C++: Can't work with simple pointers?

Question

I apologize as this is so simple, I was working with my own XOR swap method and wanted for the heck of it compare the speed difference between reference and pointer use (do not spoil it for me!)

My XOR ptr function is as follows:

 void xorSwapPtr (int *x, int *y) {
     if (x != y && x && y) {
         *x ^= *y;
         *y ^= *x;
         *x ^= *y;
     }
 }

I copied this to a xorSwapRef function which just uses refs (int &x, etc.) anywho:

I use it like this but I get the error error: invalid conversion from ‘int’ to ‘int*’

int i,x,y = 0;
for(i=0;i<=200000;i++) {
    x = rand();
    y = rand(); 
    xorSwapPtr(x, y); //error here of course
}

How would I use the pointer function with ints, like the ref one? I just wonder because the example xor function I found in a book uses pointers, hence my wanting to test.

Answer 1

The syntax can be confusing... When you want to pass a pointer to something, you usually take the address of it, like &something. However, when declaring a function signature, and you want to define one of the parameters to be a reference to a Type called somethingElse, then you would use Type &somethingElse. Both of these use the & token, but they mean 2 different things (different semantics for the token &, just like * could mean multiply, define a pointer to, or dereference a pointer, each depending upon its grammatical place in the code).

void foo(int *x, int *y); // declare a function that takes two pointers to int
void bar(int &x, int &y); // declare a function that takes two references to int

now let's use them...

int i, j;
foo(&i, &j);  // pass in the address of i and the address of j
bar(i, j);  // pass in a reference to i and a reference to j
// can't tell by the call that it is a reference - could be "by value"
// so you gotta look at the function signature

Answer 2

This bit

xorSwapPtr(x, y);

Needs to be

xorSwapPtr(&x, &y);

Hope this helps

Answer 3

x and y are ints; xorSwapPtr takes two int*s, so you need to pass the addresses of x and y, not x and y themselves:

xorSwapPtr(&x, &y);