用两个整数来模拟浮点数运算

淡淡的烟草味﹌ 2022-05-24 00:18 260阅读 0赞

一个做底数，一个做指数，实现加减乘除运算。

注意精度，已经数为0的情况。

#include <stdio.h>
    #include <malloc.h>
    #include <iostream>
    using namespace std;
    
    /*
    	int 型是4个字节
    	取值访问为 -2147483648 ~ 2147483647
    	对于10进制数来说，有9位的精度
    */
    double pow10(int exp)            //10的指数
    {
    	return pow(double(10), double(exp));
    }
    
    class  newDouble
    {
    public:
    	newDouble(void){bot=0; exp=0;};
    	newDouble( double d )
    	{
    		bot =0;
    		exp =0;
    		int flag = d>0?1:-1;  //取符号
    		d = abs(d);           //取绝对值
    
    		while( d< pow10(8) && d!=0 )  //取8位精度
    		{
    			d *= 10;
    			exp--;
    		}
    		bot = flag*(int)(d);
    	}
    
    
    	newDouble operator+( newDouble b )
    	{
    		(*this).SpecifyPrecision(7); //都指定7位的精度，相加不会越界
    		b.SpecifyPrecision(7);         
    		int dis = exp - b.exp;
    		newDouble c;
    		if( dis >0 )
    		{
    			c.bot = (int)(bot*pow((double)10,(double)dis)) + b.bot;
    			c.exp = b.exp;
    		}
    		else
    		{
    			c.bot = bot + (int)(b.bot*pow((double)10,(double)dis));
    			c.exp = exp;
    		}
    		return c;
    	}
    
    	newDouble operator-( newDouble b )
    	{
    		(*this).SpecifyPrecision(7); //都指定7位的精度，相减不会越界
    		b.SpecifyPrecision(7);     
    		int dis = exp - b.exp;
    		newDouble c;
    		if( dis >0 )
    		{
    			c.bot = (int)(bot*pow((double)10,(double)dis)) - b.bot;
    			c.exp = b.exp;
    		}
    		else
    		{
    			c.bot = bot - (int)(b.bot*pow((double)10,(double)dis));
    			c.exp = exp;
    		}
    		return c;
    	}
    
    	newDouble operator*( newDouble b )
    	{
    		(*this).SpecifyPrecision(4); //都指定4位的精度，相乘不会越界
    		b.SpecifyPrecision(3);     
    		newDouble c;
    		c.bot = bot * b.bot;
    		c.exp = exp + b.exp;
    		
    		return c;
    	}
    
    	newDouble operator/( newDouble b )
    	{
    		if( b.bot ==0 )
    		{
    			cout << "the divider can not be zero"<<endl;
    			return newDouble();
    		}
    		(*this).SpecifyPrecision(7); //被除数指定8位的精度
    		b.SpecifyPrecision(4);       //除数指定4位精度    
    		newDouble c;
    		c.bot = bot / b.bot;
    		c.exp = exp - b.exp;
    		
    		return c;
    	}
    
    	newDouble operator=( double d )
    	{
    		bot =0;
    		exp =0;
    		int flag = d>0?1:-1;  //取符号
    		d = abs(d);           //取绝对值
    
    		while( d< pow10(8)&&d!=0 )  //取8位精度
    		{
    			d *= 10;
    			exp--;
    		}
    		bot = flag*(int)(d);
    		
    		return *this;              //可以使用连等
    	}
    
    	double ToDouble( newDouble b )
    	{
    		return (double)(b.bot)*pow((double)10, (double)(b.exp));
    	}
    
    	//指定精度，即底数的位数
    	void SpecifyPrecision( int num )
    	{
    		if( bot ==0 ) return;
    		int flag = bot>0?1:-1;
    		bot = abs(bot);
    		while( bot < pow10(num) )
    		{
    			bot *= 10;
    			exp--;
    		}
    		while( bot > pow10(num+1) )
    		{
    			bot /=10;
    			exp++;
    		}
    		bot = flag*bot;
    	}
    private:
    	int bot;
    	int exp;
    
    };
    
    ostream & operator<<(ostream &out , newDouble d )
    	{
    		out << d.ToDouble(d);
    		return out;
    	}
    
    
    int main(void)
    {
    	newDouble a(50);
    	newDouble b(8);
    	cout << "a=" << a<<endl;
    	cout << "b=" << b<<endl;
    	cout <<"a+b=" << a+b <<endl;
    	cout <<"a-b=" << a-b <<endl;
    	cout <<"a*b=" << a*b <<endl;
    	cout <<"a/b=" << a/b <<endl<<endl;
    
    	a = -0.5;
    	b = -0.03;
    
    	cout << "a=" << a<<endl;
    	cout << "b=" << b<<endl;
    	cout <<"a+b=" << a+b <<endl;
    	cout <<"a-b=" << a-b <<endl;
    	cout <<"a*b=" << a*b <<endl;
    	cout <<"a/b=" << a/b <<endl<<endl;
    
    	a = 2.5;
    	b = 4;
    	cout << "a=" << a<<endl;
    	cout << "b=" << b<<endl;
    	cout << "(a*b)+(b-a)+(b/a)=" <<(a*b)+(b-a)+(b/a)<<endl<<endl;
    
    	while(1)
    	{
    		double t=0;
    		cout << "Please input a( double ): "<<endl;
    		cin >> t;
    		a =t;
    		cout << "Please input b( double ): "<<endl;
    		cin >> t;
    		b =t;
    		cout << "a=" << a<<endl;
    		cout << "b=" << b<<endl;
    		cout <<"a+b=" << a+b <<endl;
    		cout <<"a-b=" << a-b <<endl;
    		cout <<"a*b=" << a*b <<endl;
    		cout <<"a/b=" << a/b <<endl<<endl;
    	}
    
    	return 0;
    }