parent local variable acting as shared variable between three child - c

Question

How come in the below program the local variable of parent process is acting as a shared variable between three child processes.
int main()
{
int turn = 0;
int i;
for (i = 0; i < 3; i++)
{
if (fork() == 0)
{
int me = i;
while (turn != me)
/*do nothing*/ ;
// my turn
printf("Process %d ran\n", me);
turn++;
}
}
return 0;
}
Output:
Process 0 ran
Process 1 ran
Process 2 ran
But according to me the last two processes should hang as the value of turn should never change for them.
Also if i put a exit(0); after turn++, I am returned immediately to my shell prompt with only one line of output i.e:
Process 0 ran
But still the other two processes keep running in the background and no output by any of the other two processes.

Answer 1

What's going on:
main process is 0.
process 0 calls fork. i = 0;
process 0 returns from fork = 1;
process 0 calls fork. i = 1;
process 0 returns from fork = 2;
process 0 calls fork. i = 2;
process 0 returns from fork = 3;
process 0 exists;
process 1 returns from fork. i = 0;
process 1 hits the while loop turn = 0, i = 0, me = 0, while loop exists;
process 1 calls printf.
process 1 increments turn, turn = 1;
process 1 goes back to the for loop;
process 1 calls fork, i = 1;
process 1 returns from fork = 4;
process 1 calls fork, i = 2;
process 1 returns from fork = 5;
process 1 exists;
process 2 returns from fork. i = 1;
process 2 hits the while loop turn = 0, i = 1, me = 0, while loop spins forever;
process 3 returns from fork, i = 2;
process 3 hits the while loop turn = 0, i = 2, me = 0, while loop spins forever;
process 4 (spawned from process 1) returns from fork, i = 1, turn = 1 (inherited from process 1).
process 4 hits the while loop turn = 1, i = 1, me = 1, while loop exits;
process 4 calls printf;
process 4 increments turn, turn = 2;
process 4 goes back to the for loop;
process 4 calls fork, i = 2, turn = 2;
process 4 returns from fork = 6;
process 4 exits;
process 5 (spawned from process 1) return from fork, i = 2, turn = 1 (inherited from process 1)
process 5 hits the while loop turn = 1, i = 2, me = 2, while loop spins forever;
process 6 (spawned from process 4) returns from fork i = 2, turn = 2 (inherited from process 4)
process 6 hits the while loop turn = 2, i = 2, me = 2, while loop exists;
process 6 calls printf;
process 6 drops out of the for loop
process 6 exits;
Basically, you need to remember that all processes you will continue running the for loop, not just the main one. After that it's not too hard to figure out what's going on. You spawn 6 processes in total and three of them will have the right state of i and turn.

Answer 2

Refactor your code like this:
#include <unistd.h>
#include <stdio.h>
int main()
{
int turn = 0;
int i;
for (i = 0; i < 3; i++)
{
printf("forking i = %d, turn = %d\n", i, turn);
if (fork() == 0)
{
int me = i;
while (turn != me)
/*do nothing*/ ;
// my turn
printf("Process %d ran\n", me);
turn++;
}
}
return 0;
}
It may come as a shock to you but as Art's refactoring above shows pretty clearly your code will spawn 6 processes. Putting some more printfs in there should get your understand what's really going on.

Answer 3

Variable declared before fork() is shared for child processes. You may check it by printing &turn in loop. It will be same.