Perfect Forwarding with Const References

When perfectly forwarding arguments using std::forward, it's important to consider the const-ness of the argument being forwarded. If you have a const lvalue reference and want to preserve its const-ness while forwarding, you need to use the appropriate template argument for std::forward.

Here's an example:

1#include <iostream>
2
3void func(const int& x) {
4  std::cout << "func(const int&): " << x << "\n";
5}
6
7template <typename T>
8void forward_func(T&& arg) {
9  func(std::forward<T>(arg));
10}
11
12int main() {
13  const int value = 42;
14  forward_func(value);
15}

1func(const int&): 42

In this example, forward_func() takes a forwarding reference arg and forwards it to func() using std::forward<T>(arg). When calling forward_func() with a const lvalue value, the template argument T deduces to const int&. By using std::forward<T>(arg), the const-ness is preserved, and func() correctly receives a const lvalue reference.

If you were to use std::forward<int>(arg) instead, it would strip away the const-ness, and the code would not compile because func() expects a const reference.

So, when perfectly forwarding const lvalue references, make sure to use the deduced template argument T with std::forward() to maintain the const-ness of the argument.