[SOLVED] two forks and the use of wait

Issue

Currently am doing two forks to pipeline two process, but I think am doing my wait(&status) wrong because after the command my shell just hangs and does not return to my prompt. I know my pipe is working because I can see the result if I remove the wait.

Any tips?

pipe(mypipe);
pid1=fork();
if(pid1==0)
{       
    pid2=fork();
    if(pid2==0)
    {
        close(0);
        dup(mypipe[0]);
        close(mypipe[1]);
        execv(foundnode2->path_dir,arv2);
        exit(0);
    }
    close(1);
    dup(mypipe[1]);
    close(mypipe[0]);
    pid2 = wait(&status2);
    execv(foundnode1->path_dir,arv1);
    exit(0);
}
pid1 = wait(&status2);  

Solution

Rule of Thumb: if you use dup() or dup2() to map one end of a pipe to standard input or standard output, you should close() both ends of the pipe itself. You're not doing that; your waits are waiting for the programs to finish, but the programs will not finish because there is still a process with the pipe open that could write to the pipe. Also, the process which created the pipe needs to close both ends of the pipe since it is not itself using the pipe (the child processes are using it). See also C MiniShell — Adding Pipelines.

Also, you should not be waiting for the first child to finish before launching the second (so the pid2 = wait(&status2); line is a bad idea). Pipes have a fairly small capacity (traditionally it was 5 KiB, on many modern systems it is still only 64 KiB); if the total data to be transferred is too large, the writing child may block waiting for the reading child to read, but the reading child hasn't started yet because it is waiting for the writing child to exit (and it takes a long time for this deadlock to resolve itself). You're seeing the output appear without the wait() calls because the second part of the pipeline executes and processes the data from the first part of the pipeline, but it is still waiting for more data to come from the shell.

Taking those tips into account, you might end up with:

pipe(mypipe);
pid1 = fork();
if (pid1 == 0)
{        
    pid2 = fork();
    if (pid2 == 0)
    {
        close(0);
        dup(mypipe[0]);
        close(mypipe[1]);
        close(mypipe[0]);
        execv(foundnode2->path_dir, arv2);
        fprintf(stderr, "Failed to exec %s\n", foundnode2->path_dir);
        exit(1);
    }
    close(1);
    dup(mypipe[1]);
    close(mypipe[0]);
    close(mypipe[1]);
    execv(foundnode1->path_dir, arv1);
    fprintf(stderr, "Failed to exec %s\n", foundnode1->path_dir);
    exit(1);
}
close(mypipe[0]);
close(mypipe[1]);
pid1 = wait(&status1);

Notice the error reporting to standard error when the commands fail to execv(). Also, the exit status of 0 should be reserved for success; 1 is a convenient error exit status, or you can use EXIT_FAILURE from <stdlib.h>.

There is a lot of error checking omitted still; the fork() operations could fail; the pipe() might fail. One consequence is that if the second fork() fails, you still launch the second child (identified by foundnode1->path_dir).

And I note that you could save yourself a little work by moving the pipe creation into the first child process (the parent then does not need to — indeed, cannot — close the pipe):

int pid1 = fork();
if (pid1 == 0)
{
    int mypipe[2];      
    pipe(mypipe);
    int pid2 = fork();
    if (pid2 == 0)
    {
        close(0);
        dup(mypipe[0]);
        close(mypipe[1]);
        close(mypipe[0]);
        execv(foundnode2->path_dir, arv2);
        fprintf(stderr, "Failed to exec %s\n", foundnode2->path_dir);
        exit(1);
    }
    close(1);
    dup(mypipe[1]);
    close(mypipe[0]);
    close(mypipe[1]);
    execv(foundnode1->path_dir, arv1);
    fprintf(stderr, "Failed to exec %s\n", foundnode1->path_dir);
    exit(1);
}
pid1 = wait(&status1);

Answered By - Jonathan Leffler

Answer Checked By - Cary Denson (WPSolving Admin)