Function Overloading Based on Value vs. Const Reference

Question

Does declaring something like the following

void foo(int x)        { std::cout << \"foo(int)\"         << std::endl; }
void foo(const int &x)         


        

Answer 1

How would the caller be able to differentiate between them?

It cannot be differentiated in this case. Both the overloaded functions have the same type of primitive data type as the argument. And taking by reference doesn't count for a different type.