c ++ - 在取消引用的对象指针上调用成员函数

Question

template <typename T>
class Node {
    private:
    T data;
    Node<T> * next;
    public:

    Node(T _data, Node<T> * _next): data(_data), next(_next){}

    T get_data() const {
        return this->data;
    }

    Node<T> * get_next() const {
        return this->next;
    }

    void set_data(T data) {
        this->data = data;
    }

    void set_next(Node<T> * next) {
        this->next = next;
    }   
};

now I try to invoke the 'set_next()' function on the dereferenced object pointer:

Node<T> new_element(element, NULL);
Node<T> * tmp = this->get_last(); // returns a pointer to the last element
*tmp.set_next(&new_element);

when I try to compile the console prints this errer message:

error: request for member ‘set_next’ in ‘tmp’, which is of pointer 
type ‘Node<int>*’ (maybe you meant to use ‘->’ ?)
*tmp.set_next(&new_element);

I don't understand why the compiler wants me to use '->' because '.' is a perfectly fine way to invoke a public member function, right?

However when I put:

Node<T> new_element(element, NULL);
Node<T> * tmp = this->get_last();
*tmp->set_next(&new_element);

I get this:

error: void value not ignored as it ought to be
*tmp->set_next(&new_element);

I don't understand what that means, can someone help me?

Answer 1

Due to operator precedence,

*tmp.set_next(&new_element);

is the same as

*(tmp.set_next(&new_element));

which is clearly not what you want.

You may use

(*tmp).set_next(&new_element);

or

tmp->set_next(&new_element);

Answer 2

tmp->set_next(&new_element);

is the correct usage.

The error message occurs because the * tries to dereference the return value, which is void .