fork() execution in for loop

Question

int main(int argc, char** argv) {
    int i = 0;
    while (i < 2) {
        fork();
        system(\"ps -o pid,ppid,comm,stat\");
        i++;
     }
     return (EX         


        

Answer 1

I believe the answer is 6.

in the first iteration, fork() is called, splitting the process in 2, thus calling ps twice.

in the second iteration, fork is called again in each process, so you now have 4 processes running ps.

total calls to ps: 2+4=6.

Answer 2

Initial Process
i == 0
-> Fork 1
   system call
   i == 1
   -> Fork 1.1
      system call
   system call
system call
i == 1
-> Fork 2
   system call
system call

I count 6, 2 each from the initial process and the first fork (4), and one from each process forked when i == 1 from those 2 processes.

Of course that's assuming you fix the missing end brace (and define EXIT_SUCCESS), otherwise none, since it won't compile. :-)

Answer 3

6 times.

It creates a process tree like this:

A-+
  |-B-+
  |   |-C-+
  |-D

A does it twice (i=0)

B does it twice (i=0)

C does it once (i=1)

D does it once (i=1)

Note that my usage of letters is to distinguish them. There's no predictable output ordering since process switching is non-deterministic to the eyes of a programmer.