Difference between revisions of "cpp/utility/forward like"
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{{ddcl|header=utility|since=c++23| | {{ddcl|header=utility|since=c++23| | ||
template< class T, class U > | template< class T, class U > | ||
− | + | constexpr auto&& forward_like( U&& x ) noexcept; | |
}} | }} | ||
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When {{tt|m}} is an actual member and thus {{c|o.m}} a valid expression, this is usually spelled as {{c|std::forward<decltype(o)>(o).m}} in C++20 code. | When {{tt|m}} is an actual member and thus {{c|o.m}} a valid expression, this is usually spelled as {{c|std::forward<decltype(o)>(o).m}} in C++20 code. | ||
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This leads to three possible models, called ''merge'', ''tuple'', and ''language''. | This leads to three possible models, called ''merge'', ''tuple'', and ''language''. | ||
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{{eq fun|1= | {{eq fun|1= | ||
template<class T, class U> | 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>>; | constexpr bool is_adding_const = std::is_const_v<std::remove_reference_t<T>>; |
Latest revision as of 07:28, 5 July 2024
Defined in header <utility>
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template< class T, class U > constexpr auto&& forward_like( U&& x ) noexcept; |
(since C++23) | |
Returns a reference to x which has similar properties to T&&
.
The return type is determined as below:
- 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>.
- 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.
Contents |
[edit] Parameters
x | - | a value needs to be forwarded like type T
|
[edit] Return value
A reference to x of the type determined as above.
[edit] 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
|
[edit] 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); } } |
[edit] Example
#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
[edit] See also
(C++11) |
converts the argument to an xvalue (function template) |
(C++11) |
forwards a function argument and use the type template argument to preserve its value category (function template) |
(C++17) |
obtains a reference to const to its argument (function template) |