"A"之前的预期构造函数,析构函数或类型转换

Question

template <typename T>
class A {
    class B {
        typedef int INT;
        INT func(double e) {
            return INT(e * 3.6);
        }
    };
};

我的问题是func从类声明中删除函数的定义.这是我的程序的简化案例.编译器抱怨typedef.

这是我的尝试:

template <typename T>
A<T>::B::INT A<T>::B::func(double e) {
    return INT(e * 3.6);
}

编译器错误是 main.cpp:14: error: expected constructor, destructor, or type conversion before "A".

Answer 1

g ++ 4.7直接指出了这个问题:

$ g++ 3.cpp
3.cpp:12:1: error: need 'typename' before 'A<T>::B::INT' because 'A<T>::B' is a dependent scope

所以:

    template <typename T>
    typename A<T>::B::INT A<T>::B::func(double e) {
//  ^^^^^^^^
        return INT(e * 3.6);
    }