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

Not in C++. Functional languages such as Erlang and Haskell get closer by allowing you to specify function overloads based on parameter value, but most imperative languages including C++ require overloading based on method signature; that is, the number and type of each parameter and the type of the return value.

The const keyword in the signature defines not the type of the parameter, but its mutability within the function; a "const" parameter will generate a compiler error if modified by the function or passed by reference to any function that doesn't also use const.