buffering mechanism when fork is used in c

Question

Possible Duplicate:
Working of fork() in linux gcc
Why does this code print two times?

I want to know the reason behind the output of the below code:

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

int main()
{
   FILE *fp;
   int n;
   printf("%d",45);
   //fflush(stdout);
   if((n=fork())>0){
      printf("in parent\n");  
      exit(0);
   }
   else if(n==0)
       printf("%d",45);
}

Output is

45inparent
4545

If I use fflush, then output is

45inparent
45

Also, I am running on the linux platform

A similar post which explains more about the fork call:http://theunixshell.blogspot.in/2013/07/fork-system-call-and-duplicating-buffer.html — Vijay, Apr 22 '14 at 10:20

score 5 · Answer 1 · answered Jul 05 '12 at 14:25

The child process inherits the open file descriptors (stdout in this case) and the buffer associated with it.

If you don't flush the buffer before the fork, then the content of the buffer is duplicated (including "45"), and "45" is printed twice.
If you flush before the fork, the buffer is emptied and the child gets an empty copy of the buffer, therefore "45" is printed only once by the parent.

score 3 · Answer 2 · answered Jul 05 '12 at 14:17

The first printf() writes the string 45 in a memory buffer.

During the fork() call, the buffer is virtually duplicated in the child process, so both the parent and the child have 45 in stdout`s buffer.

Flushing that buffer in both processes will write 45 twice.

buffering mechanism when fork is used in c

2 Answers2