fork() branches more than expected?

Question

Consider the following piece of code:

#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>

int main(void)
{
    int i;
    for(i = 0; i < 2; i++)
    {
        fork();
        printf(".");
    }
    return 0;
}

This program outputs 8 dots. How can that be possible? Should not there be 6 dots instead?

Answer 1

The fork() primitive often stretches the imagination. Until you get a feel for it, you should trace out on paper what each operation is and account for the number of processes. Don't forget that fork() creates a near-perfect copy of the current process. The most significant difference (for most purposes) is that fork()'s return value differs between parent and child. (Since this code ignores the return value, it makes no difference.)

So, at first, there is one process. That creates a second process, both of which print a dot and loop. On their second iteration, each creates another copy, so there are four processes print a dot, and then exit. So we can easily account for six dots, like you expect.

However, what printf() really does is buffer its output. So the first dot from when there were only two processes does not appear when written. Those dots remain in the buffer—which is duplicated at fork(). It is not until the process is about to exit that the buffered dot appears. Four processes printing a buffered dot, plus the new one gives 8 dots.

If you wanted to avoid that behavior, call fflush(stdout); after printf().