为什么std :: optional :: operator =（U&）要求U是非标量类型？ [英] Why does std::optional::operator=(U&&) require U to be a non-scalar type?

问题描述

对于可选的模板< class U = T>可选的< T& operator =（U& v）; 标准要求（请参阅 [optional.assign] /3.16 ）：

For optional's template<class U = T> optional<T>& operator=(U&& v); the standard demands that (see [optional.assign]/3.16):

该函数不得参与重载解析，除非
...
conjunction_v
...

This function shall not participate in overload resolution unless ... conjunction_v<is_scalar<T>, is_same<T, decay_t<U>>> is false ...

为什么在分配标量时必须排除大小写类型 U == T ？

Why do we have to exclude case when assigning a scalar of type U == T?

推荐答案

存在的目的在于： / p>

This exists to support:

optional<int> o(42);
o = {}; // <== we want this to reset o

我们有一堆分配超载，其花费为：

We have a bunch of assignment overloads, which take:

1. nullopt_t


2. 可选const&


3. 可选&&


4. U&& ;


5. optional< U> const&


6. 可选< U>&

1. nullopt_t
2. optional const&
3. optional&&
4. U&&
5. optional<U> const&
6. optional<U>&&

对于标量，具体来说，＃4将是标准转换，而其他则将是用户定义的转换-因此这将是最佳匹配。但是，这样做的结果是将 o 分配给值为 0 的用户。这意味着 o = {} 可能根据 T 的类型可能具有不同的含义。因此，我们排除了标量。

For scalars, specifically, #4 would be a standard conversion whereas anything else would be a user-defined conversion - so it would be the best match. However, the result of that would be assigning o to be engaged with a value of 0. That would mean that o = {} could potentially mean different things depending on the type of T. Hence, we exclude scalars.

对于非标量，＃4和＃3等效（都是用户定义的转换），而＃3将成为非模板而获胜。没问题。

For non-scalars, #4 and #3 would be equivalent (both user-defined conversions), and #3 would win by being a non-template. No problem there.

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