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?
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;
}
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[]) {
...
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.
