重载运算符的嵌套类型(overloading operator for a nested type)

重载运算符的嵌套类型(overloading operator for a nested type)

以下代码构建正常，但不调用重载的模板运算符。 这是为什么？ main()函数打印120而不是x 。

#include <iostream> namespace foo { struct bar { enum Type { x = 'x', y = 'y' }; }; template<typename T> std::ostream& operator<<(std::ostream& s, typename T::Type o) { return s << char(o); } } int main() { foo::bar::Type t{ foo::bar::x }; std::cout << t; }

The following code builds fine but the overloaded template operator is not invoked. Why is that? The main() function prints 120 instead of x.

#include <iostream> namespace foo { struct bar { enum Type { x = 'x', y = 'y' }; }; template<typename T> std::ostream& operator<<(std::ostream& s, typename T::Type o) { return s << char(o); } } int main() { foo::bar::Type t{ foo::bar::x }; std::cout << t; }

最满意答案

为了实例化模板，编译器将永远无法确定类型T应该是什么（请查看此问题以获取更多信息）。 这样做：

std::ofstream & operator<<(std::ostream & s, bar::Type t){...}

您声明的enum是公开的，因此在这种情况下不必担心访问限制。

The compiler will never be able to figure what the type T is supposed to be in order to instantiate the template (check out this question for more on that). Just do this:

std::ofstream & operator<<(std::ostream & s, bar::Type t){...}

The enum you declared is public to bar so there's no access restriction to have to worry about in this case.