CODEWITHSUNDEEP
cgcc
Salil
Salil

Reputation: 1811

fork() behavior in linux

I was trying to understand forks, and tried following in C:

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

void forker()
{
    printf("%d: A\n",(int)getpid());
    fork();
    wait();
    printf("%d: B\n",(int)getpid());
    printf("%d: C\n",(int)getpid());
    fork();
    wait();
    printf("%d: D\n",(int)getpid());
}

int main(void)
{
    forker();
    return 0;
}

When I compiled and ran resultant a.out, here is what I observed:

> ./a.out
3560: A
3561: B
3561: C
3562: D
3561: D
3560: B
3560: C
3563: D
3560: D

However when I do the following:

> ./a.out > t.txt

something weird happens:

> cat t.txt
3564: A
3565: B
3565: C
3566: D
3564: A
3565: B
3565: C
3565: D
3564: A
3564: B
3564: C
3567: D
3564: A
3564: B
3564: C
3564: D

Can someone please explain this behavior? Why is the output different when it is redirected to a file?

I am using Ubuntu 10.10, gcc version 4.4.5.

Upvotes: 6

Views: 645

Answers (3)

Tudor
Tudor

Reputation: 62439

The problem is that the output of printf is passed through a library buffer before being sent to the file, which causes the strange behavior you mentioned. If you add a fflush(stdout) after each printf your output will be correct also inside the file.

You can read more about this here: http://www.pixelbeat.org/programming/stdio_buffering/

Upvotes: 3

calandoa
calandoa

Reputation: 6137

The other answers do not exactly describe what is happening, and I had to think a bit more to understand. So, in the second case (output buffered because of file redirection), and by using 1,2,3 and 4 instead of 3564, 3565, 3566 and 3567:

  • process 1 prints "A:1" in its internal stdout buffer;
  • process 1 forks and process 2 is created, this creation implies the copy of the internal stdout buffer which is still not printed;
  • process 1 prints "B:1" and "C:1" in its internal stdout buffer, process 2 "B:2" and "C:2";
  • both processes fork (in your case 1->4 and 2->3, but it could have been different), duplicating both internal buffers;
  • All 4 processes prints the D line in their buffers, then exit.

At this point, the contents of the 4 internal stdout buffers are:

- process 1:
    A:1
    B:1
    C:1
    D:1
- process 2:
    A:1
    B:2
    C:2
    D:2
- process 3:
    A:1
    B:2
    C:2
    D:3
- process 4:
    A:1
    B:1
    C:1
    D:4
  • Finally, the 4 buffers are printed in non deterministic order. In your case, the order was 3, 2, 4, 1.

This behavior is not happening when stdout is the shell, or with fflush(), because the stdout buffer is dumped before each fork(), so only empty buffers are duplicated.

Upvotes: 2

mah
mah

Reputation: 39797

The reason this happens is data buffering. At the time of the fork(), in the case of directing to a file, your output has not been flushed yet... so both the parent and the child now have outstanding output buffers.

Put a call to fflush(stdout); before each fork(); to resolve this.

Upvotes: 10

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