c++ - std::declval 和未计算的表达式

Question

参考 Proposing Standard Library Support for the C++ Detection Idiom中的以下示例：

// primary template handles types that do not support pre-increment
template< class, class = void_t<> >
struct
has_pre_increment_member : false_type { };

// specialization recognizes types that do support pre-increment
template< class T >
struct
has_pre_increment_member<T, void_t<decltype( ++declval<T&>() )>> : true_type { };

表达式如何++declval<T&>()分类为未评估？

在上面，假设declval()返回T&如Is there a reason declval 返回 add_rvalue_reference 而不是 add_lvalue_reference 中讨论的那样，表达式的结果++T&（由 产生++declval<T&>）不会变成 odr-used 而不是未评估？根据ODR 使用：

如果使用了引用并且在编译时不知道其引用对象，则该引用是 odr-used；

在上述情况下，不是在编译时不知道所指对象吗？在那种情况下，如何首先使用引用declval()？

Answer 1

how does the expression ++declval<T&>() classify as unevaluated?

Because it's within decltype():

The operand of the decltype specifier is an unevaluated operand.

A function, variable, structured binding, assignment operator or constructor, etc. must appear in a potentially evaluated expression in order to be odr-used. decltype() doesn't meet that criteria.