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)