Perfect Forwarding with Const References

How can I perfectly forward a const lvalue reference using std::forward?

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:

#include <iostream>

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

template <typename T>
void forward_func(T&& arg) {
  func(std::forward<T>(arg));
}

int main() {
  const int value = 42;
  forward_func(value);
}
func(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.

Perfect Forwarding and std::forward

An introduction to problems that can arise when our functions forward their parameters to other functions, and how we can solve those problems with std::forward

Questions & Answers

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