c – 从非成员模板函数访问私有内部类类型

#include <iostream>
using namespace std;

class Outer {
    struct Inner {
        int num;    
    };

public:
 static Inner GetInner() {
    return Inner{-101};
}
};

// void func1(Outer::Inner inner) {  // [1] Does not compile as expected
//  cout << inner.num <<endl;
//}

template <typename Dummy>
void func2(Outer::Inner inner,Dummy = Dummy()) {
    cout << inner.num << endl;
}


int main() {
    // func1(Outer::GetInner()); // [2] does not compile as expected 
    func2<int>(Outer::GetInner()); // [3] How does this compile? 
                                   // Outer::Inner should not be accessible
                                   // from outside Outer
    return 0;
}

我怎么能在非成员函数func2中使用类型为Outer :: Inner的参数,这是一个私有类型？当我尝试使用以下错误消息时,’func1正确地抱怨：

prog.cpp: In function 'void func1(Outer::Inner)':
prog.cpp:5:9: error: 'struct Outer::Inner' is private
  struct Inner {
         ^
prog.cpp:15:19: error: within this context
 void func1(Outer::Inner inner) {
               ^

我在ubuntu上使用g 4.8.2,但我也在gcc-4.9.2上看到了这一点(在www.ideone.com上)

您可以在这里试用代码：Ideone