Since we changed the value of int to 10 inside the if, why do i get 33 printed after the second fork for both child and the child it created?

Question

here is the code:

int main(int argc, char *argv[ ]){
    int x = 33;
    if(!fork()){
        int x = 10;
        printf("%d\n", x);
    }
    fork();
    printf("%d\n", x);
    return 0;
}

i would assume that because the value of int changes for the original child, it carries the value of 10 and in the second fork as well(fork();), so the original child and its child print 10 so the output is like this: 33 33 10 10 10 But it seems that although the value of int changes inside the if, the child then gets back the original value of x which is 33 and both print 33 after the second fork(so the output is like this: 33 33 10 33 33. Why is that?(i believe \n doesnt have anything to do with it) here is a photo of what i think is happening but apparently it doesnt

Answer 1

The x printed in the first printf inside the if block is defined with a local scope that ends at the }. The second printf uses the x defined in the scope of the main body that still has the initial value 33.

Redefining a variable in a local scope with the same name as one define in an enclosing scope is called shadowing. It is confusing and error prone and will be reported with a warning if you enable advanced compiler warnings with -Wall -Wextra or -Weverything.

Here is a modified version that should produce output closer to your expectations:

#include <stdio.h>
#include <unistd.h>

int main(int argc, char *argv[]) {
    int x = 33;
    if (!fork()) {
        x = 10;
        printf("%d\n", x);
    }
    fork();
    printf("%d\n", x);
    return 0;
}

This version should print 33 twice and 10 three times in an unpredictable order.

The original version posted in the question should print 33 4 times and 10 just once in an unpredictable order.