我相信这是 clang++ 中与访问 class 的 public 成员函数相关的错误

Question

#include <iostream>

void f() { std::cout << "f()\n"; }

struct S {
    typedef void(*p)();
    operator p() { return f; }
};

int main()
{
    S s;
    s.operator p()();
}

产量：

main.cpp:13:16: error: unknown type name 'p'; did you mean 'S::p'?
    s.operator p()();
               ^
               S::p

main.cpp:6:19: note: 'S::p' declared here
    typedef void(*p)();
                  ^

但它应该，因为表达式 s.operator p()() 访问对象 S::s 的 public 成员函数。我错过了什么吗？

如果我错了，我将不胜感激来自支持答案的标准的引用。

Answer 1

这似乎是 Clang 中的一个错误。我相信代码是正确的。

Clang 4.0.0 报告：

<source>:13:16: error: unknown type name 'p'; did you mean 'S::p'?
    s.operator p()();
           ^

但是，从 C++14 3.4.5/7 [basic.lookup.classref]

If the id-expression is a conversion-function-id, its conversion-type-id is first looked up in the class of the object expression and the name, if found, is used. Otherwise it is looked up in the context of the entire postfix-expression. In each of these lookups, only names that denote types or templates whose specializations are types are considered.

[ Example:
struct A { };
namespace N {
struct A {
    void g() { }
    template <class T> operator T();
};
}


int main() {
    N::A a;
    a.operator A();
       // calls N::A::operator N::A
}
— end example ]

在您的示例中，类型 p 应该已在 class 中找到而无需限定。

我相信这是 clang++ 中与访问 class 的 public 成员函数相关的错误

I believe this is a bug in clang++ related to the access to a class's public member function

c++

public

language-lawyer

clang++