Question

I am trying to pass arguments to execl() function for ls command. But when I pass

/bin/ls -l -a

as arguments to my program, the execl() function doesn’t recognise the last two arguments. Why is that?Here is the code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
int i,childpid;

if(argc!=4)
{
    printf("you didn't provide any commandline arguments!\n");
    return 0;
}

childpid=fork();

if(childpid<0)
{
    printf("fork() failed\n");
    return 0;
}
else
    if(childpid==0)
    {
         printf("My ID %d\n\n",getpid());
        execl(argv[1],argv[2], argv[3],NULL);

        printf("If you can see this message be ware that Exec() failed!\n");
    }

    while(wait(NULL)>0);

    printf("My ID %d, Parent ID %d, CHild ID %d\n", getpid(),getppid(),childpid);

    return 0;
}

I am on Ubuntu.

Regards

Answer 1

When running your program /bin/ls appears to ignore the -l argument because it is passed in the position reserved for argv[0], which is normally the (rather useless) program name.

Specifically, the first argument to execl is the program to run, and the remaining arguments get copied to argv vector as-is. Since argv[0] is expected to contain the program name and the actual arguments begin from argv[1], you must compensate by providing the program name twice:

execl(argv[1], argv[1], argv[2], argv[3], (char *) NULL);