C++ Template:error: 'Node' is not a template type



 template <class T>
struct Node{
    T val;
    Node *next;
    Node(T &t){
        this->val = t;
        this->next = NULL;
    }
};

template <class T>
class SingleList {
public:
    SingleList();
    ~SingleList();
    void push_front(T &t);

private:
    int m_nListCount;
    Node<T> *m_head;
};

template <class T>
SingleList<T>::SingleList(){
    m_head = NULL;
    m_nListCount = 0;
}

template <class T>
SingleList<T>::~SingleList(){
    Node<T> *p, *pNext;
    for (p = m_head; p != NULL ; p = pNext) {
        pNext = p->next;
        free(p);
    }
    m_nListCount = 0;
}

template <class T>
void SingleList<T>::push_front(T &t){
    Node<T> *pNode = (Node<T> *)malloc(sizeof(Node<T>));
    if(NULL != pNode){
        pNode->val = t;
        pNode->next = m_head;
        m_head = pNode;
        m_nListCount++;
    }
}

代码全部放在头文件中,但是使用Clion编译提示:error: 'Node' is not a template type,求解答!

stl 链表 C++

神隐的四叶 9 years, 5 months ago

用vs2015试了下,可以编译通过。
但是在用g++在编译 Node<T> *pNode = (Node<T> *)malloc(sizeof(Node<T>));
这句时出现 error:there are no arguments to 'malloc' that depend on a template parameter, so a declaration of 'malloc' must be available [-fpermissive]

改成


 Node<T> *pNode = new Node<T>();

就可以通过编译了(建议类和模板用new分配内存好一点,因为new可以自动调用默认的构造函数,还内置了sizeof、类型转换和类型安全检查功能)

但是没有因为Clion,不能试。。应该是编译器实现问题吧。。

阿尔托丽亚 answered 9 years, 5 months ago

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