模板类的成员函数，将模板类型作为参数 [英] Member functions of a templated class, that take a template type as argument

问题描述

我有一个节点结构和堆栈类。当我把'void Push（T data）'的定义放在类定义之外时，我得到：

I have a node struct and stack class. When I put the definition for 'void Push(T data)' outside the class definition I get:

error: 'template<class T> class Stack' used without template parameters

但是当我把它放在类定义里面它工作正常。 br>
下面是代码：

But when I put it inside the class definition it works fine.
Here is the code:

template <class T>
struct Node
{
    Node(T data, Node <T> * address) : Data(data), p_Next(address) {}
    T Data;
    Node <T> * p_Next;
};

template <class T> 
class Stack
{
public:

    Stack() : size(0) {}
    void Push(T data);
    T Pop();

private:
    int size;
    Node <T> * p_first;
    Node <T> * p_last;  
};

Push（T data）的实现是：

The implementation for Push(T data) is :

void Stack::Push(T data)
{
    Node <T> * newNode;   

    if(size==0)
        newNode = new Node <T> (data, NULL);
    else
        newNode = new Node <T> (data, p_last);

    size++;
    p_last = newNode;
}

编辑：解决方案工作除了现在我得到一个链接错误，调用函数。

The solutions worked except that now I get a linking error whenever I try to call the functions.

Stack<int>::Pop", referenced from   
_main in main.o   
symbol(s) not found.

除非Stack.h中的定义是Stack.cpp

unless the definitions are in Stack.h instead of Stack.cpp

推荐答案

您需要再次使用模板< class T> 再次使用 T 作为类的模板参数）：

You need to use the template <class T> again (and then use that T again as the template parameter for the class):

template <class T>
void Stack<T>::Push(T data)
{
    Node <T> * newNode;   

    if(size==0)
        newNode = new Node <T> (data, NULL);
    else
        newNode = new Node <T> (data, p_last);

    size++;
    p_last = newNode;
}

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