小编Lea*_*ner的帖子

我试图在C++的链表中使用overload = operator并编写下面的代码.

template<class T>
class List {

    public:
        List();
        List (T t_data);
 List& operator=(const List<T> &L);
 private:
        template<class L>
            class Node {

                public:
                    L data;
                    Node *next;
                    Node *prev;
                    Node(T t_data) {
                        data = t_data;
                        next = prev = NULL;
                    }
            };

        Node<T> *head;


};

template<class T>
List&  List<T>::operator=(const List<T> &L) {
    Node<T> *t_head = head;
    Node<T> *t_tail = head->prev;
    Node<T> *temp;
    while(t_head ! = t_tail) {
        temp = t_head;
        t_head = t_next;
        delete temp;

    }
    head = L.head; …

我正在研究运算符重载=并看到下面的例子.

class Ratio {
    public:
        Ratio(int , int  );
        Ratio(const Ratio&);
        Ratio& operator= (const Ratio&);
    private:
        int nNum, nDenum;
};

Ratio::Ratio(int n = 0, int d = 1) {
    nNum = n;
    nDenum = d;
}

Ratio::Ratio(const Ratio &T) {
    nNum = T.nNum;
    nDenum = T.nDenum;
}

Ratio& Ratio::operator= (const Ratio& R) {
    nNum = R.nNum;
    nDenum = R.nDenum;
    return *this;
}

int main() {
    Ratio r1;
    Ratio r2(2,3);
    r1 = r2;//STATEMENT 1
}

这段代码运行良好,但我想知道为什么？由于函数operator=返回对Ratio对象的引用,但在STATEMENT 1中,我们没有在任何Ratio对象中获取返回引用.