Question

I’m learning to work with fork(), and I have some questions.

Consider the following code:

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

int main()
{
   int i;

   for(i = 0; i < 5; i++)
   {
      printf("%d", i);

      if((i%2)==0)
         if(fork())
            fork();
   }
}

When I output to a terminal, I get the result I expect (i.e.: 0,1,1,1,2,2,2,...). But when I output to a file, the result is completely different:

• Case 1: (output to terminal, e.g.: ./a.out):

  Result is: 0,1,1,1,2,2,2,...

• Case 2: (output to file, e.g.: ./a.out > output_file)

  Result is: 0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,...

Why it is like this?

Answer 1

When you output to a file, the stdio library automatically block-buffers the outbound bits.

When a program calls exit(2) or returns from main(), any remaining buffered bits are flushed.

In a program like this that doesn’t generate much output, all of the I/O will occur after the return from main(), when the destination is not a tty. This will often change the pattern and order of I/O operations all by itself.

In this case, the result is further complicated by the series of fork() calls. This will duplicate the partially filled and as-yet-unflushed I/O buffers in each child image.

Before a program calls fork(), one might first flush I/O using fflush(3). If this flush is not done, then you may want all processes except one (typically: the children) to _exit(2) instead of exit(3) or return from main(), to prevent the same bits from being output more than once. (_exit(2) just does the exit system call.)