IO redirection in C shellPlease implement input output redirect.pdf
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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, \.
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![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;
}
}](https://image.slidesharecdn.com/ioredirectionincshellpleaseimplementinputoutputredirect-230708122143-cf06130f/85/IO-redirection-in-C-shellPlease-implement-input-output-redirect-pdf-1-320.jpg)
![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);](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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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(.
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 ...
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.
Shell to be modified#include stdlib.h #include unistd.h .pdf
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.
Programming For Big Data [ Submission DvcScheduleV2.cpp and StaticA.pdf
Programming For Big Data [ Submission: DvcScheduleV2.cpp and StaticArray.h and/or
DynamicArray.h ]
Assignment 5's runtime was too slow -- a couple of minutes or so. It's because of the duplicate-
checking, with over 4 billion compares.
Rewrite the duplicate-checking logic from Assignment 5, using a technique from "Techniques
For Big Data, Reading" to do fewer compares (check the term first then section number for the
duplicate check), and come up with the exact same results as Assignment 5.
You may use your StaticArray.h from Assignment 3 and/or your DynamicArray.h from
assignments 4, but you may not use any STL containers. Submit the H file(s) you use in your
solution, even if there are no changes since your previous work. Your project will be compiled
for grading using the default stack memory size of 1MB.
Since this version is supposed to be fast, there is no longer a need for a progress bar. Include one
if you wish (you may see the run time dramatically changed), or you may leave it out -- your
choice. But if you do have a progress bar, do remember to "flush"...
[Submission] - Submit the driver program (DvcScheduleV2.cpp) with the header files used
The code I wrote for previous assignment:
Main:
#include
#include
#include
#include
#include "DynamicArray.h"
using namespace std;
struct Class
{
string code;
int count;
};
int main()
{
DynamicArray sub;
DynamicArray sem;
DynamicArray sec;
int totalSubjects = 0;
int dup = 0;
int total = 0;
int counter = 0;
bool duplicate;
bool stored;
// For parsing input file
char* token;
char buf[1000];
const char* const tab = "\t";
// Open input file
ifstream fin;
fin.open("dvc-schedule.txt");
if (!fin.good())
{
cout << "I/O error. File can't be found!\n";
return 1; // Exit the program with an error code
}
// Read the input file
while (fin.good())
{
// Progress bar
if (counter % 1000 == 0)
{
cout << '.';
cout.flush();
}
duplicate = false;
stored = false;
string line;
getline(fin, line);
total++; // Total lines processed
strcpy(buf, line.c_str());
if (buf[0] == 0)
continue; // Skip blank lines
// Parse the line
const string term(token = strtok(buf, tab));
const string section(token = strtok(0, tab));
const string course((token = strtok(0, tab)) ? token : "");
const string instructor((token = strtok(0, tab)) ? token : "");
const string whenWhere((token = strtok(0, tab)) ? token : "");
if (course.find('-') == string::npos)
continue;
const string code(course.begin(), course.begin() + course.find('-'));
// Check for duplicates
for (int i = 0; i < counter; i++)
{
if (sem[i] == term && sec[i] == section)
{
dup++;
duplicate = true;
break;
}
}
if (duplicate == true)
continue;
sem[counter] = term;
sec[counter] = section;
counter++;
for (int i = 0; i < totalSubjects; i++)
{
if (sub[i].code == code)
{
sub[i].count++;
stored = true;
break;
}
}
if (stored == true)
continue;
Class y;
y.code = code;
y.count = 1;
sub[totalSubjects] = y;
totalSubjects++;
}
fin.close();
cout << endl;
for (int i = 0; i < totalSubjects; i++)
{
f.
-- 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
.
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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(.
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 ...
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.
Shell to be modified#include stdlib.h #include unistd.h .pdf
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.
Programming For Big Data [ Submission DvcScheduleV2.cpp and StaticA.pdf
Programming For Big Data [ Submission: DvcScheduleV2.cpp and StaticArray.h and/or
DynamicArray.h ]
Assignment 5's runtime was too slow -- a couple of minutes or so. It's because of the duplicate-
checking, with over 4 billion compares.
Rewrite the duplicate-checking logic from Assignment 5, using a technique from "Techniques
For Big Data, Reading" to do fewer compares (check the term first then section number for the
duplicate check), and come up with the exact same results as Assignment 5.
You may use your StaticArray.h from Assignment 3 and/or your DynamicArray.h from
assignments 4, but you may not use any STL containers. Submit the H file(s) you use in your
solution, even if there are no changes since your previous work. Your project will be compiled
for grading using the default stack memory size of 1MB.
Since this version is supposed to be fast, there is no longer a need for a progress bar. Include one
if you wish (you may see the run time dramatically changed), or you may leave it out -- your
choice. But if you do have a progress bar, do remember to "flush"...
[Submission] - Submit the driver program (DvcScheduleV2.cpp) with the header files used
The code I wrote for previous assignment:
Main:
#include
#include
#include
#include
#include "DynamicArray.h"
using namespace std;
struct Class
{
string code;
int count;
};
int main()
{
DynamicArray sub;
DynamicArray sem;
DynamicArray sec;
int totalSubjects = 0;
int dup = 0;
int total = 0;
int counter = 0;
bool duplicate;
bool stored;
// For parsing input file
char* token;
char buf[1000];
const char* const tab = "\t";
// Open input file
ifstream fin;
fin.open("dvc-schedule.txt");
if (!fin.good())
{
cout << "I/O error. File can't be found!\n";
return 1; // Exit the program with an error code
}
// Read the input file
while (fin.good())
{
// Progress bar
if (counter % 1000 == 0)
{
cout << '.';
cout.flush();
}
duplicate = false;
stored = false;
string line;
getline(fin, line);
total++; // Total lines processed
strcpy(buf, line.c_str());
if (buf[0] == 0)
continue; // Skip blank lines
// Parse the line
const string term(token = strtok(buf, tab));
const string section(token = strtok(0, tab));
const string course((token = strtok(0, tab)) ? token : "");
const string instructor((token = strtok(0, tab)) ? token : "");
const string whenWhere((token = strtok(0, tab)) ? token : "");
if (course.find('-') == string::npos)
continue;
const string code(course.begin(), course.begin() + course.find('-'));
// Check for duplicates
for (int i = 0; i < counter; i++)
{
if (sem[i] == term && sec[i] == section)
{
dup++;
duplicate = true;
break;
}
}
if (duplicate == true)
continue;
sem[counter] = term;
sec[counter] = section;
counter++;
for (int i = 0; i < totalSubjects; i++)
{
if (sub[i].code == code)
{
sub[i].count++;
stored = true;
break;
}
}
if (stored == true)
continue;
Class y;
y.code = code;
y.count = 1;
sub[totalSubjects] = y;
totalSubjects++;
}
fin.close();
cout << endl;
for (int i = 0; i < totalSubjects; i++)
{
f.
-- 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
.
filesHeap.h#ifndef HEAP_H#define HEAP_H#includ.docx
files/Heap.h
#ifndef HEAP_H
#define HEAP_H
#include <vector>
#include <stdexcept> // std::out_of_range
#include <math.h> // pow()
using namespace std;
template<typename T>
class Heap
{
private:
vector<T> _items; // Main vector of elements for heap storage
/**
* Used to take unsorted data and heapify it
*/
void buildHeap()
{
for (int i = _items.size() / 2; i >= 0; i--)
{
percolateDown(i);
}
}
/*********************************************************************/
/********************* Microassignment zone *************************/
/**
* Percolates the item specified at by index down
* into its proper location within a heap.
* Used for dequeue operations and array to heap conversions
* MA TODO: Implement percolateDown!
*/
void percolateDown(int index)
{
}
/**
* Percolate up from a given index to fix heap property
* Used in inserting new nodes into the heap
* MA TODO: Implement percolateUp
*/
void percolateUp( int current_position )
{
}
/************************** Microassigment zone DONE *********************/
public:
/**
* Default empty constructor
*/
Heap()
{
}
/**
* Constructor with a vector of elements
*/
Heap(const vector<T> &unsorted)
{
for (int i = 0; i < unsorted.size(); i++)
{
_items.push_back(unsorted[i]);
}
buildHeap();
}
/**
* Adds a new item to the heap
*/
void insert(T item)
{
int current_position = size(); // Get index location
_items.push_back(item); // Add data to end
percolateUp( current_position ); // Adjust up, as needed
}
/**
* Returns the top-most item in our heap without
* actually removing the item from the heap
*/
T& getFirst()
{
if( size() > 0 )
return _items[0];
else
throw std::out_of_range("No elements in Heap.");
}
/**
* Removes minimum value from heap and returns it to the caller
*/
T deleteMin()
{
int last_index = size() - 1; // Calc last item index
int root_index = 0; // Root index (for readability)
T min_item = _items[root_index]; // Keep item to return
_items[root_index] = _items[last_index]; // Move last item to root
_items.erase(_items.end() - 1); // Erase last element entry
percolateDown(0); // Fix heap property
return min_item;
}
/**
* Returns true if heap is empty, else false
*/
bool isEmpty() const
{
return _items.size() == 0;
}
/**
* Returns current quantity of elements in heap (N)
*/
int size() const
{
return _items.size();
}
/**
* Return heap data in order from the _items vector
*/
string to_s() const
{
string ret = "";
for(int i = 0; i < _items.size(); i++)
{
ret += to_string(_items[i]) + " ";
}
return ret;
}
/**
...
How to Write Node.js Module
Explain how to writie a Node.js Module, C/C++ Addon and upload modules to NPM registry.
In Class AssignmetzCST280W13a-1.pdfCST 280 In-Class Pract.docx
In Class Assignmetz/CST280W13a-1.pdf
CST 280 In-Class Practice – Week 13
Manually determine the configuration of the priority queue (stored as a heap) created
by the following operations. Trace the following logic and define the output:
enqueue(7);
enqueue(17);
enqueue(2);
enqueue(5);
enqueue(22);
enqueue(19);
enqueue(6);
enqueue(11);
enqueue(13);
write the queue contents
dequeue and write front item
enqueue(15);
enqueue(8);
dequeue and write front item
dequeue and write front item
enqueue(24);
enqueue(14);
write the queue contents
Part 2
Then, verify the output by implementing the algorithm by rewriting the priority
queue demonstration program discussed in class. Files needed:
testPQueue.cpp pqType.h heap.cpp
Deliverables
• This cover sheet (with your names on it)
• Driver source code and output for verification program exectution.
In Class Assignmetz/CST280W13b.pdf
CST 280 In-Class Practice – Week 13
Use this page as a worksheet to sketch the progression of the elements up to the first
split for the QuickSort algorithm. Use the middle array element as the split value:
15 34 99 42 11 41 66 23 55 93 48
Next, access the file quickSort.cpp from the course web page. Tailor the program
by entering the array values above in place of the integer values used for an in-class
demonstration. Be sure to adjust the index range to match the size of this array.
Remember that the parameters to the QuickSort algorithm are starting and ending index
values, not the size of the array.
Next, insert code to demonstrate the state of the array after the first split. This should
verify what you did by hand above. Insert the following code at various points within
the partition function to “see” the array at various stages of processing:
for (int i = start; i <= end; i++) // <== ADD
cout << set[i] << ' ';
cout << endl;
Insert the code at these positions:
int partition(int set[], int start, int end)
{
int pivotValue, pivotIndex, mid;
mid = (start + end) / 2;
swap(set[start], set[mid]);
pivotIndex = start;
pivotValue = set[start];
ç HERE
for (int scan = start + 1; scan <= end; scan++)
{
if (set[scan] < pivotValue)
{
pivotIndex++;
swap(set[pivotIndex], set[scan]);
}
ç HERE
}
swap(set[start], set[pivotIndex]);
ç HERE
return pivotIndex;
}
Finally, identify the line that matches what you concluded above.
Deliverables:
Deliver the following for this assignment:
• This work sheet with a sketch of the array first split
• Program source code with required change
• Program output demonstrating array configuration after first split .
Basic c++ 11/14 for python programmers
A short list of some common python programming patterns and their C++ equivalents. This can help programmers learn C++ in a more efficient way if he or she already knows Python.
Part of this material is used for internal training of Appier Inc, one of the leading artificial intelligence company in Asia.
Thank Appier Inc. for allowing me to share this.
#include iostream #include cstring #include vector #i.pdf
#include
#include
#include
#include
#include \"textFile.hpp\"
using namespace std;
void displayHelp();
bool isInteger(const string &s);
bool askSave();
int main(int argc, const char * argv[]) {
int resultCode = 0;
// Convert to a vector.
vector args(argv, argv + argc);
// Process arguments, must be 1 argument, otherwise exit with code 1.
if (args.size() < 2) {
cout << \"Missing command line argument!\" << endl;
cout << \"Usage: ./myEditor \" << endl;
resultCode = 1;
} else {
string fileName = args[1];
textFile dataFile(fileName);
// Check if the file is ok to use, if it is, lets go.
if (!dataFile.checkFile()) {
resultCode = 2;
} else {
// We can now assume the file is good, so we can work with it.
dataFile.loadFile();
// Display some text to help the user.
displayHelp();
// Now get user input.
bool done = false;
string userInput = \"\";
while (!done) {
cout << \"> \";
getline(cin, userInput);
char cmd = tolower(userInput.c_str()[0]);
char cmd2 = \' \';
if (userInput.size() > 1) {
cmd2 = userInput.c_str()[1];
}
// This check makes sure the command format is correct
// so that commands don\'t fire off by mistake.
if (strncmp(&cmd, \" \", 1) != 0 && strncmp(&cmd2, \" \", 1) == 0) {
switch(cmd) {
case \'q\': {
done = true;
break;
}
case \'i\': {
unsigned long lineIndex = dataFile.textLines() + 1;
string newLine;
if (userInput.size() > 1) {
string cmdArg = userInput.substr(2);
lineIndex = stol(cmdArg);
}
cout << to_string(lineIndex) << \"> \";
getline(cin, newLine);
dataFile.insertLine(lineIndex - 1, newLine);
break;
}
case \'d\': {
if (userInput.size() > 1) {
unsigned long lineIndex;
string cmdArg = userInput.substr(2);
lineIndex = stol(cmdArg);
dataFile.removeLine(lineIndex - 1);
}
break;
}
case \'l\': {
for (int x = 0; x < dataFile.textLines(); x++) {
cout << to_string(x + 1) << \"> \" << dataFile.getLine(x) << endl;
}
break;
}
case \'h\': {
displayHelp();
break;
}
case \'s\': {
dataFile.saveFile();
break;
}
}
}
}
// Ask about saving the file before exiting.
bool save = askSave();
if (save) {
dataFile.saveFile();
}
cout << \"Thank you for using myEditor.\" << endl;
}
}
return resultCode;
}
/*
Asks the user if they\'d like to save or not, returns true if they do. Anything
other than \"y\" or \"Y\" will be interpreted as a negative answer.
*/
bool askSave()
{
bool result = false;
string userInput = \"\";
cout << \"Do you wan to save the change? (Y - yes; N - no) \";
getline(cin, userInput);
userInput = tolower(*userInput.c_str());
if (strncmp(userInput.c_str(), \"y\", 1) == 0) {
result = true;
}
return result;
}
/*
Displays help text.
*/
void displayHelp()
{
cout << \"Welcome to my text editor.\" << endl;
cout << \"\\tTo insert text at the end of the file, type \'I\'\" << endl;
cout << \"followed by the text.\" << endl;
cout << \"\\tTo insert text at a certain line number, type \'I\'\" << endl;
cout << \"followed by a space and the desired line number.\" << endl;
cout << \"\\tTo delete a line, type \'D\' followed by a space and the\" << endl;
cou.
Basic C++ 11/14 for Python Programmers
Our new blog post featuring some common python programming patterns and their C++ equivalents is now up!
Leave us a comment below and let us know what you'd like to see covered in our future posts!
█ Read More
Technical Insights: Introduction to GraphQL｜goo.gl/d7PyXH
Identify the organ labelled 3.liverstomacht h e mesenteryjej.pdf
Identify the organ labelled 3.
liver
stomach
t h e mesentery
jejunoileum
large intestine
liver
stomach
t h e mesentery
jejunoileum
large intestine 56 8
Solution
Jejunoileum - the jejunum and the ileum considered as a single organ. Its main function is to
absorb nutrients from particles that pass through..
Fahrenheit ToCelsiusValue.java Done Program name Fahrenheit To Celsiu.pdf
Fahrenheit ToCelsiusValue.java Done Program name Fahrenheit To Celsiusvalue. java
//Programmer name Peter Faragalla Purpose This program takes the fahrenheit temperature as
input and will convert the same to celsius equivalent value, and will print to the screen import
java.util. class FahrenheitToCelsius Value public static void main(String args) Scanner sc new
Scanner (System. in) Reade the temner ature in fahrenheit
Solution
read value of temrature in ferenhite from user
compute celcius from ferenhite using following formula
celcius=(5/9.0)*(ferenhite-32)
print the values.
More Related Content
Similar to IO redirection in C shellPlease implement input output redirect.pdf
filesHeap.h#ifndef HEAP_H#define HEAP_H#includ.docx
files/Heap.h
#ifndef HEAP_H
#define HEAP_H
#include <vector>
#include <stdexcept> // std::out_of_range
#include <math.h> // pow()
using namespace std;
template<typename T>
class Heap
{
private:
vector<T> _items; // Main vector of elements for heap storage
/**
* Used to take unsorted data and heapify it
*/
void buildHeap()
{
for (int i = _items.size() / 2; i >= 0; i--)
{
percolateDown(i);
}
}
/*********************************************************************/
/********************* Microassignment zone *************************/
/**
* Percolates the item specified at by index down
* into its proper location within a heap.
* Used for dequeue operations and array to heap conversions
* MA TODO: Implement percolateDown!
*/
void percolateDown(int index)
{
}
/**
* Percolate up from a given index to fix heap property
* Used in inserting new nodes into the heap
* MA TODO: Implement percolateUp
*/
void percolateUp( int current_position )
{
}
/************************** Microassigment zone DONE *********************/
public:
/**
* Default empty constructor
*/
Heap()
{
}
/**
* Constructor with a vector of elements
*/
Heap(const vector<T> &unsorted)
{
for (int i = 0; i < unsorted.size(); i++)
{
_items.push_back(unsorted[i]);
}
buildHeap();
}
/**
* Adds a new item to the heap
*/
void insert(T item)
{
int current_position = size(); // Get index location
_items.push_back(item); // Add data to end
percolateUp( current_position ); // Adjust up, as needed
}
/**
* Returns the top-most item in our heap without
* actually removing the item from the heap
*/
T& getFirst()
{
if( size() > 0 )
return _items[0];
else
throw std::out_of_range("No elements in Heap.");
}
/**
* Removes minimum value from heap and returns it to the caller
*/
T deleteMin()
{
int last_index = size() - 1; // Calc last item index
int root_index = 0; // Root index (for readability)
T min_item = _items[root_index]; // Keep item to return
_items[root_index] = _items[last_index]; // Move last item to root
_items.erase(_items.end() - 1); // Erase last element entry
percolateDown(0); // Fix heap property
return min_item;
}
/**
* Returns true if heap is empty, else false
*/
bool isEmpty() const
{
return _items.size() == 0;
}
/**
* Returns current quantity of elements in heap (N)
*/
int size() const
{
return _items.size();
}
/**
* Return heap data in order from the _items vector
*/
string to_s() const
{
string ret = "";
for(int i = 0; i < _items.size(); i++)
{
ret += to_string(_items[i]) + " ";
}
return ret;
}
/**
...
How to Write Node.js Module
Explain how to writie a Node.js Module, C/C++ Addon and upload modules to NPM registry.
In Class AssignmetzCST280W13a-1.pdfCST 280 In-Class Pract.docx
In Class Assignmetz/CST280W13a-1.pdf
CST 280 In-Class Practice – Week 13
Manually determine the configuration of the priority queue (stored as a heap) created
by the following operations. Trace the following logic and define the output:
enqueue(7);
enqueue(17);
enqueue(2);
enqueue(5);
enqueue(22);
enqueue(19);
enqueue(6);
enqueue(11);
enqueue(13);
write the queue contents
dequeue and write front item
enqueue(15);
enqueue(8);
dequeue and write front item
dequeue and write front item
enqueue(24);
enqueue(14);
write the queue contents
Part 2
Then, verify the output by implementing the algorithm by rewriting the priority
queue demonstration program discussed in class. Files needed:
testPQueue.cpp pqType.h heap.cpp
Deliverables
• This cover sheet (with your names on it)
• Driver source code and output for verification program exectution.
In Class Assignmetz/CST280W13b.pdf
CST 280 In-Class Practice – Week 13
Use this page as a worksheet to sketch the progression of the elements up to the first
split for the QuickSort algorithm. Use the middle array element as the split value:
15 34 99 42 11 41 66 23 55 93 48
Next, access the file quickSort.cpp from the course web page. Tailor the program
by entering the array values above in place of the integer values used for an in-class
demonstration. Be sure to adjust the index range to match the size of this array.
Remember that the parameters to the QuickSort algorithm are starting and ending index
values, not the size of the array.
Next, insert code to demonstrate the state of the array after the first split. This should
verify what you did by hand above. Insert the following code at various points within
the partition function to “see” the array at various stages of processing:
for (int i = start; i <= end; i++) // <== ADD
cout << set[i] << ' ';
cout << endl;
Insert the code at these positions:
int partition(int set[], int start, int end)
{
int pivotValue, pivotIndex, mid;
mid = (start + end) / 2;
swap(set[start], set[mid]);
pivotIndex = start;
pivotValue = set[start];
ç HERE
for (int scan = start + 1; scan <= end; scan++)
{
if (set[scan] < pivotValue)
{
pivotIndex++;
swap(set[pivotIndex], set[scan]);
}
ç HERE
}
swap(set[start], set[pivotIndex]);
ç HERE
return pivotIndex;
}
Finally, identify the line that matches what you concluded above.
Deliverables:
Deliver the following for this assignment:
• This work sheet with a sketch of the array first split
• Program source code with required change
• Program output demonstrating array configuration after first split .
Basic c++ 11/14 for python programmers
A short list of some common python programming patterns and their C++ equivalents. This can help programmers learn C++ in a more efficient way if he or she already knows Python.
Part of this material is used for internal training of Appier Inc, one of the leading artificial intelligence company in Asia.
Thank Appier Inc. for allowing me to share this.
#include iostream #include cstring #include vector #i.pdf
#include
#include
#include
#include
#include \"textFile.hpp\"
using namespace std;
void displayHelp();
bool isInteger(const string &s);
bool askSave();
int main(int argc, const char * argv[]) {
int resultCode = 0;
// Convert to a vector.
vector args(argv, argv + argc);
// Process arguments, must be 1 argument, otherwise exit with code 1.
if (args.size() < 2) {
cout << \"Missing command line argument!\" << endl;
cout << \"Usage: ./myEditor \" << endl;
resultCode = 1;
} else {
string fileName = args[1];
textFile dataFile(fileName);
// Check if the file is ok to use, if it is, lets go.
if (!dataFile.checkFile()) {
resultCode = 2;
} else {
// We can now assume the file is good, so we can work with it.
dataFile.loadFile();
// Display some text to help the user.
displayHelp();
// Now get user input.
bool done = false;
string userInput = \"\";
while (!done) {
cout << \"> \";
getline(cin, userInput);
char cmd = tolower(userInput.c_str()[0]);
char cmd2 = \' \';
if (userInput.size() > 1) {
cmd2 = userInput.c_str()[1];
}
// This check makes sure the command format is correct
// so that commands don\'t fire off by mistake.
if (strncmp(&cmd, \" \", 1) != 0 && strncmp(&cmd2, \" \", 1) == 0) {
switch(cmd) {
case \'q\': {
done = true;
break;
}
case \'i\': {
unsigned long lineIndex = dataFile.textLines() + 1;
string newLine;
if (userInput.size() > 1) {
string cmdArg = userInput.substr(2);
lineIndex = stol(cmdArg);
}
cout << to_string(lineIndex) << \"> \";
getline(cin, newLine);
dataFile.insertLine(lineIndex - 1, newLine);
break;
}
case \'d\': {
if (userInput.size() > 1) {
unsigned long lineIndex;
string cmdArg = userInput.substr(2);
lineIndex = stol(cmdArg);
dataFile.removeLine(lineIndex - 1);
}
break;
}
case \'l\': {
for (int x = 0; x < dataFile.textLines(); x++) {
cout << to_string(x + 1) << \"> \" << dataFile.getLine(x) << endl;
}
break;
}
case \'h\': {
displayHelp();
break;
}
case \'s\': {
dataFile.saveFile();
break;
}
}
}
}
// Ask about saving the file before exiting.
bool save = askSave();
if (save) {
dataFile.saveFile();
}
cout << \"Thank you for using myEditor.\" << endl;
}
}
return resultCode;
}
/*
Asks the user if they\'d like to save or not, returns true if they do. Anything
other than \"y\" or \"Y\" will be interpreted as a negative answer.
*/
bool askSave()
{
bool result = false;
string userInput = \"\";
cout << \"Do you wan to save the change? (Y - yes; N - no) \";
getline(cin, userInput);
userInput = tolower(*userInput.c_str());
if (strncmp(userInput.c_str(), \"y\", 1) == 0) {
result = true;
}
return result;
}
/*
Displays help text.
*/
void displayHelp()
{
cout << \"Welcome to my text editor.\" << endl;
cout << \"\\tTo insert text at the end of the file, type \'I\'\" << endl;
cout << \"followed by the text.\" << endl;
cout << \"\\tTo insert text at a certain line number, type \'I\'\" << endl;
cout << \"followed by a space and the desired line number.\" << endl;
cout << \"\\tTo delete a line, type \'D\' followed by a space and the\" << endl;
cou.
Basic C++ 11/14 for Python Programmers
Our new blog post featuring some common python programming patterns and their C++ equivalents is now up!
Leave us a comment below and let us know what you'd like to see covered in our future posts!
█ Read More
Technical Insights: Introduction to GraphQL｜goo.gl/d7PyXH
Similar to IO redirection in C shellPlease implement input output redirect.pdf (20)
filesHeap.h#ifndef HEAP_H#define HEAP_H#includ.docx
filesHeap.h#ifndef HEAP_H#define HEAP_H#includ.docx
In Class AssignmetzCST280W13a-1.pdfCST 280 In-Class Pract.docx
In Class AssignmetzCST280W13a-1.pdfCST 280 In-Class Pract.docx
#include iostream #include cstring #include vector #i.pdf
#include iostream #include cstring #include vector #i.pdf
More from forecastfashions
Identify the organ labelled 3.liverstomacht h e mesenteryjej.pdf
Identify the organ labelled 3.
liver
stomach
t h e mesentery
jejunoileum
large intestine
liver
stomach
t h e mesentery
jejunoileum
large intestine 56 8
Solution
Jejunoileum - the jejunum and the ileum considered as a single organ. Its main function is to
absorb nutrients from particles that pass through..
Fahrenheit ToCelsiusValue.java Done Program name Fahrenheit To Celsiu.pdf
Fahrenheit ToCelsiusValue.java Done Program name Fahrenheit To Celsiusvalue. java
//Programmer name Peter Faragalla Purpose This program takes the fahrenheit temperature as
input and will convert the same to celsius equivalent value, and will print to the screen import
java.util. class FahrenheitToCelsius Value public static void main(String args) Scanner sc new
Scanner (System. in) Reade the temner ature in fahrenheit
Solution
read value of temrature in ferenhite from user
compute celcius from ferenhite using following formula
celcius=(5/9.0)*(ferenhite-32)
print the values.
Explain the relationship between three aspects of science hypothesi.pdf
Explain the relationship between three aspects of science: hypothesis, research/experimentation,
and theory.
Solution
The three methods of science hypothesis, research/experimentation, and theory are parts of
scientific method. Hypothesis is a rational prediction of causation among various variables.
hypothesis is an explanation which is based on limited evidence as a basis of future research.
Research is the next step where the hypothesis is tested and studied. It allows to expand the
current knowledge with a specific end goal. A refined hypothesis then becomes the theory.
Did BP respond to the oil spill in a manner that was appropriate to .pdf
Did BP respond to the oil spill in a manner that was appropriate to stockholders or did they take
a stakeholder view?
Solution
Ans:- BP will take stakeholder view because they are the owner, so stakeholders are needed to
give their view in oil spill.
The BP oil spill released 4.9 million barrels of oil into the Gulf of Mexico and caused a grave
amount of damage to the surrounding areas both environmentally and economically. The states
most impacted were Louisiana, Mississippi, Alabama, Texas, and Florida. The main stakeholders
in relation to the spill were the environment, wildlife, fisherman, the oil industry, and tourist-
driven businesses and communities. Directly following the spill, BP set up The Gulf Coast
Claims Facility (GCCF) which was later deemed not independent and later replaced by a court
supervised settlement program which took the subjectivity of the eligibility out of the equation.
Three years after the oil spill, BP announces that the compensation fund was running low but
they would continue to pay the settlements from their profits. Yet BP attempts to stem the flow
of incoming claims using several methods, including suing the court appointed administrator and
appealing claims. In January 2013, BP pleads guilty to 11 counts of felony manslaughter, one
count of felony obstruction of Congress, and violations of the Clear Water and Migratory Bird
Treaty Acts. BP was sentenced to pay $4 billion in fines and penalties. The Deepwater Horizon
Oil Spill has shown that the regulation for spill prevention and response are not sufficient. As the
oil companies will act in the best interest of themselves, it is the duty of the government to set
regulation in place in the interest of the health and safety of its citizens. The government needs to
act to create tighter controls on oil companies to decrease the likelihood of a repeat occurrence.
Stakeholder theory simply states that the stakeholders of a company are not just its direct owners
but that stakeholders are any person, group or entity that a corporation has “benefited or
burdened by its actions and those who benefit or burden the firm with their actions” (Steiner,
2012; Miles, 2012). The first stakeholders impacted by the Deepwater Horizon Oil Spill were the
workmen on the rig itself. When the rig exploded 126 people were on the platform, only 115
were evacuated (Cleveland, 2010). After a three day search covering 5,200 miles, the Coast
Guard called off the rescue operation stating that the period for “reasonable expectations of
survival” had passed. Sadly, the 11 presumed dead members of the crew would not be the only
ones impacted by this tragedy. Other major stakeholders include, the environment, the Gulf
fishing industry, and tourist-dependent businesses and communities.
The environment is always a stakeholder in oil spills. The environmental impact of an oil spill on
sea organisms and their ecosystems has been well-documented. The Oil Pollution Act of 1990
mandates that a Nat.
Consider the subset S = {0, 2, 4, 6, 8, 10, 12} of Z14Z. (a) Show t.pdf
Consider the subset S = {0, 2, 4, 6, 8, 10, 12} of Z/14Z. (a) Show that S = {0, 2, 4, 6, 8, 10, 12}
forms a subring of Z/14Z. (b) Show that S forms a unital ring. (c) Show that S is not a unital
subring of Z/14Z.
Solution
(a) Notice that all elements of S can be written as 2k
Let 2k, 2j in S
Then 2k+2j = 2(k+j) is in S, so clearly sum is in S
Also (2k)(2j) = 4kj = 2(2kj) is in S and so product is also in S
We can also see that 0, the additive identity of Z/14Z is in S by definition
Finally if 2k is in S, then -2k is in S and so S is subring of Z/14Z
(b) The ring has identity 8
Because 2.8 = 16=2, 4.8= 32 = 4, 6.8 = 48 = 6, 8.8 = 64 = 4, 10.8 = 80 = 10, 12.8 = 96 = 12
Hence S is unital
(c) As we can see that we have {2,4,6,10,12} and here 8 is missing and so S is not unital subring
of Z/14Z.
Assignment Description When dealing with a performance improvement .pdf
Assignment Description: When dealing with a performance improvement plan, there are many
things to consider. I would like for you to write a report detailing how a performance
improvement plan is written. Start with what the criteria is and how it is determined. Move on to
the action plans and then end with the formal report.
Solution
Performance improvement plan also known as performance action plan and basically thus plan
designed to enhance the performance of struggling employees or those employees who are not
meeting out the company expectations .
Plan work for employees training , communication development , personality disorder
development if employees are lacking somewhere in this area .
The plan has so many outcomes like transfer, demotion, training and termination .
Mainly report are based on employees overall analysis and understanding about different aspects
.
These are following elements are included in performance improvement plan.
A) Details of performance issues - It\'s included the information of employees ,details of instant
manager, employees job profile ,data that relevant to employees ,action plan, signature of
employee and manager etc .Basically details of employees so that decided the criteria of
performance .
B) Designing a action plan - Basically job is come under the human resource manager that
designed a action plan that mostly based on the interview with employee.
Supervisor establish a provisional action plan for improvement and supervisor must ensure that
for this what kind of training and development required for employee.
C) Review the performance plan- After all this work employee performance should be measured
by the supervisor and report the details to his human resource manager .
D) Meetings with employee - Supervisor must meet with employee and very clear about the area
of improvement and plan of actions .
Supervisor has the authority to change the action plan according to the employee feedback .
E) Follow up- Employee and supervisor meeting on monthly basis or weekly basis that mostly
based on understanding about the employee skill , training requirements , attitudes of employee
towards performance etc .
F) Conclusion - Employee improvement plans has offers several options to employees and
options basically basee on the improvement plans ,on all these basis transfer of employee
possible, demotion of employee possible , termination ,new assignment to employee possible etc
..
Answer ALL 2 QUESTIONS please.... You configured several time settin.pdf
Answer ALL 2 QUESTIONS please.... You configured several time settings in your virtual
machines in this lab. Windows and Linux handle system time settings differently. How so? If
you were to add a router to your virtual network, what would you need to do to allow your
Windows and Linux VMs to use it to communicate with the rest of the network?
Solution
Ans.9. In windows, system time is displayed in a digital format whereas in linux, it is displayed
as a combination of digits and words.
Ans.10. Adding a router to virtual network: Click on VM -> Settings -> Select network Adapter -
> Add the details of the new router..
A heritability of body weight of 0.4 means that body weight is deter.pdf
A heritability of body weight of 0.4 means that body weight is determined for 40% by genetic
factors and for 60% by environmental factors. True or false?
Solution
True. This is not to say that there are no genetic factors influencing body weight, different people
will have different tendencies, to be lean, or not so lean, etc. just like height and other physical
features are determined by many gene variants. Even the microbiome is influenced by both
genetic factors that favor some gut bacteria over other, but diet plays a huge role too. People do
come in all shapes and sizes, but the environment plays a huge role..
After reading the Cultural Competency Module, reflect on your own co.pdf
After reading the Cultural Competency Module, reflect on your own communication
competence. In what areas are you weak? How can you address your weaknesses and build a
strong communication competence?
Solution
Weakness-Different people have different kind of communication styles as because they are
from different region and culture . I for an instant have very less interaction with people i get
very embarrased while interacting with a group of people . i do not like to attend parties or
family reunions as because there will be family memebers who will try to interact with me and i
will continuosly get embarrased beacuse of my shy nature.
Improvement- I need to be confident about myshelf and stop being shy . I should start by
interacting with small group of people and move up by increasing the number of people i interact
with. And attend parties functins and have a casual converastion with family members or friends..
A variable that consists of indivisible categories with no other sco.pdf
A variable that consists of indivisible categories with no other scores existing between
neighboring categories is called a(n) _____ variable.
Solution
DISCRETE VARIABLE. [ANSWER]
***************
Discrete variables only take certain values.
Another possible answer could be \"discontinuous\" variables, because they skip values..
C++ BinaryTree Help Creating main function for Trees...Here are .pdf
C++ BinaryTree Help : Creating main function for Trees...
Here are the header files :
PrintExpressionTour.H:
#ifndef _PRINTEXPRESSIONTOUR_H_
#define _PRINTEXPRESSIONTOUR_H_
#include \"EulerTour.h\"
#include
template
class PrintExpressionTour : public EulerTour {
protected: // ...same type name shortcuts as in EvaluateExpressionTour
public:
inline void execute( const BinaryTree& T ) { // execute the tour
initialize( T );
std::cout << \"Expression: \"; eulerTour( T.root() ); std::cout << std::endl;
}
protected: // leaf: print value
virtual void visitExternal( const Position& p, Result& r ) const {
std::cout << *p;
}
// left: open new expression
virtual void visitLeft( const Position& p, Result& r ) const {
std::cout << \"(\";
}
// below: print operator
virtual void visitBelow( const Position& p, Result& r ) const {
std::cout << *p;
}
// right: close expression
virtual void visitRight( const BinaryTree::Position& p, Result& r ) const {
std::cout << \")\";
}
};
#endif // _PRINTEXPRESSIONTOUR_H_
BinaryTree.h:
#ifndef _BINARYTREE_H_
#define _BINARYTREE_H_
#include
#include
typedef std::string Elem; // base element type
class BinaryTree {
protected:
struct Node { // a node of the tree
Elem elt; // element value
Node* par; // parent
Node* left; // left child
Node* right; // right child
Node() : elt(), par( NULL ), left( NULL ), right( NULL ) {} // constructor
};
public:
class Position { // position in the tree
private:
Node* v; // pointer to the node
public:
Position( Node* _v = NULL ) : v( _v ) {} // constructor
Elem& operator*() const // get element
{
return v->elt;
}
Position left() const // get left child
{
return Position( v->left );
}
Position right() const // get right child
{
return Position( v->right );
}
Position parent() const // get parent
{
return Position( v->par );
}
bool isRoot() const // root of the tree?
{
return v->par == NULL;
}
bool isExternal() const // an external node?
{
return v->left == NULL && v->right == NULL;
}
friend class BinaryTree; // give tree access
};
typedef std::list PositionList; // list of positions
public:
BinaryTree(); // constructor
int size() const; // number of nodes
bool empty() const; // is tree empty?
Position root() const; // get the root
PositionList positions() const; // list of nodes
void addRoot(); // add root to empty tree
void expandExternal( const Position& p ); // expand external node
Position removeAboveExternal( const Position& p ); // remove p and parent
// housekeeping functions omitted...
protected: // local utilities
void preorder( Node* v, PositionList& pl ) const; // preorder utility
private:
Node* _root; // pointer to the root
int n; // number of nodes
};
#endif // _BINARYTREE_H_
EulerTour.h:
#ifndef _EULERTOUR_H_
#define _EULERTOUR_H_
#include \"BinaryTree.h\"
template // element and result types
class EulerTour { // a template for Euler tour
protected:
struct Result { // stores tour results
R leftResult; // result from left subtree
R rightResult; // result from right subtree
R finalR.
Write a short essay explaining the four primary groups of production.pdf
Write a short essay explaining the four primary groups of production. Include in your
explanation an example of each group as well as what we call the payment each group receives.
Solution
Four primary groups of production are four main factors of production which are a foundation of
any productions. these four factors are land, labor, capital, and entrepreneurship.
the first factor is Land that consists all type of natural resources used to create goods and
services. examples :
land, Gold, oil, forest, animal etc.
Labor who transform these natural resources into usable resources. the value of labor depends on
their skills and educations. to increase the productivity of firm owner trained their employees
(labors) to become more skilled.
capital includes buildings, tools, and machines that are labor usees to make outputs of the
process. some examples are trucks, automated machines, computers, etc.
entrepreneurship who create the idea and combine these all three factors of productions and
makes a profit and increases the productivity of a country.
payments to land and capital are rent, wages to labor, and profit to entrepreneurship..
Why didnt the government just arrange to have excess food commodit.pdf
Why didn\'t the government just arrange to have excess food commodities shipped to places
where people didn\'t have enough to eat during the Great Depression?1.Transportation systems
during the 1930\'s had basically shut down all over the country.
Solution
2) is the correct option
The government\'s goal, via the AAA, was to preserve the idea of a market economy for
agricultural products, not just to feed the hungry.
1) is wrong as there is water transport by that time,
3) is wrong as there r some places where people Use to die due to hunger.
4) is wrong as they have the chance to export food grains, some other which dont get spoil for
few months..
Where is Virtual Reality going, and where would it be 20 yearsS.pdf
Where is Virtual Reality going, and where would it be 20 years?
Solution
Answer:
Virtual reality is turnng towards the another world where the influence of artificial intelligence
would be at peaks. Recently t has been declared that Quantum computers will gonna be rule the
future. This computer is going to be detect and fix the errors by itself. Virtual reality is simply a
focus and hence eye tracking will be improve. It is true that Virual reality will rule but the
control would be physical . 4K X 4K resolution would be there and it would be very hard to
unthether the PCs.
What is the negative side to overusing technical communication tools.pdf
What is the negative side to overusing technical communication tools and examine how to avoid
or address the issue.
Solution
Technical communication can be defined as the process of creating something to transfer others
by using electronic gadgets and tools, like computer applications, medical procedures, or
environmental regulations, web pages, help files, or social media sites, or any other form, where
technology used for creating and delivering something. It is the high level languages, where a
common man can not be understand. The people who are professional in nature, they only able to
understand this kind of communication.
The disadvantages are:
These limitations can be covered by providing proper guideliness to the creaters and users of this
information. Otherwise, there may be chance to not understading or misunderstanding the
information..
What is the method of action of aldehydes They damage proteins .pdf
What is the method of action of aldehydes? They damage proteins They disrupt membranes
and DNA They damage nucleic acids They inhibit enzymes.
Solution
Answer will be 2. Aldehydes disrupts membranes and DNA.
Aldehydes like glutaraldehyde and formaldehyde involves a strong association with outer layer
of bacterial cells, specifically with unprotonated amines present on the cell surface, representing
reactive site. this inhibits transport through the cell leading to inhibition of substrate reaching to
rhe enzymes.
Action mechanism of aldehydes also include nucleic acid damage by alkylation of amino- ,
carboxyl- or hydroxyl- group of nucleic acid.
What are rubisco and RuBP And what do they do Briefly list 4 simil.pdf
What are rubisco and RuBP? And what do they do? Briefly list 4 similarities in how ATP is
made in mitochondria and in chloroplasts. A. How does the second law of thermodynamics
explain why diffusion occurs across a membrane? B. Describe how oxidative phosphorylation,
substrate-level phosphorylation, and photo phosphorylation differ in the way ATP is made.
Solution
RuBisCO , is an enzyme Which is used in the Calvin cycle for catalyzing the first step of carbon
fixation, a process by which the atoms of atmospheric carbon dioxide are made available to
organisms in the form of energy-rich molecules such as sucrose. RuBisCO catalyzes either the
carboxylation or oxygenation of ribulose-1,5-bisphosphate (also known as RuBP) with carbon
dioxide or oxygen.
Ribulose-1,5-bisphosphate (RuBP) is an important 5-carbon intermediate in the Calvin cycle
taking place during photosynthesis. It is the substrate, which is used by the enzyme to fix CO2 to
create a highly unstable 6 Carbon PO4 which decays into two molecules of glycerate 3-
phosphate.
In both i.e. Mitochindria as well as in Chloroplast ATP synthesis takes place via a proton
gradient-
Both have ATP synthatases
Both have Electron Transport Chains
Both have 70s ribosomes.
The second law is a trend towards the randimization or increasing entropy. When the
concentration of a substance on both sides of a membrane are equal, the distribution is more
random than when they are unequal. Diffusion of a substance to a region, where it is initially less
concentrated increases entropy, making it energetically favourable (SPONTANEOUS) process.
-Substrate level phosphorylation occurs in the cytoplasm during Glycolysis and Mitochondria
during Krebs cycle.
-2ATP/GTP are produced by conversion of ADP or GDP
OXIDATIVE PHOSPHORYLATION
-Occurs in Mitochondria
Occurs during respiration
pigment systems are not involved
ATP is produced from ADP and iP
Molecular O2 is required for terminal oxidation.
PHOTOPHOSPHORYLATION
Occurs during photosynthesis inside the chloroplasts.
Pigment systemt I & II are involved
Sunlight is the external source of energy
Molecular Oxygen is not required.
Virtual Machines Discuss what are the new innovation today in this .pdf
Virtual Machines: Discuss what are the new innovation today in this area and what effect this
area have on the computer field. Look up information on websites then post the links at the
bottom of your answer. How will this area change in the future? Will this area change society?
Solution
A virtual machine (VM) is an operating system (OS) or application environment
that is installed on software, which imitates dedicated hardware.
The end user has the same experience on a virtual machine as they would have on dedicated
hardware.
Virtualization does, however, require more bandwidth,
storage and processing capacity than a traditional server or desktop
if the physical hardware is going to host multiple running virtual machines.
VMs can easily move, be copied and reassigned between host servers to optimize hardware
resource utilization.
Several vendors offer virtual machine software, but two main vendors dominate in the
marketplace: VMware and Microsoft. VMware has a mature product portfolio, with many years
of use in the IT industry.
While a bit of a late player to virtualization, Microsoft is showing considerable progress.
A classical example of VM which we use is Java Virtual Machine (JVM).
The goal of a Java Virtual Machine (JVM) is to provide a runtime space for a set of Java
code to run on any operating system staged on any hardware platform without needing to
make code changes to accommodate the different operating systems or hardware.
Another important application of VM is in Mainframes. Hypervisor, a variant of VM is used
in providing many machines (called as Partitions in Mainframes) to outside world.
Interesting application of VM is in Cloud Computing.
A cloud is built up of numerous physical machines (the hardware).
Each of these machines then run multiple virtual machines, which is what are
presented to the end-users.
Virtual machines are only limited in the way that their specifications
cannot exceed that of their host (the underlying physical machine). There are exceptions to this
as well, for example Amazon EC2 does the storage offline,
so in that case it would be possible to request drive space that exceeds the host,
but again not exceeding the physical size of the actual place where the storage is done.
The security is the main issue in case of Virtual Machines and a lot of research is going on in this
area.
With newer technolgies like Internet Of Things(IOT), Machine Learning and huge amount of
data to analyse
VM role will further increase in the coming days.
Concept of VM has destroyed the business of big companies like EMC, Dell, IBM, HP... in the
sense
like the small startups who had no option other than purchasing big boxes from these companies
are now using the cloud service to store the data which inturn uses VM concept.
-------------------
Sorry, I cannot give links here as it is againsr the Chegg Policy.
May be you can google out more on the above points to find links and use them..
Using Java, please write the program for the following prompt in the.pdf
Using Java, please write the program for the following prompt in the first picture. The second
and third picture show a sample execution of the program. Collections and generics are
acceptable to use. You just cannot use java methods to iterate through an array list, it must be
iterated through using a for loop. Thank you! A. Objectives l Use at least one array of Strings.
(10 points) 2. Use at least one 2-dimensional array. (10 points) 3. Use sorting to arrange data. (10
points) 1. Use methods to do separate and rcpetitive work. The main method should not have
more than 20 lines of code (10 points) 5. Implement the project correctly. (50 points) 10 points
will be awarded for use of meaningful identificrs, consistent indentation, cxplanatory comments
in your code and properly formatted output. B. Description Each spring brings hope for National
Football League (NFL) teams, especially for those that had poor win-loss records in the previous
season. The NFL draft is held in April and is the opportunity for teams to improve their roster by
adding those college players considered to be the most talented in the world NFL teams develop
their player rosters through three methods: free-agent signings, trading their players for players
from other teams, and drafting college players who have doclared themselves oligiblc for the
NFL draft. The NFL draft is a two-day event in which all 32 teams take their turns selecting
these players. In this project we are going to simulate a limited version of the draft process. This
simulated draft would involve 8 teams and 32 players. Pick any 8 teams from the actual NFL and
32 players of your choice. Be creative! l. Assign a random winning percentage to each of the
teams for the previous season. 2. Based on the winning percentage, rank the teams in order
starting with the one that had the worst performance 3. Each team can now select players in the
order they were ranked 4, Assign numbers l to 4 to the players randomly indicating the round in
which they can be picked. 5. In each round, the teams in order of their numbers will choose
randomly from the players eligible for that round and select them. 6. At thc end of 4 rounds, cach
of the 8 teams will have 4 players they picked.
Solution
import java.applet.*;
import java.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import java.awt.image.*;
import java.util.*;
//Primary class for the game
public class Asteroids extends Applet implements Runnable, KeyListener{
/**
*
*/
private static final long serialVersionUID = 1L;
//the main thread becomes the game loop
Thread gameloop;
//use this as a double buffer
BufferedImage backbuffer;
//the main drawing object for the back buffer
Graphics2D g2d;
//toggle for drawing bounding boxes
boolean showBounds = false;
//create the asteroid array
int ASTEROIDS = 20;
Asteroid[] ast = new Asteroid[ASTEROIDS];
//create the bullet array
int BULLETS = 10;
Bullet[] bullet = new Bullet[BULLETS];
int currentBullet = 0;
//the player\'.
True or FalseBecause IPsec is located in the IP layer, when it is.pdf
True or False:
Because IPsec is located in the IP layer, when it is implemented in a router, there is no need to
change applications software on a user or server system.
Solution
Answer: True
when IPSec is implemented in the router ot firewall, There is no need to change software on a
user or server system because IPsec is located in the IP layer.
This is the one of the advantage of IPSec..
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