Swapping void* pointers in C++

Question

I wanted to ask if this kind of swapping implementation would work:

 void swap (void* &a, void* &b ) 
 {    
     void * tmp = a;
     a=b;
     b=tmp;
 }

int main() 
{
   void * a;
   void * b;   
   swap (a,b);
   return 0;
}

Do I need to somehow send the size of the object to the swap method and use it? If so, then what is a good implementation for this function?

Answer 1

Looks fine to me (a pointer is just a pointer, so the sizes of the things it's pointing to doesn't matter), but why not just use std::swap ?

In this example, you're messing with uninitialized variables, which is bad, but that doesn't seem to be the main point of your question.

Answer 2

It works. I'm not sure it does what you think it does: The pointer values are swapped, not whatever they point to.

Answer 3

This function will work just fine for swapping void pointers; the compiler knows their size. If you want to swap something other than void pointers you need a different function.

EDIT: but that's the general pattern for a swap template function:

template <class Ty>
void swap(Ty& lhs, Ty& rhs) {
    Ty tmp = lhs;
    lhs = rhs;
    rhs = tmp;
}