Unexpected output with fork()

Question

I read somewhere that in Unix, after fork() is executed successfully, both processes will start their execution at the next statement following the fork() call. Meanwhile when I run this code in C

#include <stdio.h>
int main()
{
   printf(" do ");
   if(fork()!=0) printf ("ma ");
   if(fork()==0) printf ("to \n");
   else printf("\n")

    return 0;
}

one possible output is

 do ma
 do ma to
 do 
 do to

"printf(" do ");" is before the fork() call so how comes "do" is repeated several times in the output?

you're forking TWICE... it should be `pid = fork(); if (pid) { ... } else { ...}` to detect if you're in the child or parent. and since your `do` output has no line break, it sits in the output buffer until a line break is encountered, which is **AFTER** your fork. so you have two buffers with `do` sitting in them. — Marc B, May 19 '15 at 20:19
Someone else find the dups - there are loads - stdout buffering thing. — Martin James, May 19 '15 at 20:20
Than you for your answer. I know I'm forking twice, it is done on purpose, it is an exercice from my teacher where I should predict the output, but I still don't understand why do is written several times as it is before any of the two fork() calls. — Artemisia, May 19 '15 at 20:28
I think the "do"s are there because they are buffered in the parent output memory which is copied to all children. — Tim3880, May 19 '15 at 20:32

dev_ankit · Accepted Answer · 2015-05-19T23:08:24.830

2

If you see the definition of fork it says that the child and parent are exact same copy and both start executing at the next instruction.
Now, Since C doesn't flush output until a newline is printed or you explicitly flush output, the "do" stays in buffer and gets copied to child too.
use:

printf(" do ");
fflush(stdout);

or

printf(" do \n"); //Notice the newline in the end

and you will get expected output

edited May 19 '15 at 23:08

answered May 19 '15 at 20:33

dev_ankit

516
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This is nothing to do with "middle-level" vs. "high-level". – Oliver Charlesworth May 19 '15 at 22:06
middle level vs high level is pretty abstract to such interpretations. Still removed it as it wasn't part of the answer anyways. – dev_ankit May 19 '15 at 23:12

rlbond · Answer 2 · 2015-05-19T21:04:57.187

   if(fork()!=0) printf ("ma ");

This line creates two processes; one of them is the parent and runs the printf statement, and one is the child and doesn't.

   if(fork()==0) printf ("to \n");

Both processes run this statement; the result is four processes:

Parent-parent; this runs the first printf and prints " do ma "
Parent-child; this runs both and prints " do ma to \n"
Child-parent; this runs neither and prints " do "
Child-child; this runs the second and prints " do to \n"

The issue you have is that you think the fork return "skips" to the next statement; rather, when fork returns, you now have two processes with different return values. Typically, printf is not flushed until the program ends or you print a newline.

score 0 · Answer 3 · answered May 19 '15 at 20:44

0

I think the "do"s are there because they are buffered in the parent output memory which is copied to all children.

If you add "\n" to the first "do":

printf(" do\n");

you will not see any "do" after the fork.

answered May 19 '15 at 20:44

Tim3880

2,563
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Unexpected output with fork()

3 Answers3