dereferencing a pointer when passing by reference

Question

what happens when you dereference a pointer when passing by reference to a function?

Here is a simple example

int& returnSame( int &example )         


        

Answer 1

A compiler doesn't "call" anything. It just generates code. Dereferencing a pointer would at the most basic level correspond to some sort of load instruction, but in the present code the compiler can easily optimize this away and just print the value directly, or perhaps shortcut directly to loading inum.

Concerning your "temporary object": Dereferencing a pointer always gives an lvalue.

Perhaps there's a more interesting question hidden in your question, though: How does the compiler implement passing function arguments as references?