Overloading on const-ness of Parameters

Yes, you can overload a function based on the const-ness of its parameters. This is because in C++, const is part of the type system, and a const type is considered distinct from its non-const counterpart.

Here's an example:

1#include <iostream>
2
3void Print(int& x) {
4  std::cout << "Non-const reference parameter";
5}
6
7void Print(const int& x) {
8  std::cout << "Const reference parameter";
9}
10
11int main() {
12  int a = 10;
13  const int b = 20;
14
15  Print(a);// calls Print(int&)
16  Print(b);// calls Print(const int&)
17}

1Non-const reference parameter
2Const reference parameter

In this code, we have two overloads of Print, one taking a non-const reference (int&) and the other taking a const reference (const int&).

When we call Print(a), a is a non-const int, so it matches Print(int&) exactly.

When we call Print(b), b is a const int. It cannot bind to int& because that would discard the const-ness, so it instead matches Print(const int&).

This allows you to provide different implementations depending on whether the parameter is const or not. For example, the non-const version might modify the parameter, while the const version would only read it.

However, note that this only applies to reference (or pointer) parameters. For parameters passed by value, the const-ness does not matter for overload resolution, because the parameter is always copied.