为什么此公共成员函数不能在类内部声明的私有struct成员上调用decltype？

Question

The following code, which loosely represents some serialization stuff I'm working on, compiles with g++ ( http://ideone.com/0rsGmt ), but Visual Studio Express 2013 RC fails with the following errors:

Error 1 error C2326: 'void foo::print(void)' : function cannot access 'foo::bar::member_'
Error 2 error C2039: 'bar' : is not a member of 'foo'

The code:

#include <iostream>

class foo
{   
    private:
        struct bar
        {
            int member_;
        };

    public:
        void print()
        {
            std::cout << sizeof(decltype(foo::bar::member_)) << std::endl;
        }
};

int main(int argc, char* argv[])
{
    foo f;
    f.print();
    return 0;
}

What's wrong? Visual Studio inadequacy or something else? Obviously I can move the struct declaration out of the class; and Daniel Frey has offered a workaround below; but I want to know why the above code won't compile as-is with Visual Studio.

Update: The accepted answer says that it should work, but as is typical for Microsoft it doesn't. I've filled a bug report here: http://connect.microsoft.com/VisualStudio/feedback/details/801829/incomplete-decltype-support-in-c-11-compiler

(If someone can suggest a better title for the question, I'd appreciate it!)

Answer 1

I think your code should work (as on GCC or Clang), according to

5 Expressions [expr]

⁸ In some contexts, unevaluated operands appear (5.2.8, 5.3.3, 5.3.7, 7.1.6.2). An unevaluated operand is not evaluated. An unevaluated operand is considered a full-expression. [ Note: In an unevaluated operand, a non-static class member may be named (5.1) and naming of objects or functions does not, by itself, require that a definition be provided (3.2). — end note ]

It seems that VC++ does not implement what the note clarifies, so you need a (faked) instance as a work-around to make VC++ happy. This should work:

void print()
{
    std::cout << sizeof(std::declval<bar>().member_) << std::endl;
}

Note that I removed the decltype as sizeof can work on expressions directly.

Answer 2

也许您的代码的问题在于结构就像类，并且您在类内部定义了结构，但是编译器直到编译整个类才知道该类（或结构）是否存在。