How this fork() code create five processes?

Question

 main()
    {
        fork() && fork() || fork();

        printf("forked\n");
        return 0;
    }

//how is it printing fork 5 times instead of 8 times?

Another one is [here](http://stackoverflow.com/q/8726965/2455888). — haccks, Feb 21 '14 at 18:01
Why 3 upvotes to a duplicate question that has been asked before many times? — haccks, Feb 21 '14 at 18:10
http://stackoverflow.com/questions/8726965/fork-system-call-in-c?lq=1 — haccks, Feb 21 '14 at 18:14
http://stackoverflow.com/questions/7091918/confused-with-output-of-fork-system-call?lq=1 — haccks, Feb 21 '14 at 18:14

score 4 · Answer 1 · edited Feb 21 '14 at 18:09

Because of short circuiting.

If I rewrite this:

main()
{
    // fork() && fork() || fork();

    if (fork()) {
        if (!fork()) {
            fork();
        }
    } else {
        fork();
    }

    printf("forked\n");
    return 0;
}

After first fork you will have one process that will go into the else branch straight away and one that will continue. The one that goes into the else loop will fork just once (right branch). The one that goes into the inner one will fork once and its child will fork again.

  f
/   \
f   f
|
f

score 1 · Answer 2 · answered Feb 21 '14 at 18:02

because of the and operator. Remember and operator will terminate as soon as it hits a false. fork will return 0 (false) for either parent or child(forgot), and it will terminate the Boolean expression which skips the other 2 forks, that is why you will only print fork 5 times instead of 8

How this fork() code create five processes?

2 Answers2