How to capture output of execvp

Question

I'm developing a program which executes a program using execvp. It needs to capture the results of the child process and parse them in the main process. It seems there is a way, using named pipes, and duping. I'm trying to hunt down a good example of this, but so far no luck. If anyone has any pointers, links and/or suggestions about this, I'd greatly appreciate it.

Answer 1

You don't need named pipes; unnamed pipes work just fine. Actually, often you can just use popen instead of doing the pipe/fork/dup/exec yourself. popen works like this (though your libc's implementation likely has more error checking):

FILE *popen(const char *command, const char *type) {
    int fds[2];
    const char *argv[4] = {"/bin/sh", "-c", command};
    pipe(fds);
    if (fork() == 0) {
        close(fds[0]);
        dup2(type[0] == 'r' ? 0 : 1, fds[1]);
        close(fds[1]);
        execvp(argv[0], argv);
        exit(-1);
    }
    close(fds[1]);
    return fdopen(fds[0], type);
}

This creates an unnamed pipe, and forks. In the child, it reattaches stdout (or stdin) to one end of the pipe, then execs the child. The parent can simply read (or write) from the other end of the pipe.

Answer 2

Can't you just use popen()?

Answer 3

Here is a simple example that demonstrates the use of popen to achieve your goal. Just put something more interesting than "echo" as the command :)

#include <stdio.h>

int main() 
{
    char buf[100];
    int i = 0;
    FILE *p = popen("echo \"Test\"","r");
    if (p != NULL )
    {
        while (!feof(p) && (i < 99) )
        {
            fread(&buf[i++],1,1,p);
        }
        buf[i] = 0;
        printf("%s",buf);
        pclose(p);
        return 0;
    }
    else
    {
        return -1;
    }
}