Here's an example of how to pipe three commands together using C. This one uses `ps aux | grep root | grep sbin`. This topic is horribly documented online so hopefully this'll help someone else out.

threePipeDemo.c

// This program is an example of how to run a command such as
// ps aux | grep root | grep sbin
// using C and Unix.

#include <stdlib.h>
#include <unistd.h>

int pid;
int pipe1[2];
int pipe2[2];

void main() {

  // create pipe1
  if (pipe(pipe1) == -1) {
    perror("bad pipe1");
    exit(1);
  }

  // fork (ps aux)
  if ((pid = fork()) == -1) {
    perror("bad fork1");
    exit(1);
  } else if (pid == 0) {
    // stdin --> ps --> pipe1
    exec1();
  }
  // parent

  // create pipe2
  if (pipe(pipe2) == -1) {
    perror("bad pipe2");
    exit(1);
  }

  // fork (grep root)
  if ((pid = fork()) == -1) {
    perror("bad fork2");
    exit(1);
  } else if (pid == 0) {
    // pipe1 --> grep --> pipe2
    exec2();
  }
  // parent

  // close unused fds
  close(pipe1[0]);
  close(pipe1[1]);

  // fork (grep sbin)
  if ((pid = fork()) == -1) {
    perror("bad fork3");
    exit(1);
  } else if (pid == 0) {
    // pipe2 --> grep --> stdout
    exec3();
  }
  // parent


}


void exec1() {
  // input from stdin (already done)
  // output to pipe1
  dup2(pipe1[1], 1);
  // close fds
  close(pipe1[0]);
  close(pipe1[1]);
  // exec
  execlp("ps", "ps", "aux", NULL);
  // exec didn't work, exit
  perror("bad exec ps");
  _exit(1);
}

void exec2() {
  // input from pipe1
  dup2(pipe1[0], 0);
  // output to pipe2
  dup2(pipe2[1], 1);
  // close fds
  close(pipe1[0]);
  close(pipe1[1]);
  close(pipe2[0]);
  close(pipe2[1]);
  // exec
  execlp("grep", "grep", "root", NULL);
  // exec didn't work, exit
  perror("bad exec grep root");
  _exit(1);
}

void exec3() {
  // input from pipe2
  dup2(pipe2[0], 0);
  // output to stdout (already done)
  // close fds
  close(pipe2[0]);
  close(pipe2[1]);
  // exec
  execlp("grep", "grep", "sbin", NULL);
  // exec didn't work, exit
  perror("bad exec grep sbin");
  _exit(1);
}

Could you please explain why it is necessary to close the stdin and stdout files within the exec calls, meaning lines 70, 84, and 87. It seems as if these would need to be open to read from and write to, but I must be misunderstanding something. Thanks! the code is very helpful.

aidanrh7, you raise a good question: Won't closing the pipe preemptively (i.e. in the child process) prevent the child processes from communicating? In short, the answer is no. I'm unfamiliar with the inner mechanisms of dup2; however, with regard to this program, it effectively makes the pipe access points in the child processes unnecessary. This is because the pipe resides in the kernel's address space. So, for example, when you redirect the standard output of ps in child process 1 using dup2, the pipe access points in the child's address space are no longer needed; the binary executed by execlp will direct its standard output to the write-end of the pipe.

It's worth noting that my understanding of this topic is limited; this is simply how it makes sense to me. Hopefully, someone more knowledgeable can come along and offer a proper explanation. I figured I'd offer a simple explanation that might be helpful to someone else who comes along wondering the same thing as you.

More about pipes in Linux: https://unix.stackexchange.com/questions/148401/how-pipes-work-in-linux
More examples of redirection: http://www.cs.loyola.edu/~jglenn/702/S2005/Examples/dup2.html
More about dup2: http://codewiki.wikidot.com/c:system-calls:dup2
Why you should close pipes: https://stackoverflow.com/questions/19265191/why-should-you-close-a-pipe-in-linux.

@aidanrh7 The answer to your question is that inside each process, only one end of one pipe will be written to or read from. The side that will be written to/ read from is duped, everything else is closed. Closing simply means that the fd that was once associated with a given object(in this case the read or write end of a pipe), will be released so that it no longer points to that object in question. The end that was duped, will already have been assigned, either 0 (for stdin) or 1(for stdout), so that either 0 or 1 will be pointing to that end. Hence, the need to release the original fd that was pointing to the end in question.
Note that instead of using dup2(), you could also use close() and dup() as and alternative
So
dup2(pipe2[0], 0);
can also be achieved by
close(0);
dup(pipe2[0]);

thz...

Appreciate the help, but not sure why you would post something that doesn't compile.

Thank you, this helped a lot :)

Thank you for this, it really makes the concept simple and understandable

For additional help, an excellent example of pipes in C in Linux is here under the "EXAMPLES" section: https://man7.org/linux/man-pages/man2/pipe.2.html

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Also, I can confirm that threePipeDemo.c does compile, contrary to what @nskoro says. Just make sure you do NOT use -Werror in your compile command, as there are a variety of warnings.

Here is my compile command. I tested this on Linux Ubuntu 18.04.

# compile
mkdir -p bin && gcc -Wall -Wextra -O3 -std=c17 threePipeDemo.c -o bin/a

# then run the program with:
bin/a 

Here's my compile command and the output with a bunch of warnings:

eRCaGuy_hello_world/c/todo$ mkdir -p bin && gcc -Wall -Wextra -O3 -std=c17 threePipeDemo.c -o bin/a
threePipeDemo.c:52:6: warning: return type of ‘main’ is not ‘int’ [-Wmain]
 void main() {
      ^~~~
threePipeDemo.c: In function ‘main’:
threePipeDemo.c:56:5: warning: implicit declaration of function ‘perror’ [-Wimplicit-function-declaration]
     perror("bad pipe1");
     ^~~~~~
threePipeDemo.c:66:5: warning: implicit declaration of function ‘exec1’; did you mean ‘execl’? [-Wimplicit-function-declaration]
     exec1();
     ^~~~~
     execl
threePipeDemo.c:82:5: warning: implicit declaration of function ‘exec2’; did you mean ‘execl’? [-Wimplicit-function-declaration]
     exec2();
     ^~~~~
     execl
threePipeDemo.c:96:5: warning: implicit declaration of function ‘exec3’; did you mean ‘execl’? [-Wimplicit-function-declaration]
     exec3();
     ^~~~~
     execl
threePipeDemo.c: At top level:
threePipeDemo.c:104:6: warning: conflicting types for ‘exec1’
 void exec1() {
      ^~~~~
threePipeDemo.c:66:5: note: previous implicit declaration of ‘exec1’ was here
     exec1();
     ^~~~~
threePipeDemo.c:118:6: warning: conflicting types for ‘exec2’
 void exec2() {
      ^~~~~
threePipeDemo.c:82:5: note: previous implicit declaration of ‘exec2’ was here
     exec2();
     ^~~~~
threePipeDemo.c:135:6: warning: conflicting types for ‘exec3’
 void exec3() {
      ^~~~~
threePipeDemo.c:96:5: note: previous implicit declaration of ‘exec3’ was here
     exec3();
     ^~~~~

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This is incredibly helpful for anyone trying to understand the super confusing execlp() cmd: I do not understand how execlp() works in Linux