Why do references occupy memory when member of a class?

Question

I have been told that references, when they are data members of classes, they occupy memory since they will be transformed into constant pointers by the compiler. Why is tha

Answer 1

The C++ standard only defines the semantics of a reference, not how they are actually implemented. So all answers to this question are compiler-specific. A (silly, but compliant) compiler might choose to store all references on the hard-disk. It's just that it proved to be the most convenient/efficient to store a reference as a constant pointer for class members, and replace the occurence of the reference with the actual thing where possible.

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As an example for a situation where it is impossible for the compiler to decide at compile time to which object a reference is bound, consider this:

#include 

bool func() {
    int i;
    std::cin >> i;
    return i > 5;
}

int main() {
    int a = 3, b = 4;
    int& r = func() ? a : b;
    std::cout << r;
}

So in general a program has to store some information about references at runtime, and sometimes, for special cases, it can prove at compile time what a reference is bound to.