14.1.解释

概念:提供一种方法顺序访问一个聚合对象中各个元素,而又不需暴露该对象的内部表示。

main(),客户

IProject,产品接口

CProject,产品类

IIterator,迭代器接口

IProjectIterator,产品迭代器接口

CProjectIterator,产品迭代器实现类

说明:CProject实现产品类,能够返回一个迭代器的指针。这个迭代器将封装产品类里的一个数组。所以迭代器在运行Next函数时,可以遍历这个数组的所有元素。

简单来说,就是用代码实现vector<int>::iterator或vector<int>::const_iterator。

我们来看代码:

//IProject.h

#pragma once
#include "IProjectIterator.h"
#include <iostream>
using std::string;
class IProject
{
public:
    IProject(void)
    {
    }
    virtual ~IProject(void)
    {
    }
    virtual void Add(string name, int num, int cost) = 0;
    virtual string GetProjectInfo() = 0;
    virtual IProjectIterator* GetIterator() = 0;
    virtual void Erase() = 0;
};

//Project.h

#pragma once
#include "iproject.h"
#include "IProjectIterator.h"
#include <iostream>
#include <vector>
using std::string;
using std::vector;
class CProject :
    public IProject
{
public:
    CProject(void);
    CProject(string name, int num, int cost);
    ~CProject(void);
    string GetProjectInfo();
    void Add(string name, int num, int cost);
    IProjectIterator * GetIterator();
    void Erase();
private:
    string m_name;
    int m_num;
    int m_cost;
    vector<IProject*> m_projectList;
};

//Project.cpp

#include "StdAfx.h"
#include "Project.h"
#include "..\CommonDeclare\Convert.h"
#include "ProjectIterator.h"
#include <iostream>
#include <vector>
using std::string;
using std::vector;
CProject::CProject( void )
{
    m_name = "";
    m_num = 0;
    m_cost = 0;
}
CProject::CProject(string name, int num, int cost) :m_name(name), m_num(num), m_cost(cost)
{
}
CProject::~CProject(void)
{
}
string CProject::GetProjectInfo()
{
    string info = "";
    info.append("项目名称是:");
    info.append(this->m_name);
    info.append("\t项目人数:");
    info.append(CConvert::ToString(m_num));
    info.append("\t项目费用:");
    info.append(CConvert::ToString(m_cost));
    return info;
}
void CProject::Add( string name, int num, int cost )
{
    this->m_projectList.push_back(new CProject(name, num, cost));
}
IProjectIterator * CProject::GetIterator()
{
    return new CProjectIterator(this->m_projectList);
}
void CProject::Erase()
{
    vector<IProject*>::reverse_iterator projectDelIt = m_projectList.rbegin();
    for (; projectDelIt != m_projectList.rend(); projectDelIt++)
    {
        delete (*projectDelIt);
        (*projectDelIt) = NULL;
    }
    m_projectList.clear();
}

//IIterator.h

#pragma once
class IProject;
class IIterator
{
public:
    IIterator(void)
    {
    }
    virtual ~IIterator(void)
    {
    }
    virtual bool HasNext() = 0;
    virtual IProject * Next() = 0;
};

//IProjectIterator.h

#pragma once
#include "iiterator.h"
class IProject;
class IProjectIterator :
    public IIterator
{
public:
    IProjectIterator(void)
    {
    }
    virtual ~IProjectIterator(void)
    {
    }
    virtual bool HasNext() = 0;
    virtual IProject * Next() = 0;
};

//ProjectIterator.h

#pragma once
#include "iprojectiterator.h"
#include "IProject.h"
#include <vector>
using std::vector;
class CProjectIterator :
    public IProjectIterator
{
public:
    CProjectIterator(vector<IProject *> pl);
    ~CProjectIterator(void);
    bool HasNext();
    IProject * Next();
private:
    vector<IProject *> m_projectList;
    size_t m_currentItem;
};

//ProjectIterator.cpp

#include "StdAfx.h"
#include "ProjectIterator.h"
CProjectIterator::CProjectIterator(vector<IProject *> pl) : m_projectList(pl)
{
    m_currentItem = 0;
}
CProjectIterator::~CProjectIterator(void)
{
}
bool CProjectIterator::HasNext()
{
    bool b = true;
    if (m_currentItem >= m_projectList.size())
        b = false;
    return b;
}
IProject * CProjectIterator::Next()
{
    IProject *pp = m_projectList.at(m_currentItem ++);
    return pp;
}

// Iterator.cpp
#include "stdafx.h"
#include "IProject.h"
#include "Project.h"
#include "..\CommonDeclare\Convert.h"
#include "ProjectIterator.h"
#include <iostream>
#include <vector>
using std::vector;
using std::cout;
using std::endl;
void DoIt()
{
    cout << "----------未使用迭代模式----------" << endl;
    vector<IProject*> projectList;

    projectList.push_back(new CProject("星球大战项目", 10, 100000));
    projectList.push_back(new CProject("扭转时空项目", 100, 10000000));
    projectList.push_back(new CProject("超人改造项目", 10000, 1000000000));

    for (int i = 4; i < 6; i ++)
    {
        string name = "";
        name.append("第");
        name.append(CConvert::ToString(i));
        name.append("个项目");
        projectList.push_back(new CProject(name, i * 5, i * 1000000));
    }

    vector<IProject*>::const_iterator projectIt = projectList.begin();
    for (; projectIt != projectList.end(); projectIt++)
        cout << (*projectIt)->GetProjectInfo().c_str() << endl;

    vector<IProject*>::reverse_iterator projectDelIt = projectList.rbegin();
    for (; projectDelIt != projectList.rend(); projectDelIt++)
    {
        delete (*projectDelIt);
        (*projectDelIt) = NULL;
    }
    projectList.clear();
}
void DoNew()
{
    cout << "----------使用迭代模式----------" << endl;
    IProject *pproject = new CProject();
    pproject->Add("星球大战项目", 10, 100000);
    pproject->Add("扭转时空项目", 100, 10000000);
    pproject->Add("超人改造项目", 10000, 1000000000);

    for (int i = 4; i < 6; i ++)
    {
        string name = "";
        name.append("第");
        name.append(CConvert::ToString(i));
        name.append("个项目");
        pproject->Add(name, i * 5, i * 1000000);
    }

    IProjectIterator *pprojectIt = pproject->GetIterator();
    while(pprojectIt->HasNext())
    {
        IProject *p = dynamic_cast<IProject*>(pprojectIt->Next());
        cout << p->GetProjectInfo().c_str() << endl;
    }
    delete pprojectIt;
    pprojectIt = NULL;
    pproject->Erase();
    delete pproject;
    pproject = NULL;
}
int _tmain(int argc, _TCHAR* argv[])
{
    //使用Iterator模式之前
    DoIt();

    //使用Iterator
    DoNew();

    _CrtSetDbgFlag(_CRTDBG_LEAK_CHECK_DF | _CRTDBG_ALLOC_MEM_DF);
    _CrtDumpMemoryLeaks();

    return 0;
}

这个模式,可能有一点绕,需要再仔细的思考一番。本来希望把模式的工作方式,用最简单易懂的语言来总结,但其实这个总结更难。简单的模式还容易些理解,也容易总结,复杂的就难了。打算在之后,学习数据结构的过程中,都总结出这样一句话来。很期待后面对数据结构的学习,发现自己越来越爱学习了。

posted on 2011-04-19 21:53  星晨_jqren  阅读(7279)  评论(2编辑  收藏  举报