c++ - 堆上没有匹配函数

Question

我创建了一个 mix 方法，它通过引用获取两个列表。我创建了混合列表 (3)，我将列表 1 和 2 中的元素分配给它，然后返回列表 3。

当我尝试实现它时，它在堆栈中工作，而不是在堆中工作。 什么是可能的问题？

(sth3 = sth3->mix(sth1,sth2);) - 我没有匹配函数问题。

不工作:

    Sequence<int,string> *sth1 = new Sequence<int,string>();
    sth1->AddNode(1,"n1");
    sth1->AddNode(2,"n2");

    Sequence<int,string> *sth2 = new Sequence<int,string>();
    sth2->AddNode(10,"n1");   

    Sequence<int,string> *sth3 = new Sequence<int,string>();
    sth3 = sth3->mix(sth1,sth2);
    sth3->Print();

工作一个:

    Sequence<int,string> st1;
    st1.AddNode(1,"n1");

    Sequence<int,string> st2;
    st2.AddNode(10,"n1");

    Sequence<int,string> st3;
    st3 = st3.mix(st1,st2);
    st3.Print();

简化函数 mix(..)

template<typename key,typename info>
Sequence<key, info> Sequence<key, info>::mix(const Sequence<key, info> &s1, const Sequence<key,info> &s2)
{
            Sequence<key,info> s;
            Node<key, info> *curr1 = s1.head;
            Node<key, info> *curr2 = s2.head;

            while (s.count < 10)
            {
                s.AddNode(curr1->GetId(), curr1->GetData())
                curr1 = curr1->GetNext();
                s.AddNode(curr2->GetId(), curr2->GetData())
                curr2 = curr2->GetNext();

                if (curr1 == NULL && curr2 == NULL)
                    break;

            }

    return s;
}

Answer 1

将 sth3 = sth3->mix(sth1,sth2) 更改为 *sth3 = sth3->mix(*sth1,*sth2)。