On cppreference, it is written that the correct way of using std::result_of
is:
template<class F, class... Args>
std::result_of_t<F&&(Args&&...)>
// instead of std::result_of_t<F(Args...)>, which is wrong
my_invoke(F&& f, Args&&... args) {
/* implementation */
}
I was wondering how std::invoke_result_t
should be used:
invoke_result_t
:
template<class F, class... Args>
std::invoke_result_t<F&&, Args&&...> my_invoke(F&& f, Args&&... args);
Or:
template<class F, class... Args>
std::invoke_result_t<F, Args...> my_invoke(F&& f, Args&&... args);
invoke_result
is defined in terms of declval
:
If the expression
INVOKE(declval<Fn>(), declval<ArgTypes>()...)
is well-formed when treated as an unevaluated operand, the member typedeftype
names the typedecltype(INVOKE(declval<Fn>(), declval<ArgTypes>()...));
otherwise, there shall be no membertype
.
and declval
is specified as:
template<class T> add_rvalue_reference_t<T> declval() noexcept;
So there's no difference between std::invoke_result_t<F&&, Args&&...>
and std::invoke_result_t<F, Args...>
. Well, the latter is 4 characters shorter, but they mean exactly the same thing (since neither F
nor Args...
could be void
).