为什么我无法使用这段代码找到sqrt？

Question

我试图找到一种方式来获得一个开方没有内置插件，来到了第i个本，不幸的是它不会工作，我不知道为什么

double num=0; 
    while ((num*num)!=this.first) 
     num=num+0.0001; 
    return num; 

Answer 1

从谷歌受骗：

What is floating point error? 
The most common situation is illustrated by the decimal number 0.1. 
Although it has a finite decimal representation, in binary it has an 
infinite repeating representation. Thus when = 2, the number 0.1 lies 
strictly between two floating-point numbers and is exactly representable 
by neither of them. 

所以，在你的9例如填充，你的循环可能看起来像：

num = 0; add 0.0001 -> num is now 0.000099999999 
add 0.0001 -> num is now 0.000199999999998 
add 0.0001 -> num is now 0.000299999999997 
etc... 
add 0.0001 -> num is now 2.9999999999953667 
add 0.0001 -> num is now 3.000099999994321 

因此，您3精确比较将不匹配。

Answer 2

你不会得到确切的平等。你可能会想到真正平方根的0.0001，但就是这样。但num*num不会完全等于this.first，除非它实际上是0.0001的倍数的正方形。

while ((num * num) < this.first)可能会更接近你想要的。

Answer 3

使用此：

public class FindSqrt { 

public static void main(String[] strings) { 

    double num = 3; 
    System.out.println(sqrt(num, 0, num)); 
} 


private static double sqrt(double num, double min, double max) { 
    if (max-min<=0.0002) min=max; 
    double middle = (min + max)/2; 
    double x = middle * middle; 
    if ((num>=x&&num-x<=0.02)||(x>=num&&x-num<=0.02)) { 
     return middle; 
    } else if (x < num) { 
     return sqrt(num, middle, max); 
    } else { 
     return sqrt(num, min, middle); 
    } 
} 
} 

如果您需要在没有递归的解决方案（但while循环，也OK，以下的作品）：

public class FindSqrt { 

public static void main(String[] strings) { 

    double num = 131072; 
    System.out.println(sqrt(num, 0, num)); 
} 


private static double sqrt(double num, double min, double max) { 
    boolean done = false; 
    double answer = 0; 

    while(!done){ 
     if (max-min<=0.0002) min=max; 
     double middle = (min + max)/2; 
     double x = middle * middle; 
     if ((num>=x&&num-x<=0.02)||(x>=num&&x-num<=0.02)) { 
      done = true; 
      answer = middle; 
     } else if (x < num) { 
      min = middle; 
     } else { 
      max = middle; 
     } 
    } 
    return answer; 



} 
} 

然而，在这两种情况下，你可以使用这个找到数字的平方根< = 131072