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C++: Perfect Forwarding or Forwarding References

8 months ago by Lukas Kerkemeier

I ran into a pretty interesting problem regarding perfect forwarding. TL;DR: Never set template parameter explicitly for a forwarding reference if you don't have to.

Reference Collapsing

By the use of using or alias it is possible to create references of references. To combat this there are reference collapsing rules:

InnerOuterResult
&&&
&&&&
&&&&
&&&&&&

Also only the inner qualifiers (const, volatile) are used. The outer ones will be ignored.

Forwarding References: why not to set parameters explicitly

First of all forwarding references will never copy. They will always take a (l- or rvalue) reference. Say you have a function like

template <typename T>
auto do_something (T&& t) {
  ...
}

If you pass a value or a lvalue reference of type int to it, T will become int&. This happens because the reference collapsing combines & with && to &. If you pass a rvalue reference of type int, T will simply become int and no collapsing will be needed. Now the problem: If you set T explicitly you might think that T=int should generate the code needed to pass a value. However as explained above it will expect a rvalue reference and most likely your code will not compile. You need to set T=int& to generate the expected code.

Why should I ever need to set a template parameter by hand?

If you have a templated class that uses these template parameter in the constructor you will probably need to set the types by hand. In that case it can be beneficial to create a free construction function like

template <typename T>
struct ExampleStruct {
  T t;
  ExampleStruct(T&& t): t(std::forward<T>(t)) {}
};
template <typename T>
auto construct(T&& t) {
  return ExampleStruct<T>(std::forward<T>(t));
}

The function can detect T and will pass the correct version to ExampleStruct.

Code example

This code was tested with clang-12.

#include <iostream>
#include <type_traits>
#include <utility>

using namespace std;

template <typename T> void print_type(T &&t) {
  using type = decltype(t);
  if (is_const_v<remove_reference_t<type>>) {
    cout << "    const ";
  } else {
    cout << "non-const ";
  }

  if (is_lvalue_reference_v<type>) {
    cout << "l-value";
  } else if (is_rvalue_reference_v<type>) {
    cout << "r-value";
  } else {
    cout << "  value";
  }
  cout << '\n';
}

int main() {
  std::cout << "Reference Collapsing: collapsing rules\n";
  int to_ref = 44;
  using l_type = int &;
  using r_type = int &&;
  l_type &l_l = to_ref;
  l_type &&l_r = to_ref;
  r_type &r_l = to_ref;
  //forward needed so the function will use the correct type
  print_type(std::forward<decltype(l_l)>(l_l));
  print_type(std::forward<decltype(l_r)>(l_r));
  print_type(std::forward<decltype(r_l)>(r_l));
  r_type &&r_r = std::move(to_ref);
  print_type(std::forward<decltype(r_r)>(r_r));

  std::cout << "Reference Collapsing: qualifier\n";
  using const_type = const int &;
  using non_const_type = int &;
  const_type &n_c = to_ref;
  non_const_type &n_n = to_ref;
  const_type const &c_c = to_ref;
  non_const_type const &c_n = to_ref;
  print_type(n_c);
  print_type(n_n);
  print_type(c_c);
  print_type(c_n);

  std::cout << "Forwarding References: why not to set parameter explicitly\n";
  print_type(42);
  const int number = 43;
  print_type(number);
  print_type<int>(42);
}

Output:

Reference Collapsing: collapsing rules
non-const l-value
non-const l-value
non-const l-value
non-const r-value
Reference Collapsing: qualifier
    const l-value
non-const l-value
    const l-value
non-const l-value
Forwarding References: why not to set parameter explicitly
non-const r-value
    const l-value
non-const r-value