Question

所以我要做的就是将double转换为有理数。我检查小数点后有多少位数，我想将数字123.456保存为123456/1000，例如。

public Rational(double d){      
    String s = String.valueOf(d);
    int digitsDec = s.length() - 1 - s.indexOf('.');        

    for(int i = 0; i < digitsDec; i++){
        d *= 10;
    }

    System.out.println((int)d); //checking purposes
}

然而，对于数字123.456，我得到一个舍入错误，结果是123455.我想有可能用BigDecimal解决这个问题，但是我无法让它工作。此外，计算了它的有理数，我想调用另一个带参数的构造函数（int numerator，int denominator），但我现在无法在println所在的行中调用构造函数。我该怎么做？

Answer 1

对于问题的第一部分，Java将.6存储为.5999999（重复）。见这个输出：

(after first multiply): d=1234.56
(after second multiply): d=12345.599999999999
(after third multiply): d=123455.99999999999

一个解决方法是在循环结束后立即使用d = Math.round（d）。

public class Rational {

     private int num, denom;

     public Rational(double d) {
          String s = String.valueOf(d);
          int digitsDec = s.length() - 1 - s.indexOf('.');        

          int denom = 1;
          for(int i = 0; i < digitsDec; i++){
             d *= 10;
             denom *= 10;
          }
          int num = (int) Math.round(d);

          this.num = num; this.denom = denom;
     }

     public Rational(int num, int denom) {
          this.num = num; this.denom = denom;
     }

     public String toString() {
          return String.valueOf(num) + "/" + String.valueOf(denom);
     }

     public static void main(String[] args) {
          System.out.println(new Rational(123.456));
     }
}

它有效 - 尝试一下。

对于你问题的第二部分......

为了从第一个构建函数中调用第二个构造函数，可以使用“this”关键字

this(num, denom)

但它必须是构造函数中的第一行...这在这里没有意义（我们必须先做一些计算）。所以我不打算尝试这样做。

Answer 2

此代码对您来说可能有点过分，但它会处理您正在经历的舍入错误，并且还会处理重复小数（4.99999999999999变为5，而0.33333333333333333333变为1/3）。< / p>

public static Rational toRational(double number){
return toRational(number, 8);
}

public static Rational toRational(double number, int largestRightOfDecimal){

long sign = 1;
if(number < 0){
    number = -number;
    sign = -1;
}

final long SECOND_MULTIPLIER_MAX = (long)Math.pow(10, largestRightOfDecimal - 1);
final long FIRST_MULTIPLIER_MAX = SECOND_MULTIPLIER_MAX * 10L;
final double ERROR = Math.pow(10, -largestRightOfDecimal - 1);
long firstMultiplier = 1;
long secondMultiplier = 1;
boolean notIntOrIrrational = false;
long truncatedNumber = (long)number;
Rational rationalNumber = new Rational((long)(sign * number * FIRST_MULTIPLIER_MAX), FIRST_MULTIPLIER_MAX);

double error = number - truncatedNumber;
while( (error >= ERROR) && (firstMultiplier <= FIRST_MULTIPLIER_MAX)){
    secondMultiplier = 1;
    firstMultiplier *= 10;
    while( (secondMultiplier <= SECOND_MULTIPLIER_MAX) && (secondMultiplier < firstMultiplier) ){
        double difference = (number * firstMultiplier) - (number * secondMultiplier);
        truncatedNumber = (long)difference;
        error = difference - truncatedNumber;
        if(error < ERROR){
            notIntOrIrrational = true;
            break;
        }
        secondMultiplier *= 10;
    }
}

if(notIntOrIrrational){
    rationalNumber = new Rational(sign * truncatedNumber, firstMultiplier - secondMultiplier);
}
return rationalNumber;
}

这提供了以下结果（测试用例的结果显示为注释）：

Rational.toRational(110.0/3.0); // 110/3
Rational.toRational(11.0/1000.0); // 11/1000
Rational.toRational(17357.0/33300.0); // 17357/33300
Rational.toRational(215.0/21.0); // 215/21
Rational.toRational(0.123123123123123123123123); // 41/333
Rational.toRational(145731.0/27100.0); // 145731/27100
Rational.toRational(Math.PI); // 62831853/20000000
Rational.toRational(62.0/63.0); // 62/63
Rational.toRational(24.0/25.0); // 24/25
Rational.toRational(-24.0/25.0); //-24/25
Rational.toRational(-0.25333333333333333333333); // -19/75
Rational.toRational(-4.9999999999999999999999); // -5
Rational.toRational(4.9999999999999999999999);  // 5
Rational.toRational(123.456); // 15432/125

Answer 3

但这并不优雅，我相信这就是你所要求的。

double a = 123.456;
String aString = Double.toString(a);        
String[] fraction = aString.split("\\.");

int denominator = (int)Math.pow(10, fraction[1].length());
int numerator = Integer.parseInt(fraction[0] + "" + fraction[1]);

System.out.println(numerator + "/" + denominator);

Answer 4

这里，d = 123.456，然后num = 123456，j = 1000

public Rational(double d){      
  double temp =d;
  int j=1, num;
  do{
     j=j*10;
     }while((temp*j)%10==0);        
  j=j/10;
  num=(int)(d*j);
  System.out.println(num);
  System.out.println(j);
}

Answer 5

尝试

for(int i = 0; i <= digitsDec; i++){

}

在Java中加倍分数

5 个答案: