Question

I’m having trouble using dup2 to make a c program execute a command such as ls /bin | grep grep | grep b. When I comment out the third command and associated pipe it executes ls /bin | grep grep fine, but with the last command it just returns right away. Also when I enter ‘ps’ the processes are still running. I think it’s due to how I close the pipes. My code is below:

int main()
{
    int pipeA[2];
    int pipeB[2];

    pipe(pipeA);
    pipe(pipeB);

    int pidA,pidB,pidC;

    if(pidA = fork())
    {
            close(pipeA[0]);
            dup2(pipeA[1],1);
            close(pipeA[1]);
            execlp("ls","ls","/bin",NULL);
            printf("error\n");
    }

    if(pidB = fork())
    {
            close(pipeA[1]);
            dup2(pipeA[0],0);
            close(pipeA[0]);

            close(pipeB[0]);
            dup2(pipeB[1],1);
            close(pipeB[1]);
            execlp("grep","grep","grep",NULL);
            printf("error\n");
    }

    if(pidC = fork())
    {
            close(pipeB[1]);
            dup2(pipeB[0],0);
            close(pipeB[0]);
            execlp("grep","grep","b",NULL);
            printf("error");
    }


    while(pidA != wait(0)){}

    return 0;
}

Answer 1

You aren’t closing enough file descriptors.

/* Semi-working code */
int main()
{
    int pipeA[2];
    int pipeB[2];

    pipe(pipeA);
    pipe(pipeB);

    int pidA,pidB,pidC;

    if (pidA = fork())
    {
            close(pipeB[0]);  // "ls" is not going to use the second pipe
            close(pipeB[1]);  // Ditto
            close(pipeA[0]);
            dup2(pipeA[1], 1);
            close(pipeA[1]);
            execlp("ls", "ls", "/bin", (char *)NULL);
            fprintf(stderr, "error executing 'ls'\n");
            exit(1);
    }

    if (pidB = fork())
    {
            close(pipeA[1]);
            dup2(pipeA[0],0);
            close(pipeA[0]);
            close(pipeB[0]);
            dup2(pipeB[1],1);
            close(pipeB[1]);
            execlp("grep", "grep", "grep", (char *)NULL);
            fprintf(stderr, "error execing 'grep grep'\n");
            exit(1);
    }

    if (pidC = fork())
    {
            close(pipeA[0]);  // The second grep is not going to use the first pipe
            close(pipeA[1]);  // Ditto
            close(pipeB[1]);
            dup2(pipeB[0],0);
            close(pipeB[0]);
            execlp("grep", "grep", "b", (char *)NULL);
            fprintf(stderr, "error execing 'grep b'\n");
            exit(1);
    }

    close(pipeA[0]);  // The parent process is not using the pipes at all
    close(pipeA[1]);
    close(pipeB[0]);
    close(pipeB[1]);

    while (pidA != wait(0))
        ;

    return 0; 
}

Because you didn’t close pipeA in the second grep, you end up with the first grep waiting for input from the pipe the second grep still has open, even though the process will not write to it. Because of that, the first grep does not finish, so the second doesn’t finish either – even though the ls does complete. These comments would apply even if the parent process closed its copies of the pipes – as the corrected code does.

Notice how you end up closing all 4 descriptors returned by the two calls to pipe() in each of the four processes – three children and the parent process.

This leaves one residual problem – the process hierarchy is upside down because of your aconventional use of if (pidA = fork()). You have a child process waiting for its parents. You need to use:

if ((pidA = fork()) == 0)
{
    /* Be childish */
}

Similarly for each of the other two processes. You should also check the pipe() calls and the fork() calls for failure, just to be sure.

#include <stdio.h>
#include <unistd.h>
#include <stdarg.h>
#include <sys/wait.h>
#include <stdlib.h>

static void err_exit(const char *format, ...);

/* Working code */
int main(void)
{
    int pipeA[2];
    int pipeB[2];

    if (pipe(pipeA) != 0 || pipe(pipeB) != 0)
        err_exit("Failed to create a pipe\n");

    int pidA,pidB,pidC;

    if ((pidA = fork()) < 0)
        err_exit("Failed to fork (A)\n");
    else if (pidA == 0)
    {
            close(pipeB[0]);  // "ls" is not going to use the second pipe
            close(pipeB[1]);  // Ditto
            close(pipeA[0]);
            dup2(pipeA[1], 1);
            close(pipeA[1]);
            execlp("ls", "ls", "/bin", (char *)NULL);
            err_exit("error executing 'ls'\n");
    }

    if ((pidB = fork()) < 0)
        err_exit("failed to fork (B)\n");
    else if (pidB == 0)
    {
            close(pipeA[1]);
            dup2(pipeA[0],0);
            close(pipeA[0]);
            close(pipeB[0]);
            dup2(pipeB[1],1);
            close(pipeB[1]);
            execlp("grep", "grep", "grep", (char *)NULL);
            err_exit("error execing 'grep grep'\n");
    }

    if ((pidC = fork()) < 0)
        err_exit("failed to fork (C)\n");
    else if (pidC == 0)
    {
            close(pipeA[0]);  // The second grep is not going to use the first pipe
            close(pipeA[1]);  // Ditto
            close(pipeB[1]);
            dup2(pipeB[0],0);
            close(pipeB[0]);
            execlp("grep", "grep", "b", (char *)NULL);
            err_exit("error execing 'grep b'\n");
    }

    close(pipeA[0]);  // The parent process is not using the pipes at all
    close(pipeA[1]);
    close(pipeB[0]);
    close(pipeB[1]);

    while (wait(0) != -1)
        ;

    printf("Continuing here...\n");
    sleep(3);
    printf("That's enough of that!\n");

    return 0; 
}

static void err_exit(const char *format, ...)
{
    va_list args;
    va_start(args, format);
    vfprintf(stderr, format, args);
    va_end(args);
    exit(1);
}

When doctored to use /usr/bin instead of /bin, this program works OK on Mac OS X 10.7.3. It lists three files and then generates the message about ‘Continuing here’:

bzegrep
bzfgrep
bzgrep
Continuing here...
That's enough of that!