Re: How to write an is_nothrow_swapable<T>?
On 2012-11-16 10:13, Zhihao Yuan wrote:
Non-throw swapable is quite useful, and it can be more useful if the
concept can be tested. If we already have an object of type T, I
think the following one is enough:
template <typename T>
constexpr bool is_nothrow_swapable(T& t) {
using std::swap;
return noexcept(swap(t, t)) or
(std::is_nothrow_move_constructible<T>::value and
std::is_nothrow_move_assignable<T>::value);
}
However, I want it to be able to work on just a type, to work as an
integral_constant. Is that doable? Comments on how to implement an
is_swapable<T> are also welcome. Thanks.
I'm not sure whether I understood your question correctly, because you
mention (a) the ability to work on just a type and (b) you refer to
is_swapable (instead of is_nothrow_swapable). I guess that (b) is just a
thinko on your side. What about a variation of your constexpr function
so that it does not depend on any argument anymore:
template <typename T>
constexpr bool is_nothrow_swappable() {
using std::swap;
return noexcept(swap(std::declval<T&>(), std::declval<T&>())) or
(std::is_nothrow_move_constructible<T>::value and
std::is_nothrow_move_assignable<T>::value);
}
template<typename T>
struct is_nothrow_swappable_trait :
std::integral_constant<bool, is_nothrow_swappable<T>()>
{
};
?
HTH & Greetings from Bremen,
Daniel Kr?gler
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