`typename std::remove_reference 之间有什么区别<t> ` 和 `constexpr 类型名 std::remove_reference<t> `?</t></t>

Question

As per the documentation( https://en.cppreference.com/w/cpp/utility/move ), there are two kinds of constructors for std::move<T> , which are posted below.

What are the differences between these constructors? What confused me most is that why there needs the keyword( typename ) in the second constructor.

I am a novice in C++. I would be thankful for any hint on this question.

template< class T >
typename std::remove_reference<T>::type&& move( T&& t ) noexcept; (since C++11)(until C++14)

template< class T >
constexpr typename std::remove_reference<T>::type&& move( T&& t ) noexcept;  (since C++14)

Answer 1

[...] there are two kinds of constructors for std::move<T> ...

No, they are not constructors, rather function signatures of std::move . One is prior to c++14 (ie since c++11 ) and the second one since C++14.

In the second one the specifier constexpr is used , meaning

constexpr - specifies that the value of a variable or function can appear in constant expressions

read more here: What are 'constexpr' useful for?

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What confused me most is that why there needs the keyword( typename ) in the second constructor.

As per the cppreference.com , there is a helper type for std::remove_reference , since c++14

template< class T >
using remove_reference_t = typename remove_reference<T>::type;  (since C++14)

therefore in the second one, it could have been

template< class T >
constexpr std::remove_reference_t<T>&& move( T&& t ) noexcept;
//        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^