C++模板类的用法

template <class T> 
class actioncontainer 
{ 
public: 
    //构造函数 
    actioncontainer() 
    { 
        m_nRedoPos = 0; 
        m_nUndoPos = 0; 
    } 
    //容器的接口函数 
    void add(T value); 
    T redo(); 
    T undo(); 
    //容器的属性 
private: 
    int m_nRedoPos; 
    int m_nUndoPos; 
    const static int ACTION_SIZE=5; 
 
    T m_RedoAction[ACTION_SIZE]; 
    T m_UndoAction[ACTION_SIZE]; 
}; 
 
template<class T> 
void actioncontainer<T>::add(T value) 
{ 
    if (m_nUndoPos >= ACTION_SIZE) 
    { 
        //如果容器已潢，刚调整添加位置 
        m_nUndoPos = ACTION_SIZE - 1; 
        for(int i = 0; i < ACTION_SIZE; i++) 
        { 
            m_UndoAction[i] = m_UndoAction[i+1]; 
        } 
    } 
    m_UndoAction[m_nUndoPos++] = value; 
} 
 
template<class T> 
T actioncontainer<T>::redo() 
{ 
    //将恢复动作复制到撤销数组中 
    m_UndoAction[m_nUndoPos++] = m_RedoAction[--m_nRedoPos]; 
 
    //返回恢复的动作 
    return m_RedoAction[m_nRedoPos]; 
} 
 
template<class T> 
T actioncontainer<T>::undo() 
{ 
    m_RedoAction[m_nRedoPos++] = m_UndoAction[--m_nUndoPos]; 
 
    return m_UndoAction[m_nUndoPos]; 
}

// test_iostream.cpp : 定义控制台应用程序的入口点。 
// 
#include "StdAfx.h" 
#include "main.h" 
using namespace std; 
 
int _tmain(int argc, _TCHAR* argv[]) 
{ 
    actioncontainer<int> intaction; 
 
    //向容器中加动作 
    intaction.add(1); 
    intaction.add(2); 
    intaction.add(3); 
    intaction.add(4); 
 
    //撤销上一步动作 
    int nUndo = intaction.undo(); 
    nUndo = intaction.undo(); 
 
    //恢复 
    int nRedo = intaction.redo(); 
    return 0; 
}