设计模式 - 策略和职责链模式 C++

策略和职责链模式: 运行时选择的算法。

策略模式的主要意图是： 对于一个问题，我们其实可以通过多种方式来解决，但最好的解决方式都是根据"具体的情况"来决定的。这就是动态的选择，是策略。 职责链是将一系列的策略串在一起，当遇到问题时从头到尾对职责链进行遍历，来找到最适合的解决方式。

这里的"具体情况"在程序中可以理解为上下文环境 context，它的输入会决定到底采取什么样的策略。

如果用UML图表示的话，策略模式与命令模式或状态模式会有些接近，所以理解它解决问题的思路和意图会更加实用。

示例程序 <来自 Thingking in C++>

场景： 假如忘了一个人的名字怎么办？ 策略模式

#include <iostream>

using namespace std;

class NameStrategy
{
public:
virtual void greet() = 0;
};

class SayHi : public NameStrategy
{
public:
void greet() {cout << "Hi, how are you" << endl;}
};

class SaySorry : public NameStrategy
{
public:
void greet() {cout << "Sorry, I forget your name" << endl;}
};

class Ignore : public NameStrategy
{
public:
void greet() {cout << "Pretend, I didn't see you" << endl;}
};

class Context
{
private:
NameStrategy&  m_NameStrategy;
public:
Context(NameStrategy& s) : m_NameStrategy(s){};
void greet() {m_NameStrategy.greet();};
};

void main()
{
int i = 0;
SayHi sayhi;
SaySorry saysorry;
Ignore ignore;

Context c1(sayhi),  c2(saysorry), c3(ignore);

c1.greet();
c2.greet();
c3.greet();

cin >> i;
}

场景： 一个小孩想和家人要礼物，他会和妈妈，爸爸，爷爷，奶奶挨个要，看看谁会满足他。 职责链模式

#include <iostream>
#include <vector>

using namespace std;

enum ANSWER {NO, YES};

class GimmeStragegy
{
public:
virtual ANSWER mayIhave() = 0;
virtual ~GimmeStragegy() {};
};

class AskMom : public GimmeStragegy
{
public:
ANSWER mayIhave() {cout << "Mom may I have this?" << endl; return NO;}
};

class AskDad : public GimmeStragegy
{
public:
ANSWER mayIhave() {cout << "Dad may I have this?" << endl; return NO;}
};

class AskGrandpa : public GimmeStragegy
{
public:
ANSWER mayIhave() {cout << "Grandpa may I have this?" << endl; return NO;}
};

class AskGrandma : public GimmeStragegy
{
public:
ANSWER mayIhave() {cout << "Grandma may I have this?" << endl; return YES;}
};

class Gimme : public GimmeStragegy
{
private:
vector<GimmeStragegy*> m_ResponseChain;
public:
Gimme()
{
m_ResponseChain.push_back(new AskMom());
m_ResponseChain.push_back(new AskDad());
m_ResponseChain.push_back(new AskGrandpa());
m_ResponseChain.push_back(new AskGrandma());
}

ANSWER mayIhave()
{
vector<GimmeStragegy*>::iterator it = m_ResponseChain.begin();
while (it != m_ResponseChain.end())
if ((*it++)->mayIhave() == YES) return YES;

cout << "Nothing ... ..." << endl;
return NO;
}

~Gimme()
{
for (vector<GimmeStragegy *>::iterator it = m_ResponseChain.begin(); it != m_ResponseChain.end(); it ++)
if (NULL != *it)
{
delete *it;
*it = NULL;
}
m_ResponseChain.clear();
}
};

void main()
{
int i = 0;
Gimme chain;
chain.mayIhave();

cin >> i;
}