How to set the Return Type of a Class Member Function as the object of a Private Struct

问题

sorry for the long and confusing title, but I couldn't think of a better way to ask this. So, what I have is a class:

template <typename T>
class Set
{
    public:
        //random member functions here
    private:
        struct Node{
            T key;
            Node *right;
            Node *left;
            int height;
        };
    public:
        Node* r_add(Node *temp);
};


Node* Set<T>::r_add(Node *temp)
{
    return temp;
}

When I try to implement the function r_add, I keep getting the error that the return type of out-of-line definition differs from that in the declaration for the r_add function. I'm not exactly clear on how to declare the return type for when I try to call a private structure in a class member function.

回答1:

The syntax for that needs to be:

template <typename T>
Set<T>::Node* Set<T>::r_add(Node *temp)
{
    return temp;
}

Please note that you don't have to use Set<T>::Node* for the argument type since the scope Set<T> will be used for Node at that point.

Another option is to use trailing return type. That will allow you to avoid having to type Set<T> any more than is necessary.

template <typename T>
auto Set<T>::r_add(Node *temp) -> Node*
{
    return temp;
}

回答2:

template <typename T>
typename Set<T>::Node* Set<T>::r_add(typename Set<T>::Node* temp)
{
    return temp;
}

here is my implementation for map, and you can find the solution to your problem

来源：https://stackoverflow.com/questions/50575836/how-to-set-the-return-type-of-a-class-member-function-as-the-object-of-a-private