Counting processes after fork in for loop

Question

I have the following code:

for(i=1; i<=2; i++)
{
    fork();
    printf("x ");   
}

I calculated that x should be printed out 6 times: twice in the first iteration and 4 times in the second.

Instead, X is printed 8 times. Why?

possible duplicate of [fork() and output](http://stackoverflow.com/questions/9364410/fork-and-output) — hmjd, May 19 '13 at 14:25

score 7 · Accepted Answer · answered May 19 '13 at 14:24

Because of buffering. Usually, stdout is line-buffered, so

printf("x ");

doesn't immediately write the "x " to the terminal but to the output buffer. That is copied when the process fork()s, so each of the four processes after the second iteration has two "x " in the output buffer [one from the parent/before forking in the first iteration, one from the second iteration] when it exits and eight xs are printed altogether.

Flush the buffer immediately after the printf("x "); and only six will be printed.

`fflush(stdout);` is the easiest way to follow Daniel's advice here. — Alex North-Keys, May 19 '13 at 15:00

Counting processes after fork in for loop

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