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C++单一职责原则示例代码浅析

北冥有鱼丶丶 人气:0

单一职责原则:

就一个类而言,应该只有一个引起它变化的原因,如果一个类承担的职责过多就等于把这些职责耦合在一起,至少会造成以下两方面的问题:

我们在设计一个类时要学会发现职责,并把那些职责相互分离,其实要去判断是否应该分离出一个类来并不难,前面说过,一个类应该只有一个引起它变化的原因,如果你能想到其它的原因也能去改变这个类,那么这个类就具有多于1个的职责,就应该考虑类的职责分离。

在之前的这篇博客中,传送门,我们实现的计算器实际上也用到了单一职责原则,这里我们选出其中最经典的3.0版本和5.0版本来学习单一职责原则。

3.0版本计算器代码如下:

#include<iostream>
using namespace std;
#include<string>
//业务逻辑
//异常类用于处理异常情况
class opeException
{
public:
	void getMessage()
	{
		cout << "您的输入有误!" << endl;
	}
};
//运算类用于处理运算
class Operation
{
public:
	Operation(string& _num1, string& _num2, string& _ope) :num1(_num1), num2(_num2), ope(_ope){}
		//获取运算结果
	int getResult()
	{
		if (!(isStringNum(num1) && isStringNum(num2) && (ope == "+" || ope == "-" || ope == "*" || ope == "/")))
			throw opeException();
		if (ope == "+")
		{
			re = stoi(num1) + stoi(num2);
		}
		else if (ope == "-")
		{
			re = stoi(num1) - stoi(num2);
		}
		else if (ope == "*")
		{
			re = stoi(num1) * stoi(num2);
		}
		else if (ope == "/")
		{
			if (stoi(num2) != 0)
			{
				re = stoi(num1) / stoi(num2);
			}
			else
				throw opeException();
		}
		return re;
	}
private:
	int re;
	string num1;
	string num2;
	string ope;
	//判断一个字符串是不是数字
	bool isStringNum(string& s)
	{
		bool flag = true;
		for (auto e : s)
			if (!(isdigit(e)))
			{
				flag = false;
				break;
			}
		return flag;
	}
};
//界面逻辑
int main()
{
	try
	{
		string _num1 = " ";
		string _num2 = " ";
		string _ope = " ";
		cout << "请输入左操作数:" << endl;
		cin >> _num1;
		cout << "请输入右操作数:" << endl;
		cin >> _num2;
		cout << "请输入操作符" << endl;
		cin >> _ope;
		Operation operation(_num1, _num2, _ope);
		cout << operation.getResult() << endl;
	}
	catch (opeException &ex)
	{
		ex.getMessage();
	}
	return 0;
}

仅仅一个运算类Operation就实现了加减乘除4种功能,很明显在这个类中我至少有4个原因去修改这个类,我修改加法算法的时候可能会影响到其它的运算算法,这个类的耦合太高且严重违反了单一职责原则。

修改后的5.0版本如下:

#include<iostream>
using namespace std;
#include<string>
//业务逻辑
//异常类用于处理异常情况
class opeException
{
public:
	void getMessage()
	{
		cout << "您的输入有误!" << endl;
	}
};
//运算类
class Operation
{
	//判断一个字符串是不是数字
	bool isStringNum(string& s)
	{
		bool flag = true;
		for (auto e : s)
			if (!(isdigit(e)))
			{
				flag = false;
				break;
			}
		return flag;
	}
protected:
	bool isError(string& _strNum1, string& _strNum2, string& _ope)
	{
		if (!(Operation::isStringNum(_strNum1) && Operation::isStringNum(_strNum2) && (_ope == "+" || _ope == "-" || _ope == "*" || _ope == "/")))
		{
			return false;
		}
	}
public:
	virtual int getResult() = 0;
};
//加法运算类
class addOperation :public Operation
{
private:
	string strNum1;
	string strNum2;
	string ope;
	int re;
public:
	addOperation(string& _strNum1, string& _strNum2, string& _ope) :strNum1(_strNum1), strNum2(_strNum2), ope(_ope), re(0) {}
	virtual int getResult() override
	{
		if (!isError(strNum1, strNum2, ope))
			throw opeException();
		else
			re = stoi(strNum1) + stoi(strNum2);
		return re;
	}
};
//减法运算类
class subOperation :public Operation
{
private:
	string strNum1;
	string strNum2;
	string ope;
	int re;
public:
	subOperation(string& _strNum1, string& _strNum2, string& _ope) :strNum1(_strNum1), strNum2(_strNum2), ope(_ope), re(0) {}
	virtual int getResult() override
	{
		if (!isError(strNum1, strNum2, ope))
			throw opeException();
		else
			re = stoi(strNum1) - stoi(strNum2);
		return re;
	}
};
//乘法运算类
class mulOperation :public Operation
{
private:
	string strNum1;
	string strNum2;
	string ope;
	int re;
public:
	mulOperation(string& _strNum1, string& _strNum2, string& _ope) :strNum1(_strNum1), strNum2(_strNum2), ope(_ope), re(0) {}
	virtual int getResult() override
	{
		if (!isError(strNum1, strNum2, ope))
			throw opeException();
		else
			re = stoi(strNum1) * stoi(strNum2);
		return re;
	}
};
//除法运算类
class divOperation :public Operation
{
private:
	string strNum1;
	string strNum2;
	string ope;
	int re;
public:
	divOperation(string& _strNum1, string& _strNum2, string& _ope) :strNum1(_strNum1), strNum2(_strNum2), ope(_ope), re(0) {}
	virtual int getResult() override
	{
		if (!isError(strNum1, strNum2, ope))
			throw opeException();
		else if (stoi(strNum2) != 0)
			re = stoi(strNum1) / stoi(strNum2);
		else
			throw opeException();
		return re;
	}
};
//运算工厂类
class OpeFactory
{
public:
	Operation& choose(string &_strNum1,string &_strNum2,string &_ope)
	{
		if (_ope == "+")
		{
			operation = new addOperation(_strNum1, _strNum2, _ope);
		}
		else if (_ope == "-")
			operation = new subOperation(_strNum1, _strNum2, _ope);
		else if (_ope == "*")
			operation = new mulOperation(_strNum1, _strNum2, _ope);
		else if (_ope == "/")
		{
			operation = new divOperation(_strNum1, _strNum2, _ope);
		}
		else
			operation = nullptr;
		return *operation;
	}
private:
	Operation* operation;
};
//界面逻辑
int main()
{
	try
	{
		string _strNum1 = " ";
		string _strNum2 = " ";
		string _ope = " ";
		cout << "请输入左操作数:" << endl;
		cin >> _strNum1;
		cout << "请输入右操作数:" << endl;
		cin >> _strNum2;
		cout << "请输入操作符:" << endl;
		cin >> _ope;
		OpeFactory factory;
		Operation* re = &factory.choose(_strNum1, _strNum2, _ope);
		if (re != nullptr)
			cout << (*re).getResult() << endl;
		else
			cout << "您的输入有误!" << endl;
	}
	catch (opeException ex)
	{
		cout << "您的输入有误" << endl;
	}
	return 0;
}

在5.0版本的计算器代码中,我们将运算类分成了4种类,分别是加法类、减法类、乘法类、除法类,还创建了一个工厂类专门用于根据不同情况实例化对象,每个类只有一个职责,我们要修改某个功能只需要去修改对应的类即可,极大降低了代码之间的耦合。

单一职责原则的核心就是控制类的粒度大小、将对象解耦、提高其内聚性。如果遵循单一职责原则将有以下优点:

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