by NSU or the instructor at any time before the first class session.
In the spring semester of Browns sixth year, he filed a complaint with NSU claiming that one of
his students, Sylvia Davis, had engaged in inappropriate and disruptive classroom conduct.
Brown gave her an incomplete grade and asked NSU to request her to apologize as a condition of
receiving a final grade. NSU claimed, and Brown denied, that he was told to assign Davis a
grade if he wanted to teach in the fall. Toward the end of the semester, Brown was told which
classes he would teach in the fall, but the parties did not sign a written contract. The Wednesday
before classes began, NSU terminated him. Brown filed a suit against NSU and others, alleging
breach of contract. Did the parties have a contract? If so, what type of contract did they have.
Did NSU breach it? Explain..
PROJECT 2Construction of a Medical ClinicMany people were severe.pdf
PROJECT 2
Construction of a Medical Clinic
Many people were severely and negatively affected by COVID-19 losing many loved ones.
There is need to construct a medical clinic to service one community in which, many people had
to travel a long distance in order to reach a clinic facility.
Pay attention to the following project information:
Cost R50 million
Return on Investment 10%
Time to market 36 months
Risk factor Low
Chances of success High
There should be a minimum of 10 resources to manage this project.
You have been given five resources. Add the remaining five resources. Resource Role Cost (per
hour)
1. Project Manager Developing project documents R500
2. Accounts Manager Drawing up a budget R500
3. Human Resources Manager Determining employee availability R500
4. Salesperson Negotiating client requirements R300
5. Production Manager Managing the entire event R500
6. IT manager R500
7. operations manager R500
8. Staff manager R500
9. emergency R200
10. Administration R500
Question 6
Using any tool of your choice , create the RESPONSIBILITY MATRIX clearly determining the
primary and support roles for each work item for your chosen project..
Procuring textiles made from recycled fibresBackgroundThe Mini.pdf
Procuring textiles made from recycled fibres
Background
The Ministry of Defence of the Kingdom of the Netherlands (MODNL) is a large public sector
buyer, employing around 58,800 people across the army, air force, military police and other
supporting roles. It is also is one of 45 public and private parties brought together by the Dutch
Government's Circular Procurement Green Deal, which tasked participants with carrying out two
circular procurement initiatives between 2013 and 2016 in order to increase knowledge and
accelerate the transition to a circular economy. One area of focus identified by the MODNL was
textiles, due to its need to equip its large numbers of personnel with uniforms and other textile
products. Textile production is linked with a number of environmental and social problems.
These impacts can vary depending on the type or mix of fibre that a product is made from, but in
general, the energy to transform raw materials and manufacture new fibres in the production of
textiles has a significant carbon footprint, and the eco-toxicity of many industrial inputs (such as
dyes) can result in water, air and terrestrial pollution, damaging ecosystems as well as risking
human health (particularly for workers in the supply-chain subject to inadequate working
conditions). As such, textiles are ranked as the product category with the fourth greatest
environmental impact, after food and drink, transport and housing.1 New textiles are normally
derived from either natural fibres such as cotton (where water-use plus the pollution and
emissions arising from fertilisers and pesticide production and use can be high) or synthetic
fibres such as polyester or nylon, which are derived from fossil fuels. However, a third option -
recycled fibres - also exists, and by seeking to procure textiles containing recycled content, the
MODNL realised it would be able to drastically reduce the environmental impacts of its
procurement, while also supporting manufacturers repurposing the vast amounts of textile waste
created each year in Europe.
Procurement Approach
The MODNL began exploring the market for recycled textiles in January 2014 by publishing a
Request for Information, and conducting an open meeting for sector-related suppliers. The aim
of this market engagement was to assess possibilities for requiring recycled fibres to be used in
the production of certain items. The market research
SCM 322 02 Case
demonstrated that manufacturers were able to meet requirements around the use of recycled
content, and that to facilitate this, the MODNL should focus on functional instead of descriptive
technical specifications i.e. focus on an items ability to perform its use rather than technical
values such as tensile strength. As a result, the MODNL decided to pilot reused fibre
requirements in three lots: towels and wash cloths, overalls, and scarves and handkerchiefs.
Criteria used in the procurement process
Subject matter of the contract
Procuring towels and over.
Read the following excerpt and answer the questions that followPle.pdf
Read the following excerpt and answer the questions that follow
Please note all questions are based on the below
The need to develop global managers has been widely recognised in literature (Holopainen and
Bjorkman, 2005; Riusala, 2000 and Suutari, 1999). For example, 85 per cent of the Fortune 500
firms surveyed do not think that they have an adequate number of global leaders (Gregersen et
all, 1998). International assignment is seen as one of the major tools and a powerful strategy for
developing global leaders. In line with this, Derr (1993) states that expatriation is the top strategy
for internationalising a young manager.
The foreign assignment is also a good management tool as the assignment is generally a higher
position than the previous domestic one and this makes the assignment an important learning
opportunity for the expatriate.
As organisations become globalised, there is an increasing challenge to use expatriates on
international assignments to complete strategically critical tasks (Gregersen & Black 1992,
Brewster 1998, Downes & Thomas 1999).
Multinational corporations (MNCs) use expatriates, not only for corporate control and expertise
reasons in vital global markets, but also to facilitate entry into new markets or to develop
international management competencies (Bird & Dunbar 1991, Boyacigiller 1991, Rosenzweig
1994, Shaffer, Harrison & Gilley 1999, Forster 2000).
While it is recognised that Human Resource Management (HRM) problems are more complex in
the international environment, there is also increased evidence to suggest that the management of
international human resources is increasingly being acknowledged as a major determinant of
success or failure in international business (Tung 1984, Dowling 1999, Hiltrop 1999). For
renowned and established MNCs, failure to be able to communicate and coordinate their
activities in international business has the potential to plunge them into a crisis. The crises
confronting MNCs includes failed assignments due to premature return of expatriates and the
loss of their returned expatriates due to poor repatriation. These crises, due to poor expatriate
management, can therefore threaten the organisations performance and capabilities in the
international arena.
Expatriates may be losing their security blanket, but the challenges and potential longterm
rewards of taking an international assignment are growing. Most global organisations see
international experience as a prerequisite for promotion to the top jobs. Although this experience
can be gained from working in crossborder teams and projects, expatriation remains the preferred
way of creating a global mindset amongst managers.
However, there is a need for new skills and abilities amongst expatriates, and for far more careful
planning and monitoring of assignments by organisational headquarters.
The literature on expatriate management tends to take one of two perspectives. First, much
literature focuses on the individual expatriate's.
Question 4 Read the case study about Ms Zandi and then answer the fo.pdf
Question 4 Read the case study about Ms Zandi and then answer the following questions. Ms
Zandi is the Grade 4 Social Science teacher at an urban school. She has 40 learners in her class
who come from diverse backgrounds. Over the first term of the year, she has acquired an in-
depth knowledge of her learners. She has become aware of the diversity amongst them in terms
of language proficiency, reading and writing competence, interests, learning preferences,
learning behaviours, attitudes to learning, motivational levels, prior knowledge and levels of
achievement in 7 INC3701/103/2023 numeracy. She has one learner, Mbali, who is hearing
impaired and underperforms in Social Sciences. Ms Zandi believes that Mbali can do better if
given support. Three learners are new to the school. From her assessments, Ms Zandi realises
that they are functioning below the grade level in Social Sciences. They have not mastered a lot
of the content and skills in the Grade 2 and 3 Social Sciences curricula. They are struggling with
reading and writing skills in the language of teaching and learning, which is English. Although
they are working hard, they need to be closely supervised especially during group work, or they
become playful and do not focus on the task at hand. Ms Zandi has four learners in her class who
are highly motivated and have a special liking for, curiosity about, and interest in Social
Sciences. Ms Zandi tries her best to stimulate them and provide them with a more enriching and
expanded curriculum. Njabulo appears to have developmental delays and, speaking to his
previous teachers, she learns that he has made progress but is below the grade level in most
subjects. In addition, she has Karabo in her class, a pleasant child who tries hard when Ms Zandi
works with him individually. He enjoys practical activities and working with manipulatives but
is easily distracted. He has gaps in Social Sciences learning and has not mastered some of the
content in the Grade 2 and 3 curricula. 4.1. What challenges does Ms Zandi face with the three
new learners in her class? [3] 4.2. Which support do you think Mbali should be given and why?
[3] 4.3. How would you support the four highly motivated learners in Ms Zandi's class who have
a special interest in Social Sciences? Mention four ways. [4] 4.4 Identify five strategies you
would suggest to Ms Zandi to ensure that all learners are included and that they are not
prejudged based on how they are grouped. [10] [20] Question 4 Read the case study about Ms
Zandi and then answer the following questions. Ms Zandi is the Grade 4 Social Science teacher
at an urban school. She has 40 leamers in her class who come from diverse backgrounds: Over
the first term of the year, she has acquired an in-depth knowledge of her learners. She has
become aware of the diversity amongst them in terms of language proficiency, reading and
writing competence, interests, learning prefertences, learning behaviours, atstudes to leaming.
moti.
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Procuring textiles made from recycled fibresBackgroundThe Mini.pdf
Procuring textiles made from recycled fibresBackgroundThe Mini.pdf
Read the following excerpt and answer the questions that followPle.pdf
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Shell to be modified#include stdlib.h #include unistd.h .pdf
- 1. Shell to be modified #include #include #include #include #include #include #include #include // Simplifed xv6 shell. #define MAXARGS 10 // All commands have at least a type. Have looked at the type, the code // typically casts the *cmd to some specific cmd type. struct cmd { int type; // ' ' (exec), | (pipe), '<' or '>' for redirection }; struct execcmd { int type; // ' ' char *argv[MAXARGS]; // arguments to the command to be exec-ed }; struct redircmd { int type; // < or > struct cmd *cmd; // the command to be run (e.g., an execcmd) char *file; // the input/output file int mode; // the mode to open the file with int fd; // the file descriptor number to use for the file }; struct pipecmd {
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- 6. char *s; int ret; s = *ps; while(s < es && strchr(whitespace, *s)) s++; if(q) *q = s; ret = *s; switch(*s){ case 0: break; case '|': case '<': s++; break; case '>': s++; break; default: ret = 'a'; while(s < es && !strchr(whitespace, *s) && !strchr(symbols, *s)) s++; break; } if(eq) *eq = s; while(s < es && strchr(whitespace, *s)) s++; *ps = s; return ret; } int peek(char **ps, char *es, char *toks)
- 7. { char *s; s = *ps; while(s < es && strchr(whitespace, *s)) s++; *ps = s; return *s && strchr(toks, *s); } struct cmd *parseline(char**, char*); struct cmd *parsepipe(char**, char*); struct cmd *parseexec(char**, char*); // make a copy of the characters in the input buffer, starting from s through es. // null-terminate the copy to make it a string. char *mkcopy(char *s, char *es) { int n = es - s; char *c = malloc(n+1); assert(c); strncpy(c, s, n); c[n] = 0; return c; } struct cmd* parsecmd(char *s) { char *es; struct cmd *cmd; es = s + strlen(s); cmd = parseline(&s, es); peek(&s, es, "");
- 8. if(s != es){ fprintf(stderr, "leftovers: %sn", s); exit(-1); } return cmd; } struct cmd* parseline(char **ps, char *es) { struct cmd *cmd; cmd = parsepipe(ps, es); return cmd; } struct cmd* parsepipe(char **ps, char *es) { struct cmd *cmd; cmd = parseexec(ps, es); if(peek(ps, es, "|")){ gettoken(ps, es, 0, 0); cmd = pipecmd(cmd, parsepipe(ps, es)); } return cmd; } struct cmd* parseredirs(struct cmd *cmd, char **ps, char *es) { int tok; char *q, *eq; while(peek(ps, es, "<>")){ tok = gettoken(ps, es, 0, 0);
- 9. if(gettoken(ps, es, &q, &eq) != 'a') { fprintf(stderr, "missing file for redirectionn"); exit(-1); } switch(tok){ case '<': cmd = redircmd(cmd, mkcopy(q, eq), '<'); break; case '>': cmd = redircmd(cmd, mkcopy(q, eq), '>'); break; } } return cmd; } struct cmd* parseexec(char **ps, char *es) { char *q, *eq; int tok, argc; struct execcmd *cmd; struct cmd *ret; ret = execcmd(); cmd = (struct execcmd*)ret; argc = 0; ret = parseredirs(ret, ps, es); while(!peek(ps, es, "|")){ if((tok=gettoken(ps, es, &q, &eq)) == 0) break; if(tok != 'a') { fprintf(stderr, "syntax errorn"); exit(-1); }
- 10. cmd->argv[argc] = mkcopy(q, eq); argc++; if(argc >= MAXARGS) { fprintf(stderr, "too many argsn"); exit(-1); } ret = parseredirs(ret, ps, es); } cmd->argv[argc] = 0; return ret; } This assignment consists of three parts. The first part asks you to trace the execution of the open () system call. The second part asks you to implement a system call that will allow a user program to obtain the number of operating system traps that the program has encountered since it started running. The third part asks you to modify the xv6 scheduling algorithm so that it considers the number of traps that a program has encountered. 1) (15 points) You will trace the open () system call starting from the time that it is called from a user program to when its trap handler is invoked. We ask you to provide two traces, one when the system call is successful and the other when it is not. For each trace, we ask you to write down in a document which lines in which files of xv6 are executed. Organize the lines of code into blocks of code and explain the purpose of each of block. The result will be a story of what happened after open ( ) is called. The story starts in the user space when the system call first gets executed and ends when the system call returns. 2) (15 points) You will implement the system call Count Traps () that will return the number of traps a user program has encountered since it begins running. 3) (a) (5 points) Describe the existing scheduling policy in xv6. (b) (20 points) You will modify the scheduling algorithm in xv6 such that the program that has encountered the most traps will be executed first. You will need to understand how xv6 handles timer interrupts and process preemption. The xv6 code for these functionalities is basically in proc.c and trap.c. Chapter 6 of the xv6 book will be helpful. (c) (15 points) Write the programs that will help you test and demonstrate your result P(b). (d) (5 points) Criticize the scheduling policy in (b). Propose how you can make it better (Hint: what does "better"