Write a C++ program 1. Study the function process_text() in file.pdf
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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.
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![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](https://image.slidesharecdn.com/writeacprogram1-230706072830-8d465a9e/85/Write-a-C-program-1-Study-the-function-process_text-in-file-pdf-1-320.jpg)
![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 );](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);
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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.
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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);
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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 ...
Lab01Filesbuild.bat@echo offclsset DRIVE_LETTER=1.docx
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cls
set DRIVE_LETTER=%1:
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Lab01Files/Laboratory 01_ Guessing Game.pdf
8/28/2014 Laboratory 01: Guessing Game
http://mboshart.dyndns.org/boshart/2111Labs/Lab01.html 1/3
Laboratory 01: Guessing Game
Download the following files and place them in your working directory:
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Lab01Files.zip //provided, completed files, refer to header files for method
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PlayerGuessDriver.cpp //most of the work is in this file
range.txt //contains the two integers that define the range for the secret
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build.bat //convenience file for compiling and linking
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IO redirection in C shellPlease implement input output redirect.pdf
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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.
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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 ...
Lab01Filesbuild.bat@echo offclsset DRIVE_LETTER=1.docx
Lab01Files/build.bat
@echo off
cls
set DRIVE_LETTER=%1:
set PATH=%DRIVE_LETTER%\MinGW\bin;%DRIVE_LETTER%\MinGW\msys\1.0\bin;c:\Windows;c:\Windows\system32
g++ -c Keyboard.cpp
g++ -c Random.cpp
g++ -c String.cpp
g++ -c ReadFile.cpp
g++ -c WriteFile.cpp
g++ -c PlayerGuessDriver.cpp
g++ -o PlayerGuess.exe PlayerGuessDriver.o ReadFile.o WriteFile.o String.o Keyboard.o Random.o
Lab01Files/Keyboard.cpp
#include "Keyboard.h"
#include <iostream>
int readInt(string prompt)
{
cout << prompt;
int val = 0;
cin >> val;
return val;
}
int getValidatedInt(string prompt, int min, int max)
{
int validated = readInt(prompt);
cout << validated << endl;
while(validated < min || validated > max)
{
validated = readInt(prompt);
cout << validated << endl;
}
return validated;
}
double readDouble(string prompt)
{
cout << prompt;
double val = 0;
cin >> val;
return val;
}
double getValidatedDouble(string prompt, double min, double max)
{
double validated = readDouble(prompt);
cout << validated << endl;
while(validated < min || validated > max)
{
validated = readDouble(prompt);
cout << validated << endl;
}
return validated;
}
String* readString(string prompt)
{
cout << prompt;
string text;
getline(cin, text);
String* str = createString(text.c_str());
return str;
}
Lab01Files/Keyboard.h
#if !defined KEYBOARD
#define KEYBOARD
#include "Text.h"
#include <string>
using namespace std;
//pre: the string (character literal) that will prompt the user for input
//post: the input read from the keyboard interpreted as an int is returned
int readInt(string prompt);
int getValidatedInt(string prompt, int min, int max);
//pre: the string that will prompt the user for input
//post: the input read from the keyboard interpreted as a double is returned
double readDouble(string prompt);
double getValidatedDouble(string prompt, double min, double max);
//pre: the string that will prompt the user for input
// the string to store the user input and the length of the input storage string
//post: the text read from the keyboard is copied into the storage string
String* readString(string prompt);
#endif
Lab01Files/Laboratory 01_ Guessing Game.pdf
8/28/2014 Laboratory 01: Guessing Game
http://mboshart.dyndns.org/boshart/2111Labs/Lab01.html 1/3
Laboratory 01: Guessing Game
Download the following files and place them in your working directory:
Starting code for this lab:
Lab01Files.zip //provided, completed files, refer to header files for method
signatures
PlayerGuessDriver.cpp //most of the work is in this file
range.txt //contains the two integers that define the range for the secret
number
build.bat //convenience file for compiling and linking
Guessing Game
In this lab, you will write a simple guessing game. The computer will generate a
random integer within a valid spe ...
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, \.
#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.
Rust-lang
Rust
Why do you care about Rust? Who has the time to learn all these new languages? It seems like a new one is popping up every other week and this trend is growing at an exponential rate. Good news, a fair number of them are crafted really well and efficiently solve specific problems. Bad news, how do you keep up with all of this, let alone decide which languages to include in your companies technology portfolio.
Despite the challenges of all these new languages, a majority of developers are intrigued about the idea of becoming a polyglot, but don't know where to begin or don't have the time. In my polyglot travels, there is one language of late that is the sure-fire answer to the above questions, Rust.
In this talk I’ll explore the value behind becoming more polyglotic as a developer, how to pick languages to learn, and then dive deep in the the language of Rust. Which in my opinion, is hands down the best up and coming languages to learn.
About the Presenter
Anthony Broad-Crawford has been a developer since the year 2000 with a short side stint as a semi-professional poker player. Since his transition to software development Anthony has...
1. Built 8 patent receiving technologies
2. Founded two global companies
3. Been a CTO (3x), CPO (1x), and CEO (1x)
and is currently the CTO at Fooda where he manages product, user experience, and engineering. Fooda is predominantly web and mobile technology company focused on bringing great & healthy food from the best restaurant's to people while at the office.
Through his career, in production applications Anthony has used Ruby, Java, Jave (Android), Objective-C and Swift, .NET, Erlang, Scala, Node.JS, LISP, Smalltalk, and even assembly, with his recent favorite, Rust . No, not all at the same time in the same application.
Anthony now spends his time building great teams, that leverage great technology, to build great products, but still looks to codes every chance he can get :)
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.
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.
Find an LCS of X = and Y = Show the c and b table. Attach File .docx
Find an LCS of X = and Y = Show the c and b table. Attach File
IN C++ languageWrite a simple word processor that will accept text.pdf
IN C++ language
Write a simple word processor that will accept text and commands as input. The commands it
should respond to are:
Substitute/oldstring/newstring/
Copy #
Locate/string/
Delete #
Move #
Type #
Pastes Insert #
Replace #
Quit
The program should be able to open an existing file or create a new one. It should keep track of
where the user is in the file at any time. Each command is followed by a number which indicates
the number of line (the current line & those following) upon which the command should act.
Each command should operate on the specified line or lines of the file.
Solution
#include
#include
#include
#include
using namespace std;
const bool DEBUG = false;
const int MAX = 100;
struct file // This contains the file information and contents
{
string contents[MAX];
string fileName;
int currentLineIndex;
int total;
bool markupHighlighting;
};
int NumberInputReader(string); // Reads the numbers from user input
void Help(); // Prints out a help menu
void Markup(file&, int); // Does HTML highlighting
void MarkupApply(file&, string, int); // Applies the HTML highlighting to the file
//void Substitute(file&, string); // substitutes portions of lines of text
//void Type(file&, int); // prints text to the screen
//void Copy(file, string[], int&); // copies text
//void Paste(file&, string[], int, int); // pastes text
//void Locate(file &, string); // finds text
//void Insert(file &, int); // inputs text to the array
//void Delete(file &, int); // deletes text
//void Replace(file &, int); // replaces while lines of text text
//void Move(file &, int); // moves the current line
//void Quit(file); // quits the program, asks the users if they want to save.
void Open(file &); // allows the user to open a file
void New(file &); // allows the user to create a file
void Save(file); // saves the array to a file.
void IndexPrint(file); // for debugging, this prints the contents of the array.
int main() // Main program
{
file txtFile;
txtFile.currentLineIndex=-1;
txtFile.total=0;
int inputValue=0, copyNumber;
string inputString, openYN, copyQue[MAX];
char inputChar, openYNtemp;
// ANIMATION LOOP GOES HERE
cout<<\"~~~\\033[1;32m Welcome \\033[0m~~~\ \"; // first use of colors in the program
cout<<\"Would you like to open an existing file? \"
<\'
Markup(txtFile, inputValue);
break;
case \'>\': // Unhighlight
txtFile.markupHighlighting=false; // if false, it\'ll go through the motions
Markup(txtFile, inputValue); // again, but revert to the original array
break; // afterward.
case \'*\': // * saves file to disk
Save(txtFile);
break;
case \'?\':
Help(); // \"I seem to be having difficulty with my lifestyle.\" -Arthur Dent
break;
case\'.\': // default for re-looping
break;
default:
cout<<\"\\033[1;31m\'\"< total: \"< currentLineIndex: \"<=80) // If it takes up the
entire default width of
{ // the window.
cout<<\"\\033[1;31m-That\'s a very long command, \"
<<\"I\'m just but a humble Line Editor.-\\033[0m \"<=8)
{
inputColor=0; // Default color
cout<<\.
Lecture 15_Strings and Dynamic Memory Allocation.pptx
strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation , strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation, strings and Dynamic Memory Allocation ,strings and Dynamic Memory Al
Object Oriented Programming using C++: Ch12 Streams and Files.pptx
Object Oriented Programming using C++: Ch12 Streams and Files.pptx
Object Oriented Programming Using C++: Ch12 Streams and Files.pptx
Object Oriented Programming Using C++: Ch12 Streams and Files.pptx
String Manipulation Function and Header File Functions
String Manipulation Function and Header File FunctionsBosco Technical Training Society, Don Bosco Technical School (Aff. GGSIP University, New Delhi)
TO UNDERSTAND about stdio.h in C.
TO LEARN ABOUT Math.h in C.
To learn about ctype.h in C.
To understand stdlib.h in c.
To learn about conio.h in c.
To learn about String.h in c.
TO LEARN ABOUT process.h in C.
array, function, pointer, pattern matching
As a part of the course material of CSC 391- Data structure and algorithm. Review of previous concept of array, pointer and functions.
Frequency .java Word frequency counter package frequ.pdf
Frequency .java
/**
* Word frequency counter
*/
package frequency;
import java.util.Iterator;
/**
*
* @author UMD CS
*/
public class Frequency implements Iterable{
private Node first;
private int N;
Frequency(){
N = 0;
first = null;
}
@Override
public Iterator iterator() {
return new ListIterator();
}
/**
*
* List iterator
*
*/
private class ListIterator implements Iterator{
private Node current;
private int index ;
ListIterator(){
current = first;
index = 0;
}
@Override
public boolean hasNext() {
return current != null;
}
public String next() {
if(!hasNext()){
return null;
}
String word = current.key;
int count = current.count;
String r = \"(\"+word + \",\" + Integer.toString(count)+\")\";
current = current.next;
return r;
}
@Override
public void remove() {
}
}
/**
*
* Node class
*
*/
private class Node {
private String key;
private int count;
private Node next;
Node(String item){
key = item;
count = 1;
next = null;
}
@Override
public String toString(){
return \"(\"+key +\",\"+count+\")\";
}
}
/*
* Inserts a word into the linked list. If the word exists, increment the
* count by q.
*/
public void insert(String word){
if(word.equals(\"\")){
return;
}
//TODO
/*
* implement here
*/
}
/**
*
* @param str input string
* This method splits a string into words and pass the words to insert method
*
*/
public void insertWords(String str){
String delims = \"[ .,?!\'\\\"()}{;/<>&=#-:\\\\ _]+\";
String[] words = str.split(delims);
for(String s: words){
s = s.toLowerCase();
insert(s);
}
}
/**
* prints the word frequency list
*/
public void print(){
Node c = first;
while(c != null){
System.out.print(\"(\"+c.key + \",\" + c.count+\")\");
c = c.next;
}
System.out.print(\"\ \");
}
}
--------------------------------------------------------------------------------------------------------------------
------------------------------------------------------
WordFrequency.java
package frequency;
import utils.In;
import utils.Stopwatch;
/**
*
* @author UMD CS
*/
public class WordFrequency {
/**
* @param input source
* @param
* This method receives input resource and return the word frequency string
* DO NOT CHANGE THIS METHOD.
*/
public static String freq(String inputSource, int maxLines){
In in;
// Frequency class can only count the frequency of strings. DO NOT CHANGE THIS.
Frequency freq = new Frequency();
int MAX = 100;
String inputSourceName = inputSource;
try {
in = new In(inputSourceName);
while (!in.isEmpty()) {
String s = in.readLine();
//System.out.println(s);
freq.insertWords(s);
}
}
catch (Exception e) {
e.printStackTrace();
}
StringBuilder strFrequency = new StringBuilder();
int cnt = 0;
for(String s: freq){
strFrequency.append(s);
strFrequency.append(\",\");
cnt++;
if(cnt >= maxLines){break;}
}
return strFrequency.toString();
}
/**
*
*/
public static void main(String[] args) {
In in;
Frequency freq = new Frequency();
int MAX = 100;
// String inputSourceName =
\"http://www.cs.umd.edu/class/summer2015/cmsc132/projects/P3_WordFrequency/test1.html\";
String .
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.
C Programming Project
C Programming Projects -
1. Sort an array in ascending order.
2. Display sum of all odd values stored in an array.
3. Display number of even values stored in an array.
--
1. A file name is command line argument. Display the contents of the file where each word will be displayed on a new line. Display proper message if file does not exist.
2. Display no. of vowels stored in the file.
3. Display no. of “the” stored in the file.
4. Copy contents of the file to another file.
C programming language tutorial
Are you C programming beginners? Learn C programming language via javatpoint.com
Can anyone fix this code, I use Visual StudiosThe Errors im getti.pdf
Can anyone fix this code, I use Visual Studios:
The Errors im getting is that it is not reverseing correctly,
And the line if (arr[i] == strRev[i])
gives me a error code C4700 uninitialized local variable \'strRev\' used
This is the program that neeeds to be created:
In C++, with a myString.h, myStringImp.cpp, and main.cpp:
Design and implement a class myString:
-Use a dynamic array of char to store the characters of myString.
-Use an int to keep track of the length of myString.
-Provide a parameterized constructor that takes an initial string, a default constructor, destructor,
assignment operator, and copy constructor. All need to correctly handle the dynamic array.
-Also provide these member functions:
--length() //the length of myString
--isEqual() //if it is equal to another myString
--print() //output to a terminal cout or outFile
--reverse() //reverse the order of characters in myString
--isPalindrome() //if myString reads the same after being reversed
Here is the code that is broken:
C++ Code:
Header File:myString.h
#ifndef MYSTRING_H_INCLUDED
#define MYSTRING_H_INCLUDED
#include
using namespace std;
//Class definition
class myString
{
public:
//Char Array pointer
char *arr;
//Integer to hold length
int len;
//Default constructor
myString();
//Parameter constructor
myString(string str);
//Copy constructor
myString(const myString &obj);
//Overloading assignment operator
void operator=(const myString &obj);
//Destructor
~myString();
//Returns length
int length();
//Compares two string for equality
bool isEqual(myString &obj);
//Printing string to console
void print();
//Reversing the string
void reverse();
//Check for palindrome of current string and its reverse
bool isPalindrome();
};
#endif // MYSTRING_H_INCLUDED
Implementation File:myString.cpp
/* Implementation File */
#include
#include
#include \"myString.h\"
using namespace std;
//Default Constructor
myString :: myString()
{
//Initializing length to 0
len = 0;
arr = new char[0];
}
//Parameter constructor
myString :: myString(string str)
{
int i;
//Taking length
len = strlen(str.c_str());
//Allocating space
arr = new char[len+1];
//Copying characters
for(i=0; i arr[i] = str[i];
arr[i] = \'\\0\';
}
//Copy constructor
myString :: myString(const myString &obj)
{
//Fetching length
len = obj.len;
//Allocating memory
arr = new char[len];
//Assigning dynamic character arrays
*arr = *obj.arr;
}
//Assignment operator overloading
void myString :: operator=(const myString &obj)
{
//Fetching length
len = obj.len;
//Allocating memory
arr = new char[len];
//Assigning dynamic character arrays
*arr = *obj.arr;
}
//Destructor
myString :: ~myString()
{
//deallocates space
delete arr;
}
//Function that returns length of string
int myString ::length()
{
return len;
}
//Compares two strings and returns true if they are equal else false
bool myString::isEqual(myString &obj)
{
int i, cnt=0;
//Comparing lengths
if(len != (obj.len))
return false;
//Comparing character by character
for(i=0; i {
if(arr[i] ==.
Project Communications Management Case StudySeveral issues have a.pdf
Project Communications Management: Case Study
Several issues have arisen on the Recreation and Wellness Intranet Project. The person from the
HR department who was supporting the project left the company, and now the team needs more
support from that group. A member of the user group that supports the project is extremely vocal
and hard to work with, and other users can hardly get a word in at meetings. The project
manager, Tony, is getting weekly status reports from all of his team members, but many of them
do not address obvious challenges that people are facing. The team is having difficulty deciding
how to communicate various project reports and documents and where to store all of the
information being generated. Recall that the team members include you, a programmer/analyst
and aspiring project manager; Patrick, a network specialist; Nancy, a business analyst; and
Bonnie, another programmer/analyst.
Assignment – Project Communications Management: 3 Tasks
1. Prepare a partial communications management plan to address some of the challenges
mentioned in the previous paragraph.
2. Prepare a template and sample of a good weekly progress report that could be used for this
project. Include a list of tips to help team members provide information on these reports.
3. Write a one-page paper describing two suggested approaches to communicating with the hard-
to-work-with user.
Below I have provided the templates that I was given:
Communications Management Plan
Prepared by: Date:
1. Introduction
2. Collection and filing structure for gathering and storing project information
3. Distribution structure (what information goes to whom, when, and how)
4. Format, content, and level of detail of key project information
5. Production schedule and resources for producing key project information
6. Technologies, access methods, and frequency of communications
7. Method for updating the communications management plan
8. Escalation procedures
9. Stakeholder communications analysis
Stakeholders
Document Name
Document Format
Contact Person
Due Date
Comments:
10. Glossary of terms
Status/Progress Report
Project Name:
Team Member Name:
Date:
Reporting Period:
Work completed this reporting period:
Work to complete next reporting period:
What’s going well and why:
What’s not going well and why:
Suggestions/Issues:
Project changes
Stakeholders
Document Name
Document Format
Contact Person
Due Date
Comments:
Solution
1. Communications Management Plan
Prepared by: Joshua (me) Date: 3rd September 2015
This plan gives a detailed outline regarding communication and information exchange on the
Recreation and Wellness Intranet Project.
For collection and filing of information, all data will be stored on the intranet of the company, in
a seperate folder named Recreation and Wellness. Only members of the project and business
managers will have access to the folder.
Distribution structure - there should be a free and transparent flow of information amongst the
project members. .
Priority encoders are much more common than non-priority encoders. W.pdf
Priority encoders are much more common than non-priority encoders. What do you think the
reasons are for that?
Solution
consider an example.
nonpriority encoders are not used to detect interrupts in controllers only Priority encoders are
used to detecting interrupts in microcontrollers. since we have many interrupts and depends on
situation and no. of priorities the controller has to respond. the priority of the interrupts dettected
using priority encoder..
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C Exam Help
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C Homework Help
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Find an LCS of X = and Y = Show the c and b table. Attach File .docx
Find an LCS of X = and Y = Show the c and b table. Attach File
IN C++ languageWrite a simple word processor that will accept text.pdf
IN C++ language
Write a simple word processor that will accept text and commands as input. The commands it
should respond to are:
Substitute/oldstring/newstring/
Copy #
Locate/string/
Delete #
Move #
Type #
Pastes Insert #
Replace #
Quit
The program should be able to open an existing file or create a new one. It should keep track of
where the user is in the file at any time. Each command is followed by a number which indicates
the number of line (the current line & those following) upon which the command should act.
Each command should operate on the specified line or lines of the file.
Solution
#include
#include
#include
#include
using namespace std;
const bool DEBUG = false;
const int MAX = 100;
struct file // This contains the file information and contents
{
string contents[MAX];
string fileName;
int currentLineIndex;
int total;
bool markupHighlighting;
};
int NumberInputReader(string); // Reads the numbers from user input
void Help(); // Prints out a help menu
void Markup(file&, int); // Does HTML highlighting
void MarkupApply(file&, string, int); // Applies the HTML highlighting to the file
//void Substitute(file&, string); // substitutes portions of lines of text
//void Type(file&, int); // prints text to the screen
//void Copy(file, string[], int&); // copies text
//void Paste(file&, string[], int, int); // pastes text
//void Locate(file &, string); // finds text
//void Insert(file &, int); // inputs text to the array
//void Delete(file &, int); // deletes text
//void Replace(file &, int); // replaces while lines of text text
//void Move(file &, int); // moves the current line
//void Quit(file); // quits the program, asks the users if they want to save.
void Open(file &); // allows the user to open a file
void New(file &); // allows the user to create a file
void Save(file); // saves the array to a file.
void IndexPrint(file); // for debugging, this prints the contents of the array.
int main() // Main program
{
file txtFile;
txtFile.currentLineIndex=-1;
txtFile.total=0;
int inputValue=0, copyNumber;
string inputString, openYN, copyQue[MAX];
char inputChar, openYNtemp;
// ANIMATION LOOP GOES HERE
cout<<\"~~~\\033[1;32m Welcome \\033[0m~~~\ \"; // first use of colors in the program
cout<<\"Would you like to open an existing file? \"
<\'
Markup(txtFile, inputValue);
break;
case \'>\': // Unhighlight
txtFile.markupHighlighting=false; // if false, it\'ll go through the motions
Markup(txtFile, inputValue); // again, but revert to the original array
break; // afterward.
case \'*\': // * saves file to disk
Save(txtFile);
break;
case \'?\':
Help(); // \"I seem to be having difficulty with my lifestyle.\" -Arthur Dent
break;
case\'.\': // default for re-looping
break;
default:
cout<<\"\\033[1;31m\'\"< total: \"< currentLineIndex: \"<=80) // If it takes up the
entire default width of
{ // the window.
cout<<\"\\033[1;31m-That\'s a very long command, \"
<<\"I\'m just but a humble Line Editor.-\\033[0m \"<=8)
{
inputColor=0; // Default color
cout<<\.
Lecture 15_Strings and Dynamic Memory Allocation.pptx
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Object Oriented Programming using C++: Ch12 Streams and Files.pptx
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String Manipulation Function and Header File Functions
String Manipulation Function and Header File FunctionsBosco Technical Training Society, Don Bosco Technical School (Aff. GGSIP University, New Delhi)
TO UNDERSTAND about stdio.h in C.
TO LEARN ABOUT Math.h in C.
To learn about ctype.h in C.
To understand stdlib.h in c.
To learn about conio.h in c.
To learn about String.h in c.
TO LEARN ABOUT process.h in C.
array, function, pointer, pattern matching
As a part of the course material of CSC 391- Data structure and algorithm. Review of previous concept of array, pointer and functions.
Frequency .java Word frequency counter package frequ.pdf
Frequency .java
/**
* Word frequency counter
*/
package frequency;
import java.util.Iterator;
/**
*
* @author UMD CS
*/
public class Frequency implements Iterable{
private Node first;
private int N;
Frequency(){
N = 0;
first = null;
}
@Override
public Iterator iterator() {
return new ListIterator();
}
/**
*
* List iterator
*
*/
private class ListIterator implements Iterator{
private Node current;
private int index ;
ListIterator(){
current = first;
index = 0;
}
@Override
public boolean hasNext() {
return current != null;
}
public String next() {
if(!hasNext()){
return null;
}
String word = current.key;
int count = current.count;
String r = \"(\"+word + \",\" + Integer.toString(count)+\")\";
current = current.next;
return r;
}
@Override
public void remove() {
}
}
/**
*
* Node class
*
*/
private class Node {
private String key;
private int count;
private Node next;
Node(String item){
key = item;
count = 1;
next = null;
}
@Override
public String toString(){
return \"(\"+key +\",\"+count+\")\";
}
}
/*
* Inserts a word into the linked list. If the word exists, increment the
* count by q.
*/
public void insert(String word){
if(word.equals(\"\")){
return;
}
//TODO
/*
* implement here
*/
}
/**
*
* @param str input string
* This method splits a string into words and pass the words to insert method
*
*/
public void insertWords(String str){
String delims = \"[ .,?!\'\\\"()}{;/<>&=#-:\\\\ _]+\";
String[] words = str.split(delims);
for(String s: words){
s = s.toLowerCase();
insert(s);
}
}
/**
* prints the word frequency list
*/
public void print(){
Node c = first;
while(c != null){
System.out.print(\"(\"+c.key + \",\" + c.count+\")\");
c = c.next;
}
System.out.print(\"\ \");
}
}
--------------------------------------------------------------------------------------------------------------------
------------------------------------------------------
WordFrequency.java
package frequency;
import utils.In;
import utils.Stopwatch;
/**
*
* @author UMD CS
*/
public class WordFrequency {
/**
* @param input source
* @param
* This method receives input resource and return the word frequency string
* DO NOT CHANGE THIS METHOD.
*/
public static String freq(String inputSource, int maxLines){
In in;
// Frequency class can only count the frequency of strings. DO NOT CHANGE THIS.
Frequency freq = new Frequency();
int MAX = 100;
String inputSourceName = inputSource;
try {
in = new In(inputSourceName);
while (!in.isEmpty()) {
String s = in.readLine();
//System.out.println(s);
freq.insertWords(s);
}
}
catch (Exception e) {
e.printStackTrace();
}
StringBuilder strFrequency = new StringBuilder();
int cnt = 0;
for(String s: freq){
strFrequency.append(s);
strFrequency.append(\",\");
cnt++;
if(cnt >= maxLines){break;}
}
return strFrequency.toString();
}
/**
*
*/
public static void main(String[] args) {
In in;
Frequency freq = new Frequency();
int MAX = 100;
// String inputSourceName =
\"http://www.cs.umd.edu/class/summer2015/cmsc132/projects/P3_WordFrequency/test1.html\";
String .
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.
C Programming Project
C Programming Projects -
1. Sort an array in ascending order.
2. Display sum of all odd values stored in an array.
3. Display number of even values stored in an array.
--
1. A file name is command line argument. Display the contents of the file where each word will be displayed on a new line. Display proper message if file does not exist.
2. Display no. of vowels stored in the file.
3. Display no. of “the” stored in the file.
4. Copy contents of the file to another file.
C programming language tutorial
Are you C programming beginners? Learn C programming language via javatpoint.com
Can anyone fix this code, I use Visual StudiosThe Errors im getti.pdf
Can anyone fix this code, I use Visual Studios:
The Errors im getting is that it is not reverseing correctly,
And the line if (arr[i] == strRev[i])
gives me a error code C4700 uninitialized local variable \'strRev\' used
This is the program that neeeds to be created:
In C++, with a myString.h, myStringImp.cpp, and main.cpp:
Design and implement a class myString:
-Use a dynamic array of char to store the characters of myString.
-Use an int to keep track of the length of myString.
-Provide a parameterized constructor that takes an initial string, a default constructor, destructor,
assignment operator, and copy constructor. All need to correctly handle the dynamic array.
-Also provide these member functions:
--length() //the length of myString
--isEqual() //if it is equal to another myString
--print() //output to a terminal cout or outFile
--reverse() //reverse the order of characters in myString
--isPalindrome() //if myString reads the same after being reversed
Here is the code that is broken:
C++ Code:
Header File:myString.h
#ifndef MYSTRING_H_INCLUDED
#define MYSTRING_H_INCLUDED
#include
using namespace std;
//Class definition
class myString
{
public:
//Char Array pointer
char *arr;
//Integer to hold length
int len;
//Default constructor
myString();
//Parameter constructor
myString(string str);
//Copy constructor
myString(const myString &obj);
//Overloading assignment operator
void operator=(const myString &obj);
//Destructor
~myString();
//Returns length
int length();
//Compares two string for equality
bool isEqual(myString &obj);
//Printing string to console
void print();
//Reversing the string
void reverse();
//Check for palindrome of current string and its reverse
bool isPalindrome();
};
#endif // MYSTRING_H_INCLUDED
Implementation File:myString.cpp
/* Implementation File */
#include
#include
#include \"myString.h\"
using namespace std;
//Default Constructor
myString :: myString()
{
//Initializing length to 0
len = 0;
arr = new char[0];
}
//Parameter constructor
myString :: myString(string str)
{
int i;
//Taking length
len = strlen(str.c_str());
//Allocating space
arr = new char[len+1];
//Copying characters
for(i=0; i arr[i] = str[i];
arr[i] = \'\\0\';
}
//Copy constructor
myString :: myString(const myString &obj)
{
//Fetching length
len = obj.len;
//Allocating memory
arr = new char[len];
//Assigning dynamic character arrays
*arr = *obj.arr;
}
//Assignment operator overloading
void myString :: operator=(const myString &obj)
{
//Fetching length
len = obj.len;
//Allocating memory
arr = new char[len];
//Assigning dynamic character arrays
*arr = *obj.arr;
}
//Destructor
myString :: ~myString()
{
//deallocates space
delete arr;
}
//Function that returns length of string
int myString ::length()
{
return len;
}
//Compares two strings and returns true if they are equal else false
bool myString::isEqual(myString &obj)
{
int i, cnt=0;
//Comparing lengths
if(len != (obj.len))
return false;
//Comparing character by character
for(i=0; i {
if(arr[i] ==.
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More from jillisacebi75827
Project Communications Management Case StudySeveral issues have a.pdf
Project Communications Management: Case Study
Several issues have arisen on the Recreation and Wellness Intranet Project. The person from the
HR department who was supporting the project left the company, and now the team needs more
support from that group. A member of the user group that supports the project is extremely vocal
and hard to work with, and other users can hardly get a word in at meetings. The project
manager, Tony, is getting weekly status reports from all of his team members, but many of them
do not address obvious challenges that people are facing. The team is having difficulty deciding
how to communicate various project reports and documents and where to store all of the
information being generated. Recall that the team members include you, a programmer/analyst
and aspiring project manager; Patrick, a network specialist; Nancy, a business analyst; and
Bonnie, another programmer/analyst.
Assignment – Project Communications Management: 3 Tasks
1. Prepare a partial communications management plan to address some of the challenges
mentioned in the previous paragraph.
2. Prepare a template and sample of a good weekly progress report that could be used for this
project. Include a list of tips to help team members provide information on these reports.
3. Write a one-page paper describing two suggested approaches to communicating with the hard-
to-work-with user.
Below I have provided the templates that I was given:
Communications Management Plan
Prepared by: Date:
1. Introduction
2. Collection and filing structure for gathering and storing project information
3. Distribution structure (what information goes to whom, when, and how)
4. Format, content, and level of detail of key project information
5. Production schedule and resources for producing key project information
6. Technologies, access methods, and frequency of communications
7. Method for updating the communications management plan
8. Escalation procedures
9. Stakeholder communications analysis
Stakeholders
Document Name
Document Format
Contact Person
Due Date
Comments:
10. Glossary of terms
Status/Progress Report
Project Name:
Team Member Name:
Date:
Reporting Period:
Work completed this reporting period:
Work to complete next reporting period:
What’s going well and why:
What’s not going well and why:
Suggestions/Issues:
Project changes
Stakeholders
Document Name
Document Format
Contact Person
Due Date
Comments:
Solution
1. Communications Management Plan
Prepared by: Joshua (me) Date: 3rd September 2015
This plan gives a detailed outline regarding communication and information exchange on the
Recreation and Wellness Intranet Project.
For collection and filing of information, all data will be stored on the intranet of the company, in
a seperate folder named Recreation and Wellness. Only members of the project and business
managers will have access to the folder.
Distribution structure - there should be a free and transparent flow of information amongst the
project members. .
Priority encoders are much more common than non-priority encoders. W.pdf
Priority encoders are much more common than non-priority encoders. What do you think the
reasons are for that?
Solution
consider an example.
nonpriority encoders are not used to detect interrupts in controllers only Priority encoders are
used to detecting interrupts in microcontrollers. since we have many interrupts and depends on
situation and no. of priorities the controller has to respond. the priority of the interrupts dettected
using priority encoder..
Prove that a reversible adiabatic process cannot also be isothermal f.pdf
Prove that a reversible adiabatic process cannot also be isothermal for T > 0.
Solution
When a system with two independent thermodynamic co-ordinates undergoes a reversible
adiabatic change of state then we can specify one thermodynamic co-ordinate for the final state
but not two , because change in entropy = S2-S1= 0, i.e. S is fixed . Thus a reversible adiabatic
process can\'t be isothermal because such a process will fix both T and S and it is impossible..
Passing arguments by-Value and using return types.#includeiost.pdf
// Passing arguments by-Value and using return types.
#include
using namespace std;
// Constant factor for gravitational acceleration
const double GRAVITATION_CONSTANT = 9.8; // meters/sec^2
// TODO: Declare 3 function prototypes
// - Name: getFallingTime
// Parameters:
// Returns: a double for the amount of time an object falls (to be entered by the user)
// - Name: fallingDistance
// Parameters:
// - 1 double for time in seconds (pass by value)
// Returns: a double for the number of meters the object falls
// - Name: displayResults
// Parameters:
// - 1 double for time (in seconds - pass by value)
// - 1 double for distance (in meters - pass by value)
// Returns:
// ********************************************
int main()
{
// Declare program variables.
double timeInSeconds, distanceInMeters;
// TODO: Call function to get the amount of time the object falls.
// A double value is returned - assign to the correct variable
// TODO: Call function to calculate the distance the object falls,
// given the seconds that it falls
// Pass one argument representing the amount of time the object falls.
// Assign the returned value to the correct variable
// TODO: Call function to print the results.
// Pass the correct two arguments to the function
// Nothing is returned, so do NOT provide a variable assignment
cout << \"End program - \";
return 0;
}
// ********************************************
// TODO: Define getFallingTime()
// Refer to the function prototype above for parameter and return type info
//
// Repeat
// - Prompt the user to enter the seconds the object falls
// - Read the user\'s keyboard response
// Until a value more than 0 is entered
// ********************************************
// TODO: Define fallingDistance()
// Refer to the function prototype above for parameter and return type info
//
// The formula for the distance an object falls due to gravity is:
// d = 1/2 gt^2 (^2 means squared)
// where:
// d = the distance the object falls (in meters)
// g = the gravitational acceleration constant (in meters per second squared)
// t = the amount of time the object falls (in seconds)
// ********************************************
// TODO: Define displayResults()
// Refer to the function prototype above for parameter and return type info
// Output statement to format results as shown in Sample Output
// using values passed to parameters.
/* Sample Output
Sample Run #1:
=====================
Enter seconds the object falls: 0
Enter seconds the object falls: 5.75
In 5.75 seconds, the object falls 162.006 meters.
End program - Press any key to continue . . .
Sample Run #2:
=====================
Enter seconds the object falls: -10
Enter seconds the object falls: -1
Enter seconds the object falls: 0.75
In 0.75 seconds, the object falls 2.75625 meters.
End program - Press any key to continue . . .
*/
Solution
#include
using namespace std;
// Constant factor for gravitational acceleration
const double GRAVITATION_CONSTANT = 9.8; // meters/sec^2
/.
Organophosphates area. herbicidesb. insecticidesDithiocarbama.pdf
Organophosphates are:
a. herbicides
b. insecticides
Dithiocarbamates are herbicides? True or False
Solution
True, because organophosphates are the reasons for many insecticides, herbicides, and nerve
agents. Organophosphate pesticides degrade rapidly by hydrolysis on experience of sunlight, air,
and soil, although small amounts is usually detected in food and drinking water. Their ability to
degrade made them a good looking substitute for the persistent organochloride pesticides.
Match the functions with the plots. The following pictures are image.pdf
Match the functions with the plots. The following pictures are images of the region U = {z = x +
i y: pi/6 lessthanorequalto x lessthanorequalto pi/3, pi/6 lessthanorequalto y lessthanorequalto
pi/3}. Match the plot with the corresponding function. (Notice that the plots have different
scales.)
Solution
Function Plot tan (z) D sin (z) C cos (z) B Log (z) A.
July 4 Paid utilities expense of $370 Cash Accounts Payable Service .pdf
July 4: Paid utilities expense of $370 Cash Accounts Payable Service Revenue u 2 14,000
Accounts Receivable Notes Payable Utilities Expense Office Supplies Common Stock 14,000
Jul. 2 Equipment Dividends
Solution
Cash
July 2
$14000
$370
July 4
July 12
$7100
$200
July 19
$840
July 21
$1600
July 27
Accounts Payable
July 27
$1600
$1600
July 5
Service Revenue
$2900
July 10
Accounts Receivable
July 10
$2900
Notes Payable
$7100
July 12
Utilities Expense
July 4
$370
Office Supplies
July 21
$840
Common Stock
$14000
July 2
Equipment
July 5
$1600
Dividends
July 19
$200
Cash
July 2
$14000
$370
July 4
July 12
$7100
$200
July 19
$840
July 21
$1600
July 27.
Income statements and balance sheets follow for The New York Times C.pdf
Income statements and balance sheets follow for The New York Times Company. Refer to these
financial statements to answer the requirements.
The New York Times Company
Consolidated Statements of Income
Fiscal year ended
(in thousands)
Dec. 29, 2016
Dec. 30, 2015
Revenues
Circulation
$ 880,543
$ 851,790
Advertising
580,732
638,709
Other
94,067
88,716
Total revenues
1,555,342
1,579,215
Production costs
Wages and benefits
363,051
354,516
Raw materials
72,325
77,176
Other
192,728
186,120
Total production costs
628,104
617,812
Selling, general and administrative costs
721,083
713,837
Depreciation and amortization
61,723
61,597
Total operating costs
1,410,910
1,393,246
Restructuring charge
14,804
0
Multiemployer pension plan withdrawal expense
6,730
9,055
Pension settlement charges
21,294
40,329
Early termination charge
0
0
Operating profit
101,604
136,585
Loss from joint ventures
(36,273)
(783)
Interest expense, net
34,805
39,050
Income from continuing operations before income taxes
30,526
96,752
Income tax expense/(benefit)
4,421
33,910
Income from continuing operations
26,105
62,842
Loss from discontinued operations, net of income taxes
(2,273)
0
Net income
23,832
62,842
Net loss attributable to the noncontrolling interest
5,236
404
Net income attributable to The New York Times Company common stockholders
$29,068
$63,246
Continued next page
The New York Times Company
Consolidated Balance Sheets
As of
(in thousands)
Dec. 29, 2016
Dec. 30, 2015
Cash and cash equivalents
$ 100,692
$ 105,776
Short-term investments
449,535
507,639
Accounts receivable, net
197,355
207,180
Prepaid assets
15,948
19,430
Other current assets
32,648
22,507
Total current assets
796,178
862,532
Long-term marketable securities
187,299
291,136
Investments in joint ventures
15,614
22,815
Property plant and equipment, net
596,743
632,439
Goodwill
134,517
109,085
Deferred income taxes
301,342
309,142
Miscellaneous assets
153,702
190,541
Total assets
$2,185,395
$2,417,690
Accounts payable
$ 104,463
$ 96,082
Accrued payroll and other related liabilities
96,463
98,256
Unexpired subscriptions
66,686
60,184
Current portion of long-term debt
0
188,377
Accrued expenses and other
131,125
120,686
Total current liabilities
398,737
563,585
Long-term debt and capital lease obligations
246,978
242,851
Pension benefits obligation
558,790
627,697
Postretirement benefits obligation
57,999
62,879
Other
78,647
92,223
Total other liabilities
942,414
1,025,650
Stockholders’ equity
Common stock of $0.10 par value
Class A common stock
16,921
16,826
Class B convertible stock
82
82
Additional paid-in capital
149,928
146,348
Retained earnings
1,331,911
1,328,744
Common stock held in treasury, at cost
(171,211)
(156,155)
Accumulated other comprehensive loss, net of tax
(479,816)
(509,094)
Total New York Times Company stockholders’ equity
847,815
826,751
Noncontrolling interest
(3,571)
1,704
Total stockholders’ equity
844,244
828,455
Total liabilities and stockholders’ equity
$2,185,395.
$2,185,395
d. Compu.
How do contemporary economic geographers view the economy From a ge.pdf
How do contemporary economic geographers view the economy? From a geographers point of
view,
Solution
The geographers view the distribution of economic resources as the means of economic activity.
Industrial locations today are dependent on the geographical distribution of resources whether
natural or man-made. Though these resources, which impact the economy are highly unequally
distributed. Their point is that the whole of economy is dependent of the geographical
distribution of resources and ease of use and doing business in the geographical location.
The geographers view also consists of service based sectors in the economy such as those
commercial spaces where people uses internet and technology. The geographers view is to
change the conventional economy into a digital economy, where there are robust and dynamic
technological changes. Geographers are interested in the impact of the economic processes on
the geographic structure. For example impact of economic processes on spatial structures..
I am solving Rational Inequalities and I am a little confused about .pdf
I am solving Rational Inequalities and I am a little confused about when I am supposed to change
the direction of the inequality sign. Is it when I divide by a negative number? If so, a negative
numerator? a negative denominator or both?
Solution
If we want to change direction of inequlity sign then we mulitply or divide by a negative number.
If we divide then either divide numerator or denominator ( any one) with negative sign..
How would your plates appear if you incubated agar when they were co.pdf
How would your plates appear if you incubated agar when they were contaminated? What about
if they weren\'t?
Solution
Ans. Incubating a contaminated plate shows microbial growth. Incubation of a sterile plate (no
contamination, no inoculation) shows no microbial growth.
In case of an inoculated plate- the contaminated plate (contaminated with microbes other than
those present in inoculum) show much rapid and outgrowth of many undesirable colonies. When
not contaminated, the same inoculated plate shows less number of colonies and clearly separated
(most often) colonies.
Sometimes it’s preferred to incubate inoculated plates as a control. The control plate serves the
following purpose-
I. Since this plate is not inoculated, there shall be no microbial growth on the plate after
incubation period. If so, absence of microbial growth on plate indicates that the all steps, from
media preparation to inoculation chamber, follows aseptic techniques and are reliable and
reproducible.
II. If an uninoculated plate show microbial growth, contamination has occurred at one or more
steps. Moreover, the cultural techniques adopted need to be recalibrated thoroughly..
Hormones are distributed throughout the body in blood and other body.pdf
Hormones are distributed throughout the body in blood and other body fluids. Why don’t all
tissues/cells
respond to all hormones?
Solution
The “endocrine glands” are ductless glands, which release the hormones into the blood and are
transported to a site distant from its origin where it produces its action. It is known as endocrine
signalling. Though the endocrine hormones travel throughout the blood, only specific tissues
respond to each hormone because hormones bind to specific receptors to produce physiological
effects. For example, the insulin released by pancreas binds to the insulin receptor. The
luteinizing hormone (LH) and follicle-stimulating hormone (FSH) acts on receptors prominent in
ovaries and uterus..
Explain the internet infrastructure stack. SolutionInternet inf.pdf
Explain the internet infrastructure stack.
Solution
Internet infrastructure includes all the hardware and software that helps in the functioning of the
internet ecosystem. In simpler words, all hardware and software that helps a web page appear on
a browser can be defined as internet infrastructure.
It includes:.
Each hormone is known to have a specific target tissue. For each of t.pdf
Each hormone is known to have a specific target tissue. For each of the following hormones, list
its target tissue and describe its specific action. thyroxine estrogen thyroid stimulating hormone
(TSH) insulin follicle stimulating hormone (FSH) What is the role of the hypothalamus in the
production of thyroxine and TsH? How does thyrotropin releasing hormone (TRH travel from
the hypothalamus to the pituitary gland? It travels through the bleed sream
Solution
2.
Hormone
Specific target tissue
Function
Thyroxine
All through the body especially liver and kidneys
Increases the basal metabolic rate and promotes growth as well as development
Estrogen
Female reproductive system
Helps in sexual growth and development and helps in good functioning of the female
reproductive system
TSH
Thyroid gland
Stimulates the thyroid hormone secretion
Insulin
Throughout the body
Helps in regulating levels of blood glucose levels; facilitates the body cells to intake glucose;
helps in increasing glycogem storage
FSH
Sex gland
Stimulates the maturation of female eggs and also production of sperms
3. when T3 and T4 levels are low in the blood, the hypothalamus senses it and secretes TRH
(thyrotropin releasing hormone) which in turn triggers the pituitary to release TSH (thyroid
stimulating hormone) which stimulates the thyroid gland to release thyroxine.
Hormone
Specific target tissue
Function
Thyroxine
All through the body especially liver and kidneys
Increases the basal metabolic rate and promotes growth as well as development
Estrogen
Female reproductive system
Helps in sexual growth and development and helps in good functioning of the female
reproductive system
TSH
Thyroid gland
Stimulates the thyroid hormone secretion
Insulin
Throughout the body
Helps in regulating levels of blood glucose levels; facilitates the body cells to intake glucose;
helps in increasing glycogem storage
FSH
Sex gland
Stimulates the maturation of female eggs and also production of sperms.
Describe the differences in the appearance of the sterile tube of Try.pdf
Describe the differences in the appearance of the sterile tube of Tryptic soy broth and the one in
which microorganisms are growing. Do the same for the slants.
Solution
Uninoculated or sterile Tryptic soy broth appears light amber and clear with no precipitates. If
microorganisms are growing in the broth, then it appears turbid with visible precipitates or
flocculates. Trypticase soy broth is an enrichment medium. It cannot be used for growing
fastidious organisms.
Sterile slants appear clear without any visible colonies. Whereas if microbes are growing, then
the colonies of the microbes can be seen on the slant..
As we entered the 21st century, the Human Genome Project, a massive .pdf
As we entered the 21st century, the Human Genome Project, a massive collaborative effort by
scientists all over the world, reached the first stage of completion; we now have the complete
sequence of human DNA. What is the value of this information?
Solution
Human Genome Project (HGP) is one of the biggest genome projects completed. In this project,
the whole human genome is sequenced to identify all the genes. From the data interpreted till
now, it was concluded that humans share some set of genes with all other species.
Also, humans contain certain unique set of genes that are necessary for intellectual functioning
and the other unique characteristic of humans. However, sharing the genome with the all other
species indicates that all living organisms arose from a common ancestor.
The major goals of Human Genome Project are,
The data provided by the human genome project directs the scientists to identify the genes
associated with several of the diseases. Identifying the disease-causing genes and their
inheritance would be useful in designing treatment for these diseases..
ActiveX is used by which of the following technologiesA. JavaB..pdf
ActiveX is used by which of the following technologies?
A. Java
B. CORBA
C. EJB
D. DCOM
Solution
D. The Distributed Object Component Model (DCOM) allows applications to be divided into
pieces and objects to be run remotely over the network. Potential vulnerabilities exist because of
the way ActiveX is integrated with DCOM. Answer A is incorrect answer B is incorrect because
CORBA is a set of standards that addresses the need for interoperability between hardware and
software. Answer C is incorrect because Enterprise JavaBeans (EJB) is designed for enterprise
networks.
CBE 20. The following information is from financial statements (in mi.pdf
CBE 20. The following information is from financial statements (in millions of dollars Total
assets 150S Net financial assets 90 financial obligations 15S common equity 75S What was the
value of the operating assets? A. 45 B. 60 C. 75 D. 135 21.At the end of the fiscal year ending
June 30, 2003, Microsot equity of $64.9 billion on its balance sheet, with $4 assets (in the form
of cash equivalents and short term investmentancial debt. For fiscal year 2004, the firm reported
S7.4 billio of which $1.1 billion was after-tax earnings on the financial assets. This month
Microsoft is distributing $34 billion of financial assets to shareholders in the of a special
dividend. 9.0 billion invested in and no financing n in comprehensive income, form a. Calculate
Microsoft\'s return on common equity (ROCE) for 2004. Holding all else constant what would
Microsoft\'s ROCE be S34 billion? after the payout of b.
Solution
PART-1).
Below is a simple Role Play Game (RPG) Simulator program.importjav.pdf
Below is a simple Role Play Game (RPG) Simulator program.
importjava.util.Random;
// Student, Programmer
publicclassRPGSimulator {
publicstaticvoidmain(Stringargs[])throwsInterruptedException
{
Randomrand1=newRandom();
Randomrand2=newRandom();
Stringmission1=\"Capture the Castle\";
Stringmission2=\"Enter the Hidden City\";
Stringmission3=\"Rescue the Princess\";
Stringmission4=\"Traverse the Forest\";
Stringmission5=\"Locate the Tunnel Entrance\";
Stringplayer1=\"\",player2=\"\";
charletter=\'\\0\',role=\'\\0\';
// initial health and treasures for the two players
intnumber= 0,mission= 0,health1= 100,health2= 100;
inttreasure1= 100,treasure2= 100,encounter= 0;
// define your role
role= (char)(rand1.nextInt(26) +\'a\');
if(role>=\'a\'&&role<=\'m\')
{
player1=\"protagonist\";
player2=\"opponent\";
}
else
{
player1=\"opponent\";
player2=\"protagonist\";
}
System.out.println(\"you are the \"+player1);
// define your mission
number=rand2.nextInt(5) + 1;
System.out.print(\"your mission is: \");
switch(number)
{
case1: System.out.println(mission1);mission= 1;break;
case2: System.out.println(mission2);mission= 2;break;
case3: System.out.println(mission3);mission= 3;break;
case4: System.out.println(mission4);mission= 4;break;
case5: System.out.println(mission5);mission= 5;break;
}
System.out.println(\"\");
// let the game simulation commence (limit the # of encounters)
for(encounter= 1;encounter<= 20;encounter++)
{
letter= (char)(rand1.nextInt(3) +\'a\');
number=rand2.nextInt(10) + 1;
// randomly encounter the enemy
if(letter==\'a\'||letter==\'b\')
//if(mission == 1 || mission == 2 || mission == 3)
{
health1-= 10;
health2+= 15;
treasure1+= 10;
treasure2-= 10;
//Thread.sleep(5000);
System.out.print(\"battle: \"+encounter);
System.out.println(\"\\t\"+\"health\"+\"\\t\"+\"treasure\");
System.out.println(player1+\"\\t\"+health1+\"\\t\"+treasure1);
System.out.println(player2+\"\\t\"+health2+\"\\t\"+treasure2);
System.out.println(\"\");
//encounter++;
}
elseif(letter==\'c\'||letter==\'d\')
{
health1+= 20;
health2-= 30;
treasure1+= 40;
treasure2-= 20;
//Thread.sleep(500);
System.out.print(\"battle: \"+encounter);
System.out.println(\"\\t\"+\"health\"+\"\\t\"+\"treasure\");
System.out.println(player1+\"\\t\"+health1+\"\\t\"+treasure1);
System.out.println(player2+\"\\t\"+health2+\"\\t\"+treasure2);
System.out.println(\"\");
//break;
}
else
{
health1+= 30;
health2-= 20;
treasure1-= 30;
treasure2+= 20;
//Thread.sleep(500);
System.out.print(\"battle: \"+encounter);
System.out.println(\"\\t\"+\"health\"+\"\\t\"+\"treasure\");
System.out.println(player1+\"\\t\"+health1+\"\\t\"+treasure1);
System.out.println(player2+\"\\t\"+health2+\"\\t\"+treasure2);
System.out.println(\"\");
//break;
}
health1-= 5;
health2-= 5;
treasure1-= 5;
treasure2-= 5;
if(health1<= 0 ||health2<= 0 ||treasure1<= 0 ||treasure2<= 0)
{
//System.out.print(\"encounter: \" + encounter + \" action status: \");
break;
}
}
System.out.println(\"simulation has ended\ \");
System.out.println(\"results:\ \".
A swim bladder allows a fish to Regulate its water content in sea wa.pdf
A swim bladder allows a fish to Regulate its water content in sea water. Regulate its salt
content in sea water. Regulate its salt content in fresh water. Control its buoyancy. Tolerate
changes in water pressure. Monotremes are classified as mammals because they Are carnivores.
Give birth to live young rather than lay eggs. Have mammary glands. Are omnivores. Are
endoderms. In plants which of the following are produced by meiosis? Haploid sporophytes.
Diploid gametes. Haploid gametes. Haploid spores. Diploid spores. Which of the following is
a feature that sets primates apart from all other mammals? Placental embryonic development.
Opposable thumb. Ability to produce milk. Naked faces. Hairy bodies. Fertilization in the
earthworm is accomplished by the use of the Nephridium. Clitellium. Flame cells. Tracheal
system Typhosole. Humans and apes are presently classified in the same category as all of the
following levels except Class. Genus. Kingdo
Solution
question
Answer
explanation
16
d. controls its buoyancy
The swim bladder or air bladder: it is an inner gas-filled organ that adds to the capacity of
numerous hard fish (yet not cartilaginous fish to control their lightness, and in this way to remain
at their ebb and flow water profundity without wasting vitality in swimming.
17
c.have mammary glands
It is an organ situated in the bosoms of females that is in charge of lactation, or the creation of
drain. Both guys and females have glandular tissue inside the bosoms; be that as it may, in
females the glandular tissue starts to create after adolescence in light of estrogen discharge
18
d. haploid spores
Spores are normally haploid and unicellular and are delivered by meiosis in the sporangium of a
diploid sporophyte
19
B, opposable thumb
Primates : they are with completely opposable thumbs incorporate the Great primates (people,
chimpanzees, gorillas, and orangutans) and Old World monkeys
20
B . clitellum
clitellum: it is a thickened glandular and non-portioned area of the body divider close to the head
in worms and bloodsuckers, that secretes a viscid sac in which the eggs are stored. It is available
around 2 cm (0.79 in) behind the foremost end of the body (around the fourteenth, fifteenth and
sixteenth sections)
21
B .genus
question
Answer
explanation
16
d. controls its buoyancy
The swim bladder or air bladder: it is an inner gas-filled organ that adds to the capacity of
numerous hard fish (yet not cartilaginous fish to control their lightness, and in this way to
remain at their ebb and flow water profundity without wasting vitality in swimming.
17
c.have mammary glands
It is an organ situated in the bosoms of females that is in charge of lactation, or the creation of
drain. Both guys and females have glandular tissue inside the bosoms; be that as it may, in
females the glandular tissue starts to create after adolescence in light of estrogen discharge
18
d. haploid spores
Spores are normally haploid and unicellular and are delivered by meios.
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- 1. 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
- 2. 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 );
- 3. 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 ); }
- 4. 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, fstream &fs, char *mode ) { if ( mode == "in" ) fs.open ( s, ios::in ); else if ( mode == "out" ) fs.open ( s, ios::out ); if ( fs.fail() ) { cout << "error opening file " << s << endl; return -1; } else return 1; } //grace_open int main( int argc, char *argv[] ) { const int max_len = 200; char s[max_len+1]; char target[100]; if ( argc < 4 ) { usage( argv[0] ); return 1;
- 5. } char *pin = argv[1]; //pointing to input filename char *pout = argv[2]; //pointing to output filename strcpy ( target, argv[3] ); //target for search fstream ifs, ofs; //input output filestream if ( grace_open ( pin, ifs, "in" ) < 0 ) return 1; //fail if ( grace_open ( pout, ofs, "out" ) < 0 ) return 1; //fail process_text( ifs, ofs, target ); return 0; } 2. Modify ( or write your own ) the program so that it does the following. 1. It displays a detailed help menu when one executes find_pat --help 2. It asks for an additional input argument, "replacing string" in addition to the three arguments described above. Namely, the four input arguments are: input filename, output filename, target string, replacing string. Instead of replacing the target strings by "XXX", replace them by the replacing string in the output file. For example find_pat text_processing.cpp out.txt in abc will replace all "in" by "abc" in "out.txt". 3. Highlight the "replacing string" on the screen instead of the "target string". ( Do not need to display the "target string" any more. ) Solution 1. int find(const string & s, int pos = 0) it searches the invoking string forsstarting at indexposand returns the index wheresis first located. substr() return a substring from the invoking string. c_str() it converts string to c styled string. it returns a const char* that points to a null-terminated c styled basic string. 2. #include #include #include #include
- 6. #include #include #include #include #include using namespace std; void usage ( char program_name[] ) { cout << "Usage: " << program_name << " infile outfile search_string replaced_string" << endl; return; } 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();
- 7. } int process_text( fstream &ifs, fstream &ofs, char target[], char rep_str[] ) { string replace_str( rep_str ); //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 ( rep_str ); //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
- 8. } //while ( !ifs.eof() ) restore_screen(); //restore the old screen return 1; } int grace_open( char *s, fstream &fs, char *mode ) { if ( mode == "in" ) fs.open ( s, ios::in ); else if ( mode == "out" ) fs.open ( s, ios::out ); if ( fs.fail() ) { cout << "error opening file " << s << endl; return -1; } else return 1; } //grace_open int main( int argc, char *argv[] ) { const int max_len = 2000; char s[max_len+1]; char target[100]; char replace[100]; if ( argc < 5 ) { usage( argv[0] ); return 1; } char *pin = argv[1]; //pointing to input filename char *pout = argv[2]; //pointing to output filename strcpy ( target, argv[3] ); //target for search strcpy ( replace, argv[4] ); //target for search fstream ifs, ofs; //input output filestream if ( grace_open ( pin, ifs, "in" ) < 0 ) return 1; //fail if ( grace_open ( pout, ofs, "out" ) < 0 ) return 1; //fail
- 9. process_text( ifs, ofs, target, replace ); return 0; }