c++ - typeid 不适用于非静态成员函数

Question

clang 不会编译下面对 typeid 的第三次调用(请参阅 live example )。但是我在 §5.2.8 中看不到任何不允许这样做的内容，特别是当我们认为表达式 B::f 不是多态类类型的泛左值时(参见第 3 段)。此外，根据本段，表达式 B::f 是未计算的操作数，因此，调用 typeid(B::f) 应该编译。请注意，GCC 不会编译以下对 typeid 的任何调用:

#include <iostream>
#include <typeinfo>

struct A{ int i; };
struct B{ int i; void f(); };

int main()
{
    std::cout << typeid(A::i).name() << '\n';
    std::cout << typeid(B::i).name() << '\n';
    std::cout << typeid(B::f).name() << '\n';
}

Answer 1

据我所知clang是正确的，如果它是数据成员，则使用非静态成员仅在未评估的上下文中有效。所以它看起来像 gcc前两种情况不正确，但 gcc在 sizeof 的情况下正常工作和 decltype也有未计算的操作数。

来自 draft C++11 standard栏目5.1.1 [expr.prim.general]:

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

并包括以下项目符号:

if that id-expression denotes a non-static data member and it appears in an unevaluated operand. [ Example:

struct S {
    int m;
};
int i = sizeof(S::m); // OK
int j = sizeof(S::m + 42); // OK 

—end example ]

其余项目符号不适用，如下:

• as part of a class member access (5.2.5) in which the object expression refers to the member’s class⁶¹ or a class derived from that class, or
• to form a pointer to member (5.3.1), or
• in a mem-initializer for a constructor for that class or for a class derived from that class (12.6.2), or
• in a brace-or-equal-initializer for a non-static data member of that class or of a class derived from that class (12.6.2), or

我们知道 5.2.8 部分未计算操作数其中说:

When typeid is applied to an expression other than a glvalue of a polymorphic class type, [...] The expression is an unevaluated operand (Clause 5).

从语法中我们可以看出，一个id-expression要么是一个unqualified-id，要么是一个qualified-id:

id-expression:
    unqualified-id
    qualified-id

更新

提交了 gcc bug report: typeid does not allow an id-expression that denotes a non-static data member .