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 ) { return example; }  int main() {   int inum = 3;   int *pinum = & inum;    std::cout << "inum: " <<  returnSame(*pinum) << std::endl;    return 0;            } 

Is there a temporary object produced?

Answer 1

Dereferencing the pointer doesn't create a copy; it creates an lvalue that refers to the pointer's target. This can be bound to the lvalue reference argument, and so the function receives a reference to the object that the pointer points to, and returns a reference to the same. This behaviour is well-defined, and no temporary object is involved.

If it took the argument by value, then that would create a local copy, and returning a reference to that would be bad, giving undefined behaviour if it were accessed.