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 functi

Answer 1

Changes made by called function does not get reflected by the caller because you are overriding the pointer address in the called function i.e ptr = &y;.

Initially, you passed the address of x but you are changing it with the address of y.

If you really want to implement the concept of call by address then change value instead of address.

Example:

void change_by_add(int *ptr) {
    *ptr = y;  //changing value
    printf("\nInside change_by_add\t %d",*ptr);
}

void main(){
    int *p;
    p = &x;
    change_by_add(p);
    printf("\nInside main\t %d \n", *p);
    return 0;
}

Output

Inside change_by_add     20
Inside main  20