Why does this code prints 20 instead of 30?

Question

#include <stdio.h>

void fun (int x){
x = 30;
}

int main(){
int y = 20;
fun(y);
printf("%d", y);
return 0;
}

Answer 1

You are passing the value of *y* to the function instead of variable *y* itself, this means when inside function, no matter whatever you do with *x*, no changes will be reflected back to *y* thus the output 20.