CODEWITHSUNDEEP
c++c++11decltype
VP.
VP.

Reputation: 16695

Is it possible to obtain a type from decltype?

decltype returns a full type of an expression/entity. Is it possible to get only type?

For example, is it possible to make p to have type T in this case?

class T;
T t;
const T& tt = t;
decltype(tt) p; //decltype makes this const T& too as same as tt

Upvotes: 0

Views: 807

Answers (1)

Barry
Barry

Reputation: 302757

It depends on entirely on what you want to do in the cases of cv T* and cv T[N]. If in all of those cases you just want T, then you'll need to write a type trait:

template <typename T>
struct tag { using type = T; };

template <typename T>
struct just_t
: std::conditional_t<std::is_same<std::remove_cv_t<T>,T>::value,
                     tag<T>,
                     just_t<std::remove_cv_t<T>>>
{ };                   

template <typename T>
struct just_t<T*> : just_t<T> { };

template <typename T>
struct just_t<T&> : just_t<T> { };

template <typename T, size_t N>
struct just_t<T[N]> : just_t<T> { };

template <typename T>
struct just_t<T[]> : just_t<T> { };

If you're okay with pointers staying as they are and arrays decaying into pointers, then simply:

template <typename T>
using just_t = std::decay_t<T>;

Upvotes: 3

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