Fork concept, dont quite grasp how fork works

Question

I try to draw out the process according to the code but I really need some explanation in why, here is the question:

B() {
     pid_t pid;
     if ((pid = fork()) != 0)
         waitpid(pid,NULL,0);
     printf("2");
     if(fork() ==0)
       { printf("3"); exit(0); }
     printf("5");
     exit(0);
}

Which one are illegals output?
232553, 235325, 232355, 235253, 252533...

This is the process I draw out according to the code and my understanding of fork.

            ___3 (exit so no more here)
           |
      __2__|___5 (I guess 5 should be here)
     |          |        
     |          |
 ____|____(wait)|(start again since printf 3 process end)

So I'm stuck right there... Any help is appreciated.

You shouldn't expect to see the output in any particular order ... besides "2" being first — paulsm4, Nov 25 '13 at 07:54
That is true, what I mean though, is how the rest play out? The drawing is my interpretation of the code and it is not finish. What I want to ask though, is that what else would appear after wait? — Rozen, Nov 25 '13 at 08:01

Dietrich Epp · Accepted Answer · 2013-11-25T08:23:20.743

Okay, there are two forks. Here is control flow, from left to right, with parents on top and children on the bottom:

    +-- B ---- waitpid() --+                  +-- "5" -- E
    |                      |                  |
A --+ fork()               +-- C -- "2" -- D -+ fork()
    |                      |                  |
    +----------------------+                  +-- "3" -- F

So, what do we know?

"2", "5", and "3" each appear twice (90 possibilities)
No prefix may contain more "3" than "2" (30 possibilities)
No prefix may contain more "5" than "2" (16 possibilities)
The second "2" must be preceded by the first "5" (7 possibilities)

The 7 possibilities are:

2,3,5,2,3,5
2,3,5,2,5,3
2,5,2,3,3,5
2,5,2,3,5,3
2,5,2,5,3,3
2,5,3,2,3,5
2,5,3,2,5,3

score 2 · Answer 2 · answered Nov 25 '13 at 08:14

This should be the program path (x denotes termination):

---+--(wait)-2-+-5-x
   |           |
   +-2-+-5-x   +-3-x
       |   
       +-*3-x

After the first fork, the parent waits for the child to finish. But in case of second fork, it does not wait. So, * marked 3 can be printed anywhere after first 2. The order of printing 5 and 3 after second fork also can not be determined. Therefore, the possible outputs are:

25235*3
25253*3
2523*35
2525*33
252*335
252*353
25*3235
25*3253
2*35235
2*35253

score 1 · Answer 3 · answered Nov 25 '13 at 08:09

Since the order of execution with fork() is not deterministic, you can only expect that each integer (i.e., 2, 3 and 5) will be printed twice. The order depends on how the scheduler chooses to schedule the processes.

You can force a particular order using sleep commands or some other synchronization primitive.

score 1 · Answer 4 · answered Nov 25 '13 at 08:17

The first fork() splits the process into 2 parts (parent and child). The parent waits and the child prints 2.

Then the child does fork(); and then child prints 5 and exits (which allows the parent to start running again) while the child's child (grandchild?) prints 3. This can happen in any order.

The parent continues and prints 2 (this may happen before or after the grandchild prints 3; but after the now terminated child printed 5).

Then the parent does fork(); and the parent prints 5 and it's second child prints 2 (which can happen in any order, and may happen before the grandchild prints 5).

score 0 · Answer 5 · answered Nov 25 '13 at 08:17

The first fork creates a child and waits for it. So the main program only goes on running as soon as the first child exits.

So the first output is "the child's" 2. Then a 3 and a 5 is printed, in any order. Only after the 3 or the 5 the second 2 can occur, and then the second 3 and/or 5.

So

232553 is ok
235325 is not ok, as the 2nd 3 comes before the 2nd 2
232355 is ok
235253 is ok
252533 is ok

Fork concept, dont quite grasp how fork works

5 Answers5