C++ 指针引用转换的问题

[Restart_2013]

报错：
“CContext::CContext(CStrategy *&)”: 不能将参数 1 从“CConcreteStrategyA *”转换为“CStrategy *&”

如果CContext(CStrategy *&)构造函数写成 CContext(CStrategy *)这种形式则会出现安全隐患。

所以要如何解决这个问题？？

#include "stdafx.h"
#include <iostream>
using std::cout;
using std::endl;



// 抽象类 -- 无法创建对象
class CStrategy
{
public:
	virtual void AlgorithmInterface() = 0;
};


//具体算法A
class CConcreteStrategyA:public CStrategy
{
public:
	CConcreteStrategyA()
	{ }
	virtual~CConcreteStrategyA()
	{
		
	}
	//算法A实现方法
public:
	virtual void AlgorithmInterface()
	{
		cout << "算法A实现" << endl;
	}
};

//具体算法B
class CConcreteStrategyB:public CStrategy
{
public:
	CConcreteStrategyB()
	{ }
	virtual~CConcreteStrategyB()
	{

	}
	//算法B实现方法
public:
	virtual void AlgorithmInterface()
	{
		cout << "算法B实现" << endl;
	}
};

//具体算法C
class CConcreteStrategyC:public CStrategy
{
public:
	CConcreteStrategyC()
	{ }
	virtual~CConcreteStrategyC()
	{

	}
	//算法C实现方法
public:
	virtual void AlgorithmInterface()
	{
		cout << "算法C实现" << endl;
	}
};

//上下文
class CContext
{
public:
	// 存在安全隐患
	// CContext(CStrategy *strategy)
	CContext(CStrategy* &strategy)
	{
		m_strategy = strategy;
	}
	virtual~CContext()
	{
		if (m_strategy)
		{
			delete m_strategy;
			m_strategy = NULL;
		}
	}

public:
	//上下文接口
	virtual void ContextInterface()
	{
		m_strategy->AlgorithmInterface();
	}

private:
	CStrategy *m_strategy;
};


int _tmain(int argc, _TCHAR* argv[])
{
	CConcreteStrategyA *pTmp = NULL;
	pTmp = new CConcreteStrategyA();
	CContext *pContextA = new CContext(pTmp);
	pContextA->ContextInterface();
	delete pTmp;
	delete pContextA;

// 	CContext *pContextB = new CContext(new CConcreteStrategyB());
// 	pContextB->ContextInterface();
// 	delete pContextB;
// 
// 	CContext *pContextC = new CContext(new CConcreteStrategyC());
// 	pContextC->ContextInterface();
// 	delete pContextC;

	return 0;
}

[Restart_2013]

引用 2 楼 xihu1364 的回复:

CConcreteStrategyA *pTmp = NULL; //这里为啥不写成CStrategy* pTmp = new CConcreteStrategyA(); 呢？
pTmp = new CConcreteStrategyA();
CContext *pContextA = new CContext(pTmp);
pContextA->ContextInterface();
delete pTmp;
delete pContextA;

哦，原来如此，为什么我那样就不行呢?

[版主大哥]

“CContext::CContext(CStrategy *&)”: 不能将参数 1 从“CConcreteStrategyA *”转换为“CStrategy *&” 不是报错了？

[zcdabing]

CStrategy *& 我在想你这个想表达什么 指针还加个引用？

[ye667788]

引用 5 楼 zcdabing 的回复:

CStrategy *& 我在想你这个想表达什么 指针还加个引用？

加上引用可以修改指针本身

[Restart_2013]

我不想写成下面这种强转的调用方式，该怎么写？

CContext *pContextA = new CContext((CStrategy*&)pTmp);

[版主大哥]

CConcreteStrategyA *pTmp = NULL; //这里为啥不写成CStrategy* pTmp = new CConcreteStrategyA(); 呢？
pTmp = new CConcreteStrategyA();
CContext *pContextA = new CContext(pTmp);
pContextA->ContextInterface();
delete pTmp;
delete pContextA;

[zcdabing]

引用 6 楼 ye667788 的回复:

[quote=引用 5 楼 zcdabing 的回复:] CStrategy *& 我在想你这个想表达什么 指针还加个引用？

加上引用可以修改指针本身[/quote] 求详细点的解释

[ye667788]

引用 7 楼 zcdabing 的回复:

[quote=引用 6 楼 ye667788 的回复:] [quote=引用 5 楼 zcdabing 的回复:] CStrategy *& 我在想你这个想表达什么 指针还加个引用？

加上引用可以修改指针本身[/quote] 求详细点的解释[/quote]

#include "stdafx.h"
#include <iostream>
using namespace std;

int a = 3;
int b = 4;

void change1(int *&p)
{
	p = &b;
}

void change2(int *p)
{
	p = &b;
}

int _tmain(int argc, _TCHAR* argv[])
{
	int *p1 = &a;
	change1(p1);
	cout<<*p1<<endl;//输出4

	int *p2 = &a;
	change2(p2);
	cout<<*p2<<endl;//输出3

	system("pause");
	return 0;
}