问题
Here is my implementation of is_destructible_v:
template<class T>
struct is_unknown_bound_array : std::false_type
{};
template<class T>
struct is_unknown_bound_array<T[]> : std::true_type
{};
template<typename T, typename U = std::remove_all_extents_t<T>>
using has_dtor = decltype(std::declval<U&>().~U());
template<typename T>
constexpr bool is_destructible_v
= (std::experimental::is_detected_v<has_dtor, T> or std::is_reference_v<T>)
and not is_unknown_bound_array<T>::value
and not std::is_function_v<T>;
template<typename T>
struct is_destructible : std::bool_constant<is_destructible_v<T>>
{};
clang compiled happily and passed all libstdcxx's testsuite, while gcc failed to compile:
prog.cc:177:47: error: 'std::declval<int&>()' is not of type 'int&'
177 | using has_dtor = decltype(std::declval<U&>().~U());
| ~~~~~~~~~~~~~~~~~~~~^
prog.cc: In substitution of 'template<class T, class U> using has_dtor = decltype (declval<U&>().~ U()) [with T = int&&; U = int&&]':
So, gcc cannot do SFINAE on using has_dtor = decltype(std::declval<U&>().~U());
.
Question:
- Which compiler object to standard here?
- What's the most elegant solution/workaround here? The ways I can think of is a little ugly
回答1:
GCC seems to be broken when handling ~T()
where T
is a reference of scalar type.
It accepts the following code, which is clearly buggy per [expr.pseudo]/2:
template<typename T> using tester = decltype(int{}.~T(), char{});
tester<int&> ch;
int main() {}
I would use if constexpr
to implement:
template<class T>
constexpr bool my_is_destructible() {
if constexpr (std::is_reference_v<T>) {
return true;
} else if constexpr (std::is_same_v<std::remove_cv_t<T>, void>
|| std::is_function_v<T>
|| is_unknown_bound_array<T>::value ) {
return false;
} else if constexpr (std::is_object_v<T>) {
return std::experimental::is_detected_v<has_dtor, T>;
} else {
return false;
}
}
It works with GCC too.
来源:https://stackoverflow.com/questions/53456848/implement-is-destructible-with-detected-idiom