Hello Members,

I am trying to understand the pipe() system call. I found this program online:

#include <unistd.h>
#include <stdio.h>

#define MSGSIZE  16

char msg1[]  = "hello #1";
char msg2[] =  "hello #2";
char msg3[]  = "hello #3";

{  char inbuf[MSGSIZE];

   int p[2], j;

   /* open pipe */

   if(pipe(p) == -1)
   {    perror("pipe call error");

   /* write down pipe */
   write(p[1], msg1, MSGSIZE);
   write(p[1], msg2, MSGSIZE);
   write(p[1], msg3, MSGSIZE);

   /* read pipe */
   for(j=0; j<3; j++)
   {   read(p[0], inbuf, MSGSIZE);
       printf("%s\n", inbuf);


What I am unable to understand is :

a) How is a string(msg1, msg2, msg3) written to an integer array(p[1])?
b) Are the above strings written one after the other at the same location(p[1])?
c) If all the three strings are written one after the other to p[1]( a single location), how is the program able to read all three of them?

Any answers will be much appreciated.

Thank you!!

6 Years
Discussion Span
Last Post by sciprog1

As per my operating system knowledge, PIPE is a file type which resides in RAM.
Pipe has two ends one is read and another is write.
when u call system call pipe u ll have two typea of disriptor in user file descriptor table.
while writting it will open a file and ll write as we are writting in a file, the only difference is that while reading we have to read in sequence that is first in first out.

This question has already been answered. Start a new discussion instead.
Be sure to adhere to our posting rules.