Question

I try to repeat and learn more advanced uses and options when cutting trees with forks in the jungle of C. But foolishly I find an example which should be very easy as I have worked with forks before and even written some code, but i can’t understand it fully.

Here comes :

main() {
 if (fork() == 0) {
  if (fork() == 0) {
   printf("3");
  }
  else if ((wait(NULL)) > 0) {
   printf("2");
  }
 }
 else {
  if (fork() == 0) {
   printf("1");
   exit(0);
  }
  if (fork() == 0) {
   printf("4");
  }
 }
 printf("0");
 return 0;
}

Possible solutions are :

1. 3201040
2. 3104200
3. 1040302
4. 4321000
5. 4030201
6. 1403020

where 2, 5 and 6 are correct answers.

First of all, shouldn’t there be four zeroes in the output?
Second… How does one come to the solution at all? Been doing this on paper for almost an hour and I’m not even close to understanding why the given solution are more correct than the false ones (except for nr3 as it can’t end with 2 since a 0 must follow).

Anyone with his forks in check who can offer some good explanation?

EDIT:

Found this here look at pdf’s from 2009. Can people now stop making posts about this being a homework and actually try to help? If not please find some other topics to spend your time. Thanks !

Answer 1

I think there should be 4 zeroes, and that is what I see if I run your code…

A good way to analyse this is to draw a diagram like this one – I’ve shown the forks as * with the parent continuing horizontally, and the child below, so that each individual process is on a separate line:

----*----*----*----0----exit (return from main)
    |    |    |
    |    |    +----4----0----exit (return from main)
    |    |
    |    +----1----exit (explicitly)
    |
    +-----*----wait----2----0----exit (return from main)
          |
          +----3----0----exit (return from main)

Now it’s easy to see that, because of the wait(), you must see 3 followed some time later by 0, before seeing 2 followed by 0.