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java - Java BigDecimal.remainder() 的问题 - 不合理和不准确的返回

转载 作者:行者123 更新时间:2023-12-04 10:21:30 25 4
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我正在做一个更改返回程序,这个看似无法解决的问题真的很令人沮丧。

我正在使用 BigDecimal,所以我可以处理精确的结果,但是我的九个 BigDecimal.remainder() 方法中有两个返回了不合理和不准确的值(其他七个工作正常),这使我的最终结果不准确。

例如,如果输入是 70.70 和 100,则输出将是一个额外的 5 美分硬币。

我在我的代码中强调了这个问题。非常感谢任何帮助。先感谢您。

import java.math.BigDecimal;
import java.math.*;
import java.util.Scanner;

public class SP010Main {

public static void main(String[] args) {

// Declaring the values of notes and coins
BigDecimal fifty = new BigDecimal(50.0);
BigDecimal twenty = new BigDecimal(20.0);
BigDecimal ten = new BigDecimal(10.0);
BigDecimal five = new BigDecimal(5.0);
BigDecimal two = new BigDecimal(2.0);
BigDecimal one = new BigDecimal(1.0);
BigDecimal fiftyC = new BigDecimal(0.5);
BigDecimal twentyC = new BigDecimal(0.2);
BigDecimal tenC = new BigDecimal(0.1);
BigDecimal fiveC = new BigDecimal(0.05);


// Getting input of cost and $ received
Scanner scanner = new Scanner(System.in);
System.out.print("Cost: ");
BigDecimal cost = scanner.nextBigDecimal();
System.out.print("$ Received: ");
BigDecimal received = scanner.nextBigDecimal();
BigDecimal totalC = received.subtract(cost);
System.out.println("Total change returned: " + totalC);


// Checking how many of each value is needed
BigDecimal fiftyI = (totalC.divide(fifty, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(fifty);
BigDecimal twentyI = (totalC.divide(twenty, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(twenty);
BigDecimal tenI = (totalC.divide(ten, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(ten);
BigDecimal fiveI = (totalC.divide(five, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(five);
BigDecimal twoI = (totalC.divide(two, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(two);
BigDecimal oneI = (totalC.divide(one, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(one);
BigDecimal fiftyCI = (totalC.divide(fiftyC, 0, RoundingMode.FLOOR));
totalC = totalC.remainder(fiftyC);

// What should be happening with the problem----------------------------
// E.g. if input is 70.70 and 100,
// Following line will return 0.30
System.out.println(totalC);
// ---------------------------------------------------------------------


BigDecimal twentyCI = (totalC.divide(twentyC, 0, RoundingMode.FLOOR));


// The problem ---------------------------------------------------------
// E.g. if input is 70.70 and 100,
// Following outputs will both be 0.0999999999999999888977697537484345
// 95763683319091796875
totalC = totalC.remainder(twentyC);
System.out.println(totalC);
BigDecimal tenCI = (totalC.divide(tenC, RoundingMode.FLOOR));
totalC = totalC.remainder(tenC);
System.out.println(totalC);
// End of the problem --------------------------------------------------


BigDecimal fiveCI = (totalC.divide(fiveC, 0, RoundingMode.FLOOR));

// Display output
System.out.printf("$50: %.0f \n$20: %.0f \n$10: %.0f \n$5: %.0f \n$2: %.0f \n$1: %.0f \n$0.50: %.0f \n$0.20: %.0f \n$0.10: %.0f \n$0.05: %.0f \n",fiftyI, twentyI, tenI, fiveI, twoI, oneI, fiftyCI, twentyCI, tenCI, fiveCI);

}

}

最佳答案

即使您正在使用 BigDecimal对象以获得精确的结果,你在声明代表纸币和硬币的对象时引入了不精确性:

BigDecimal fifty = new BigDecimal(50.0);
BigDecimal twenty = new BigDecimal(20.0);
// ...

传递给构造函数的值被解释为 double 值,但其中一些值无法以 64 位 double 格式准确捕获。

您应该使用 String基于的构造函数代替:
BigDecimal fifty = new BigDecimal("50.0");
BigDecimal twenty = new BigDecimal("20.0");
// ...

这将为您提供正确的输出:

Cost: 70.70
$ Received: 100
Total change returned: 29.30
0.30
0.10
0.00
$50: 0
$20: 1
$10: 0
$5: 1
$2: 2
$1: 0
$0.50: 0
$0.20: 1
$0.10: 1
$0.05: 0

关于java - Java BigDecimal.remainder() 的问题 - 不合理和不准确的返回,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60823631/

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