Re: How to write an is_nothrow_swapable<T>?
[2nd attempt after 1 day]
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>()>
{
};
?
Further, I think that the additional parts
or
(std::is_nothrow_move_constructible<T>::value and
std::is_nothrow_move_assignable<T>::value)
are not really helpful, because they should automatically be deduced
correctly from
#include <type_traits>
#include <utility> // std::swap templates
template <typename T>
constexpr bool is_nothrow_swappable() {
using std::swap;
return noexcept(swap(std::declval<T&>(), std::declval<T&>()));
}
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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