fork() and changing local variables?

Question

I am trying to understand the fork() concept and there's is one thing I can't seem to understand.

In the following code - why does the parent still print i=0 even when child process changes it to 5?

The wait(NULL) blocks parent process until child finishes first.

int main(int argc, char *argv[]) {
  int i = 0;
  if (fork() == 0) {
    i = 5;
  } else {
    wait(NULL);
    printf("i = %d\n", i);
  }
  return 0;
}

Can somebody explain why my assumption is incorrect?

What do you imagine `fork` does? – Kerrek SB May 20 '16 at 12:50 — Kerrek SB, May 20 '16 at 12:50

score 6 · Answer 1 · answered May 20 '16 at 12:48

6

Variables are not shared between processes. After the call to fork, there are two completely separate processes. fork returns 0 in the child, where the local variable is set to 5. In the parent, where fork returns the process ID of the child, the value of i is not changed; it still has the value 0 set before fork was called. It's the same behavior as if you had two programs run separately:

int main(int args, char *argv[]) {
    int i=0;
    printf("i = %d\n", i);
    return 0;
}

and

int main(int argc, char *argv[]) {
  int i = 0;
  i = 5;
  return 0;
}

answered May 20 '16 at 12:48

chepner

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So if I wanted i=5 to print when the parent executes, how exactly could I do that without adding the statement i=5 in the parent? In short, is there a way to share variables between child and parent? – RandomMath May 20 '16 at 12:51
You need to pass the value to/from the child using some IPC (Inter-Process Communication) mechanism (file, pipe, socket, shared memory, ...) – AhmadWabbi May 20 '16 at 12:52
Some similar topic you have [here](http://stackoverflow.com/questions/10684499/sharing-same-variable-between-more-than-one-independent-programs-in-linux). Yet you have to read about Inter Process Communication (IPC). – Marcin Kajzler May 20 '16 at 12:54

score 4 · Answer 2 · answered May 20 '16 at 12:55

Processes are not threads! When you fork, you create a full cloned process, with independant memory allocation that simply contains same values (except for the result of the fork call) at the time of the fork.

If you want a child to update some data in the parent process, you will need to use a thread. A thread shares all static and dynamic allocated memory with its parent, and simply has independant automatic variables. But even there, you should use static allocation for i variable:

int i = 0;
int main(int argc, char *argv[]) {
    ...

score 1 · Answer 3 · answered May 20 '16 at 12:52

When you fork the child process gets a copy of the address space of the parent address space, they don't share it, so when the child changed i the parent won't see it.

This copying of the address space it typically done using copy on write to avoid allocating memory that will never change.

fork() and changing local variables?

3 Answers3

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