std::result_of
From cppreference.com
| Defined in header
<type_traits>
|
||
| template< class >
class result_of; //not defined |
(1) | (since C++11) |
| template< class F, class... ArgTypes >
class result_of<F(ArgTypes...)>; |
(2) | (since C++11) |
Deduces the return type of a function call expression at compile time.
F must be a callable type, reference to function, or reference to callable type. Invoking F with ArgTypes... must be a well-formed expression |
(since C++11) |
F and all types in ArgTypes can be any complete type, array of unknown bound, or (cv-qualified) void |
(since C++14) |
Contents |
[edit] Member types
| Member type | Definition |
type
|
the return type of the callable type F if invoked with the arguments ArgTypes.... Only defined if F can be called with the arguments ArgTypes... in unevaluated context.(since C++14)
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[edit] Helper types
| template< class T >
using result_of_t = typename result_of<T>::type; |
(since C++14) | |
[edit] Possible implementation
template<class> struct result_of; // C++11 implementation, does not satisfy C++14 requirements template<class F, class... ArgTypes> struct result_of<F(ArgTypes...)> { typedef decltype( std::declval<F>()(std::declval<ArgTypes>()...) ) type; }; |
[edit] Notes
As formulated in C++11, std::result_of would fail to compile when F(ArgTypes...) is ill-formed (e.g. when F is not a callable type at all). C++14 changes that to a SFINAE (when F is not callable, std::result_of<F(Args...)> simply doesn't have the type member).
[edit] Examples
std::result_of can be used to determine the result of invoking a functor, in particular if the result type is different for different sets of arguments:
Run this code
#include <type_traits> struct S { double operator()(char, int&); float operator()(int); }; struct C { double Func(char, int&); }; int main() { // the result of invoking S with char and int& arguments is double std::result_of<S(char, int&)>::type f = 3.14; // f has type double static_assert(std::is_same<decltype(f), double>::value, ""); // the result of invoking S with int argument is float std::result_of<S(int)>::type d = 3.14; // f has type float static_assert(std::is_same<decltype(d), float>::value, ""); // result_of can be used with a pointer to member function as follows std::result_of<decltype(&C::Func)(C, char, int&)>::type g = 3.14; static_assert(std::is_same<decltype(g), double>::value, ""); }
demonstrates the C++14 changes to result_of requirements
Run this code
#include <type_traits> #include <iostream> template<class T> typename std::result_of<T(int)>::type f(T& t) { std::cout << "overload of f for callable T\n"; return t(0); } template<class T, class U> int f(U u) { std::cout << "overload of f for non-callable T\n"; return u; } struct S {}; int main() { f<S>(1); // fails to compile in C++11, calls the non-callable overload in C++14 } output= overload of f for non-callable T
[edit] See also
| (C++11)
|
obtains the type of expression in unevaluated context (function template) |