Confusion about execve parameters

Question

I have a program written by my professor that prints the working directory (pwd) by using execve(), but I don't understand the parameters.

pid_t pid = fork();

if(pid <0)
   perror(NULL);
else if(pid == 0)
{
   char*argv[] = {"pwd",NULL};
   execve("/bin/pwd",argv,NULL);
   perror(NULL);
}
else
    printf("Im the parent!");
return 0;
}

"/bin/pwd" gives the path to the executable that will be executed.

This means that it will call the pwd function, doesn't it? Then why do I need to have the parameter pwd?

Couldn't the program run without that parameter?

Answer 1

By convention, the first argument passed to a program is the file name of the executable. However, it doesn't necessarily have to be.

As an example, take the following program:

#include <stdio.h>

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

    printf("number of arguments: %d\n", argc);
    printf("program name: %s\n", argv[0]);
    for (i=1; i<argc; i++) {
        printf("arg %d: %s\n", argv[i]);
    }
    return 0;
}

If you run this program from another like this:

char*argv[] = {"myprog", "A", "B", NULL};
execve("/home/dbush/myprog",argv,NULL);

The above will output:

number of arguments: 3
program name: myprog
arg 1: A
arg 2: B

But you could also run it like this

char*argv[] = {"myotherprog", "A", "B", NULL};
execve("/home/dbush/myprog",argv,NULL);

And it will output:

number of arguments: 3
program name: myotherprog
arg 1: A
arg 2: B

You can use the value of argv[0] as a way to know how your program was called and perhaps expose different functionality based on that.

The popular busybox tool does just this. A single executable is linked with different file names. Depending on which link a user used to run the executable, it can read argv[0] to know whether it was called as ls, ps, pwd, etc.