Question

I am confused between these two functions:

void Swap_byPointer1(int *x, int *y){
    int *temp=new int;
    temp=x;
    x=y;
    y=temp;
}

---

void Swap_byPointer2(int *x, int *y){
    int *temp=new int;
    *temp=*x;
    *x=*y;
    *y=*temp;
}

Why Swap_byPointer2 succeeds to swap between x and y, and Swap_byPointer1does not?

Answer 1

The first snippet swaps the memory addresses which are the values of the pointers. Since the pointers are local copies, this has no effect for the caller.

Rewritten without a memory leak:

void Swap_byPointer1(int *x, int *y){
    //e.g x = 0xDEADBEEF and y = 0xCAFEBABE;
    int *temp=x;
    x=y;
    y=temp;
    //now x = 0xCAFEBABE and y = 0xDEADBEEF
}

The second swaps the pointees (objects that the pointers point to).

Rewritten without a memory leak:

void Swap_byPointer2(int *x, int *y){
    //e.g *x = 100 and *y = 200
    int temp =*x;
    *x=*y;
    *y=temp;
    //now *x = 200 and *y = 100
    //there are now different values at the original memory locations
}

(Pointers can point to dynamically allocated objects, but don’t have to. Using a pointer does not mean there has to be a new-allocation. Pointers can point to objects with automatic lifetime as well.)