c++ - 非静态成员函数的 decltype 格式不正确吗？

Question

我不确定是否完全理解[dcl.type]/4.3 :

For an expression e, the type denoted by decltype(e) is defined as follows:

• [...]
• (4.3) otherwise, if e is an unparenthesized id-expression or an unparenthesized class member access, decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;
• [...]

对我来说，强调的部分同时适用于id-expression 和class member access，对吧？

使用我最喜欢的编译器，我得到以下结果。

✓ 被编译器接受

namespace N { void f() {} }
using type = decltype(N::f);
type* pf = N::f;

好吧，我想； N::f 是一个未加括号的 id 表达式，没有命名一组重载函数。

✗ 被编译器拒绝

namespace N { void f() {} void f(int) {} }
using type = decltype(N::f); // error: decltype cannot resolve address of overloaded function
type* pf = N::f;

好的； N::f 确实命名了一组重载函数。

✗ 被编译器拒绝

struct S { void f(){} };
using type = decltype(S::f); // error: invalid use of non-static member function 'void S::f()'
type* pf = &S::f;

嗯？ S::f 会命名一组重载函数吗？

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总而言之，我对 [dcl.type]/4.3 的理解是不是很糟糕？ gcc 主干错了吗？两个都？没有任何？卡穆洛克斯？

Answer 1

原因很简单，S::f 的使用对类成员有限制。

[expr.prim.id]

2 An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member's class or a class derived from that class, or
• to form a pointer to member ([expr.unary.op]), or
• if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

最后一个与您的代码相关的项目符号仅适用于非静态数据成员。没有提供功能。

我只能推测为什么不允许这样做，though I previously asked that question .