Specific pipe command in Ubuntu's shell handling in C

Question

I'm trying to simulate a pipe behavior on Ubuntu's Terminal, for example the command:
"echo hello | wc".
Please assume I got the tokens from stdin, handled everything correctly and now These are the commands I "received" from the user who typed them in the shell for me to handle.
I'm trying to create two processes. Using a pipe, in the first process, I point the file descriptor of the writing edge of the pipe to stdout. The second process should read into stdin with the reading edge of the pipe what execvp(..) returned.?
Here is the code I wrote:

#include <stdio.h>
#include <stdlib.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>

int main()
{
  char* fcmd[] = {"echo", "hello", NULL};
  char* scmd[] = {"wc", NULL};
  pid_t pid;
  int pipe_des[2];
  int i;
  pipe(pipe_des);
  for(i = 0; i < 2; i++)
  {
      pid = fork(); 
      if (pid ==0)
      {
          switch (i)
          {
              case 0: // FIRST CHILD
              {
                  dup2(pipe_des[1], STDOUT_FILENO);
                  close(pipe_des[0]);
                  execvp(fcmd[0], fcmd);
                  exit(0);
              }
              case 1: //SECOND CHILD
              {
                  dup2(pipe_des[0], STDIN_FILENO);
                  close(pipe_des[1]);
                  execvp(scmd[0], scmd);
                  exit(0);
              }
            }
        }
        else if (pid < 0)
            exit(EXIT_FAILURE);
return EXIT_SUCCESS;
}

I get: " amirla@ubuntu:~/Desktop/os/class/ex4$ 1 1 6 "
Like it should, but why he's printing the bash cwd first? The pipe seems to work because I get what I should, according to the length of the word I'm sending with the echo command(in the main()). After that the cursor just waits on the line below for another command without showing me the bash pwd. (maybe stdin is waiting?)
I've looked in many posts on here as well as on other websites and I still can't seem to find a solution to my problem. Any help would be appreciated. Thanks in advance. Note: Please Ignore checking for errors, I've delete them to make the code shorter so assume they exist.

score 1 · Accepted Answer · edited May 23 '17 at 10:24

1

Why do I get a prompt before the output?

Your main process doesn't wait for the children to finish. What you see is:

Main starts
Main creates children
Main exits
BASH prints prompt
Children start their work

To prevent this, you need to wait for the children. See How to wait until all child processes called by fork() complete?

In your case, it's enough to add

 waitpid(-1, NULL, 0);

after the loop.

edited May 23 '17 at 10:24

Community

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answered May 13 '15 at 13:05

Aaron Digulla

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What you said solved everything, I've just needed to wait. I've added "waitpid(-1, NULL, 0);" after the for loop and it works! Thank you! – Amir Lavi May 13 '15 at 13:31

Specific pipe command in Ubuntu's shell handling in C

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