Visually what happens to fork() in a For Loop

Question

I have been trying to understand fork() behavior. This time in a for-loop. Observe the following code:

#include <stdio.h>  void main() {    int i;     for (i=0;i<3;i++)    {       fork();        // This printf statement is for debugging purposes       // getppid(): gets the parent process-id       // getpid(): get child process-id        printf("[%d] [%d] i=%d\n", getppid(), getpid(), i);    }     printf("[%d] [%d] hi\n", getppid(), getpid()); } 

Here is the output:

[6909][6936] i=0 [6909][6936] i=1 [6936][6938] i=1 [6909][6936] i=2 [6909][6936] hi [6936][6938] i=2 [6936][6938] hi [6938][6940] i=2 [6938][6940] hi [1][6937] i=0 [1][6939] i=2 [1][6939] hi [1][6937] i=1 [6937][6941] i=1 [1][6937] i=2 [1][6937] hi [6937][6941] i=2 [6937][6941] hi [6937][6942] i=2 [6937][6942] hi [1][6943] i=2 [1][6943] hi 

I am a very visual person, and so the only way for me to truly understand things is by diagramming. My instructor said there would be 8 hi statements. I wrote and ran the code, and indeed there were 8 hi statements. But I really didn’t understand it. So I drew the following diagram:

Diagram updated to reflect comments :)

Observations:

1. Parent process (main) must iterate the loop 3 times. Then printf is called
2. On each iteration of parent for-loop a fork() is called
3. After each fork() call, i is incremented, and so every child starts a for-loop from i before it is incremented
4. At the end of each for-loop, "hi" is printed

Here are my questions:

• Is my diagram correct?
• Why are there two instances of i=0 in the output?
• What value of i is carried over to each child after the fork()? If the same value of i is carried over, then when does the "forking" stop?
• Is it always the case that 2^n - 1 would be a way to count the number of children that are forked? So, here n=3, which means 2^3 - 1 = 8 - 1 = 7 children, which is correct?

Answer 1

Here's how to understand it, starting at the for loop.

1. Loop starts in parent, i == 0

2. Parent fork()s, creating child 1.

3. You now have two processes. Both print i=0.

4. Loop restarts in both processes, now i == 1.

5. Parent and child 1 fork(), creating children 2 and 3.

6. You now have four processes. All four print i=1.

7. Loop restarts in all four processes, now i == 2.

8. Parent and children 1 through 3 all fork(), creating children 4 through 7.

9. You now have eight processes. All eight print i=2.

10. Loop restarts in all eight processes, now i == 3.

11. Loop terminates in all eight processes, as i < 3 is no longer true.

12. All eight processes print hi.

13. All eight processes terminate.

So you get 0 printed two times, 1 printed four times, 2 printed 8 times, and hi printed 8 times.

Answer 2

1. Yes, it's correct. (see below)
2. No, i++ is executed after the call of fork, because that's the way the for loop works.
3. If all goes successfully, yes. However, remember that fork may fail.

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A little explanation on the second one:

for (i = 0;i < 3; i++) {    fork(); } 

is similar to:

i = 0; while (i < 3) {     fork();     i++; } 

So i in the forked processes(both parent and child) is the value before increment. However, the increment is executed immediately after fork(), so in my opinion, the diagram could be treat as correct.