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

You could do this with a template:

template void foo(T x) { ... }

Then you can call this template by value or by reference:

int x = 123;
foo(x);  // by value
foo(x);  // by refernce