pointers - Passing address, but it is working like call by value in C?

Question

Hello I am a beginner in C programming language. Recently I read about call by value and call by address. I have learned that in call by address changes in the called functions reflects the callee. However the following code does not work like that.

int x = 10,y = 20;
void change_by_add(int *ptr) {
    ptr = &y;
    printf("
 Inside change_by_add %d",*ptr);
    // here *ptr is printing 20
}

void main(){
    int *p;
    p = &x;
    change_by_add(p);
    printf("
Inside main %d", *p);
    // here *p is still pointing to address of x and printing 10
}

When I am passing address then why the changes made by called function does not reflect caller?

Answer 1

You are simply setting the value of the pointer in the function, not the value of the pointed to variable. The function should use the following code:

*ptr = y;

This derefences the pointer (exposing the value pointed to), and therefore when you use the equals operator, the memory pointed at is modified, not the pointer itself. I hope this helps to clarify things.