Does dereferencing a pointer make a copy of it?

Question

Does dereferencing a pointer and passing that to a function which takes its argument by reference create a copy of the object?

Answer 1

In this case the value at the pointer is copied (though this is not necessarily the case as the optimiser may optimise it out).

int val = *pPtr; 

In this case however no copy will take place:

int& rVal = *pPtr; 

The reason no copy takes place is because a reference is not a machine code level construct. It is a higher level construct and thus is something the compiler uses internally rather than generating specific code for it.

The same, obviously, goes for function parameters.

Answer 2

In the simple case, no. There are more complicated cases, though:

void foo(float const& arg); int * p = new int(7); foo(*p); 

Here, a temporary object is created, because the type of the dereferenced pointer (int) does not match the base type of the function parameter (float). A conversion sequence exists, and the converted temporary can be bound to arg since that's a const reference.