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    std::forward_like

    From cppreference.com
     
    C++
     
    Utilities library
    General utilities
    Relational operators (deprecated in C++20)
     
    Defined in header <utility>
    template< class T, class U > constexpr auto&& forward_like( U&& x ) noexcept;
    		 (since C++23)
    See More

    Returns a reference to x which has similar properties to T&&.

    The return type is determined as below:

    1. If std::remove_reference_t<T> is a const-qualified type, then the referenced type of the return type is const std::remove_reference_t<U>. Otherwise, the referenced type is std::remove_reference_t<U>.
    2. If T&& is an lvalue reference type, then the return type is also an lvalue reference type. Otherwise, the return type is an rvalue reference type.

    If T is not a referenceable type, the program is ill-formed.

    Parameters

    x - a value needs to be forwarded like type T

    Return value

    A reference to x of the type determined as above.

    Notes

    Like std::forward, std::move, and std::as_const, std::forward_like is a type cast that only influences the value category of an expression, or potentially adds const-qualification.

    When m is an actual member and thus o.m a valid expression, this is usually spelled as std::forward<decltype(o)>(o).m in C++20 code.

    This leads to three possible models, called merge, tuple, and language.

    • merge: merge the const qualifiers, and adopt the value category of the Owner.
    • tuple: what std::get<0>(Owner) does, assuming Owner is a std::tuple<Member>.
    • language: what std::forward<decltype(Owner)>(o).m does.

    The main scenario that std::forward_like caters to is adapting “far” objects. Neither the tuple nor the language scenarios do the right thing for that main use-case, so the merge model is used for std::forward_like.

    Feature-test macro Value Std Feature
    __cpp_lib_forward_like 202207L (C++23) std::forward_like

    Possible implementation

    template<class T, class U>
constexpr auto&& forward_like(U&& x) noexcept
{
    constexpr bool is_adding_const = std::is_const_v<std::remove_reference_t<T>>;
    if constexpr (std::is_lvalue_reference_v<T&&>)
    {
        if constexpr (is_adding_const)
            return std::as_const(x);
        else
            return static_cast<U&>(x);
    }
    else
    {
        if constexpr (is_adding_const)
            return std::move(std::as_const(x));
        else
            return std::move(x);
    }
}

    Example

    Run this code
    #include <cstddef>
#include <iostream>
#include <memory>
#include <optional>
#include <type_traits>
#include <utility>
#include <vector>

struct TypeTeller
{
    void operator()(this auto&& self)
    {
        using SelfType = decltype(self);
        using UnrefSelfType = std::remove_reference_t<SelfType>;
        if constexpr (std::is_lvalue_reference_v<SelfType>)
        {
            if constexpr (std::is_const_v<UnrefSelfType>)
                std::cout << "const lvalue\n";
            else
                std::cout << "mutable lvalue\n";
        }
        else
        {
            if constexpr (std::is_const_v<UnrefSelfType>)
                std::cout << "const rvalue\n";
            else
                std::cout << "mutable rvalue\n";
        }
    }
};

struct FarStates
{
    std::unique_ptr<TypeTeller> ptr;
    std::optional<TypeTeller> opt;
    std::vector<TypeTeller> container;
    
    auto&& from_opt(this auto&& self)
    {
        return std::forward_like<decltype(self)>(self.opt.value());
        // It is OK to use std::forward<decltype(self)>(self).opt.value(),
        // because std::optional provides suitable accessors.
    }
    
    auto&& operator[](this auto&& self, std::size_t i)
    {
        return std::forward_like<decltype(self)>(self.container.at(i));
        // It is not so good to use std::forward<decltype(self)>(self)[i], because
        // containers do not provide rvalue subscript access, although they could.
    }
    
    auto&& from_ptr(this auto&& self)
    {
        if (!self.ptr)
            throw std::bad_optional_access{};
        return std::forward_like<decltype(self)>(*self.ptr);
        // It is not good to use *std::forward<decltype(self)>(self).ptr, because
        // std::unique_ptr<TypeTeller> always dereferences to a non-const lvalue.
    }
};

int main()
{
    FarStates my_state
    {
        .ptr{std::make_unique<TypeTeller>()},
        .opt{std::in_place, TypeTeller{}},
        .container{std::vector<TypeTeller>(1)},
    };
    
    my_state.from_ptr()();
    my_state.from_opt()();
    my_state[0]();

    std::cout << '\n';
    
    std::as_const(my_state).from_ptr()();
    std::as_const(my_state).from_opt()();
    std::as_const(my_state)[0]();
    
    std::cout << '\n';
    
    std::move(my_state).from_ptr()();
    std::move(my_state).from_opt()();
    std::move(my_state)[0]();
    
    std::cout << '\n';
    
    std::move(std::as_const(my_state)).from_ptr()();
    std::move(std::as_const(my_state)).from_opt()();
    std::move(std::as_const(my_state))[0]();
    
    std::cout << '\n';
}

    Output: 

    mutable lvalue
mutable lvalue
mutable lvalue

const lvalue
const lvalue
const lvalue

mutable rvalue
mutable rvalue
mutable rvalue

const rvalue
const rvalue
const rvalue

    See also

    (C++11)
    		 converts the argument to an xvalue
    (function template) [edit]
    (C++11)
    		 forwards a function argument and use the type template argument to preserve its value category
    (function template) [edit]
    (C++17)
    		 obtains a reference to const to its argument
    (function template) [edit]