getting input from keyboard, passing to execvp

jumpyg · 11-01-2003, 10:53 AM

Hi, I'm trying to create a simple shell program that reads in a command from the keyboard and passes it to execvp to execute. I'm working in a Linux environment. I'm splitting the input line into a command (the first token) and arguments (the remaining tokens). For example, ls -l has ls as command and -l as the sole argument. The problem is that execvp does not recognize the first argument. If I input "ls -l", it executes just "ls" and drops the "-l." If I input "ls -l -a", then it executes "ls -a." If I input "ls," it doesn't execute anything. Any thoughts? Thanks! Here is my code:

Code:

#include <unistd.h>
#include <iostream>
#include </usr/include/errno.h>
#include </usr/include/sys/errno.h>
#include <string>

using namespace std;

int main(int argc, char *argv[])
{
        // issue prompt
        cout << "> ";

        // get input from user
        string inputStr;
        char inputStrArr[128];
        char * argArr[128];
        char separators[]   = " ,\t\n";
        char *token, *command;
        int numArgs = -1;
        int k;
        int i;

        getline(cin, inputStr);
        for(i=0; i<inputStr.length(); i++)
                inputStrArr[i] = inputStr[i];
	inputStrArr[i] = '\0';
	token = strtok(inputStrArr, separators);
	command = token;

        while( token != NULL )
        {
                token = strtok( NULL, separators );

                argArr[++numArgs] = token;
        }

        argArr[++numArgs] = '\0';
 
        // fork
        int childpid, flag;

        if ((childpid = fork()) == 0) // child code
        {
                cout << "Process ID: child: " << getpid();
                flag = execvp(command, argArr);

                if (flag < 0)
                {
                        perror("execvp failed");
                }
        }
        else if (childpid > 0) // parent code
    {
        cout << "Process ID: " << getpid();
    }
    else // fork unsuccessful
    {
        cout << "\nError: fork failed\n";
    }

return 0;
}

**Salem** · 11-01-2003, 11:00 AM

> command = token;
Follow with

argArr[++numArgs] = token;

argv[0] is the name of the program, not the first parameter

> argArr[++numArgs] = '\0';
This is a pointer, not a character, so
argArr[++numArgs] = NULL;
is more usual

jumpyg · 11-02-2003, 01:32 PM

Thanks Salem, I missed that one!

Now I have another question about comparing the value in a char pointer.
I need to see if the last token on the line is "&" -- which indicates a background process. If so, it needs to be excluded from the argument array and a bool flag set. The problem is I don't know how to compare the value in this last token with "&". Here is my code:

Code:

#include <unistd.h>
#include <iostream>
#include </usr/include/errno.h>
#include </usr/include/sys/errno.h>
#include <string>

using namespace std;

int main(int argc, char *argv[])
{
        cout << "program has started\n";

        // issue prompt
        cout << "> ";

        // get input from user
        bool background = false;
        string inputStr;
        string commandStr;
        char inputStrArr[128];
        char * argArr[128];
        char separators[]   = " ,\t\n";
        char *token, *command;
        int numArgs = -1;
        int k;
        int i;

        getline(cin, inputStr);
        while (inputStr.length() == 0)
        {
                cout << "> ";
                getline(cin, inputStr);
        }

        for(i=0; i<inputStr.length(); i++)
                inputStrArr[i] = inputStr[i];
        inputStrArr[i] = '\0';
                                                                   
	token = strtok( inputStrArr, separators );
        command = token;

        for (i=0; token[i] != '\0'; i++)
        {
                commandStr += token[i];

        }

        argArr[++numArgs] = token;

        while( token != NULL )
        {
                token = strtok( NULL, separators );
                argArr[++numArgs] = token;
        }

        if (argArr[numArgs] == '&')
        {
                background = true;
                argArr[numArgs] = NULL;
        }
        else
                argArr[++numArgs] = NULL;

    // is command built-in?
        if(commandStr == "quit")
        {
                cout << "============= quit called\n";
                return 0;
        }
        else
        {
	    // execute external command
	        int childpid, flag;
	        childpid = fork();
	        cout << "forking\n";
	        if (childpid == 0) // child code
	        {
	                cout << "Process ID: child: " << getpid();
	                flag = execvp(command, argArr);
	
	                if (flag < 0)
	                {
	                        perror("execvp failed");
	                }
	        }
	        else if (childpid > 0) // parent code
	        {
	                cout << "Process ID: " << getpid();
	        }
	        else // fork unsuccessful
	        {
	                cout << "\nError: fork failed\n";
	        }
	} // end else
return 0;
}

**Salem** · 11-02-2003, 01:38 PM

> if (argArr[numArgs] == '&')
Try
if (argArr[numArgs][0] == '&')

> #include </usr/include/errno.h>
> #include </usr/include/sys/errno.h>
1. You shouldn't specify the paths
2. you should only need errno.h
3. perhaps it should be called
#include <cerrno>
if you're following current C++ conventions for including standard C header files

jumpyg · 11-02-2003, 08:49 PM

Thanks again. I've got my shell working well at this point. However, I would like to create a class and split the main pieces of code into functions. The problem I have is with the get_command function. It takes the command from the user and does everything as before. The client then should be able to access the command and arguments. However, I must not be passing the argments to get_command properly, because I'm getting a segmentation error when the client tries to access "command" and "args." Here is the function delaration:
void get_command(char* command, char* args[]);
and the client code:

Code:

        char * argArr[128];
        char * command;
        shell1.get_command(command, argArr);

I tried passing the arguments as pointer references:
void get_command(char*& command, char*& args[]);
but no luck there, either. Then the program won't even compile. Any ideas? Thanks!

11-01-2003, 11:00 AM	#2
Salem and the hat of vanishing Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,214	> command = token; Follow with argArr[++numArgs] = token; argv[0] is the name of the program, not the first parameter > argArr[++numArgs] = '\0'; This is a pointer, not a character, so argArr[++numArgs] = NULL; is more usual __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Up to 8Mb PlusNet broadband from only £5.99 a month!
Salem is offline

11-02-2003, 01:38 PM	#4
Salem and the hat of vanishing Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,214	> if (argArr[numArgs] == '&') Try if (argArr[numArgs][0] == '&') > #include </usr/include/errno.h> > #include </usr/include/sys/errno.h> 1. You shouldn't specify the paths 2. you should only need errno.h 3. perhaps it should be called #include <cerrno> if you're following current C++ conventions for including standard C header files __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Up to 8Mb PlusNet broadband from only £5.99 a month!
Salem is offline

11-02-2003, 08:49 PM	#5
jumpyg Registered User Join Date: Oct 2003 Posts: 9	Thanks again. I've got my shell working well at this point. However, I would like to create a class and split the main pieces of code into functions. The problem I have is with the get_command function. It takes the command from the user and does everything as before. The client then should be able to access the command and arguments. However, I must not be passing the argments to get_command properly, because I'm getting a segmentation error when the client tries to access "command" and "args." Here is the function delaration: void get_command(char* command, char* args[]); and the client code: Code: char * argArr[128]; char * command; shell1.get_command(command, argArr); I tried passing the arguments as pointer references: void get_command(char& command, char& args[]); but no luck there, either. Then the program won't even compile. Any ideas? Thanks!
jumpyg is offline

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