Let's try explaining it differently... When the function starts there's 1 process, this process has a int x = 3
Once you hit this line of code:
if (fork() != 0)
Now, assuming the fork() worked, we have two processes. They both have the same execution space, they both are going to run the same code (to a point), but the child will get its own copy of x
to play with.
fork() will return a 0 to the child process, so from the child processes' prospective, the rest of the function is this:
printf(("x=%d\n", --x);
exit(0);
The parent process on the other hand, will get a valid value back from the fork() command, thus it will execute:
printf("x=%d\n", ++x);
printf("x=%d\n", --x);
exit(0);
What the output at this point will be is anyone's guess... You can't tell if parent or child will run first
But if we assume the parent hits the ++x
is the next operation then the output is:
x=4
x=3
x=2
As both parent and child will hit the --x
. (the parent's x was 4 after the ++x, will be 3 at the end. The child's x was 3, will be 2 at the end)