I don't understand this example of fork()

Question

I have this example of code, but I don't understand why this code creates 5 processes plus the original. (6 process total)

#include <unistd.h>

int main(void) {
    int i;
    for (i = 0; i < 3; i++) {
        if (fork() && (i == 1)) {
            break;
        }
    }
}

Answer 1

fork() splits a process in two, and returns either 0 (if this process is the child), or the PID of the child (if this process is the parent). So, this line:

if (fork() && (i == 1)) break;

Says "if this is the parent process, and this is the second time through the loop, break out of the loop". This means the loop runs like this:

• i == 0: The first time through the loop, i is 0, we create two processes, both entering the loop at i == 1. Total now two processes

• i == 1: Both of those processes fork, but two of them do not continue to iterate because of the if (fork() && (i == 1)) break; line (the two that don't continue are both of the parents in the fork calls). Total now four processes, but only two of those are continuing to loop.

• i == 2: Now, the two that continue the loop both fork, resulting in 6 processes.

• i == 3: All 6 processes exit the loop (since i < 3 == false , there is no more looping)

Answer 2

If your main process have pid A, and B - F are subprocesses pids, then:

A    spawns  B          i=0
A    spawns  C          i=1
C    run from 'fork'    i=1
C    spawns  D          i=2
B    run from 'fork'    i=0
B    spawns  E          i=1
D    run from 'fork'    i=2
E    run from 'fork'    i=1
E    spawns  F          i=2
F    run from 'fork'    i=2

Where i is the value of i of the (sub)process' context. Since fork creates an exact copy of the running process, the variable i will also be copied. When A spawns B, i is 0. When A spawns C, i is 1 . Process A now exits the for-loop since i == 1.

Now, subprocess C starts to run, with i == 1. Note that it will not break inside the for-loop since fork(), at C's spawning point, returns 0. Instead it will loop, increasing i to 2, spawn D, and exit because of the for-loop's condition.

The subprocess B have i == 0, when it starts. It spawn subprocess E, and breaks inside the for-loop. (i == 1)

And so on...

When you are tyring to find out of thing like these, I can give you an advice:

Make intermediate variables and print them.

I modified your code, so that it prints out the things I just described:

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

int main(void) {
    int i;
    for (i= 0; i < 3; ++i) {
        int f = fork();
        printf("%i\tspawns\t%i\ti=%i\n", getpid(), f, i);
        if (f && (i == 1))
            break;
    }

    getchar();
}