不可预测的双倍

Question

我知道双重价值的内部代表0.2是这样的0.199999。但是下面的代码仍然让我感到困惑。

码：

public static void main(String[] args) {
    double d= 0.3d;
    double f= 0.1d;
    System.out.println(d+f);
    System.out.println(d*f);
    System.out.println(d);
    System.out.println(f);
    System.out.println(d-f);
    System.out.println(d/f);
    System.out.println((d-f)*(d-f));
}

输出：

0.4
0.03
0.3
0.1
0.19999999999999998
2.9999999999999996
0.039999999999999994

到底是怎么回事？加法，乘法进行得很好，但减法，除法则不好。谁能解释一下 加法为何与减法不同 ？

Answer 1

简短的答案是，对于浮点运算，您存在表示错误和舍入错误。在toString()“知道”表示错误，所以如果没有舍入误差，你看不到它。但是，如果舍入误差太大，则可以。

解决方案是使用BigDecimal或舍入结果。

如果使用BigDecimal，它将显示您真正拥有的确切值。

double d = 0.3d;
double f = 0.1d;
System.out.println("d= " + new BigDecimal(d));
System.out.println("f= " + new BigDecimal(f));
System.out.println("d+f= " + new BigDecimal(d + f));
System.out.println("0.4= " + new BigDecimal(0.4));
System.out.println("d*f= " + new BigDecimal(d * f));
System.out.println("0.03= " + new BigDecimal(0.03));
System.out.println("d-f= " + new BigDecimal(d - f));
System.out.println("0.2= " + new BigDecimal(0.2));
System.out.println("d/f= " + new BigDecimal(d / f));
System.out.println("(d-f)*(d-f)= " + new BigDecimal((d - f) * (d - f)));

版画

d= 0.299999999999999988897769753748434595763683319091796875
f= 0.1000000000000000055511151231257827021181583404541015625
d+f= 0.40000000000000002220446049250313080847263336181640625
0.4= 0.40000000000000002220446049250313080847263336181640625
d*f= 0.0299999999999999988897769753748434595763683319091796875
0.03= 0.0299999999999999988897769753748434595763683319091796875
d-f= 0.1999999999999999833466546306226518936455249786376953125
0.2= 0.200000000000000011102230246251565404236316680908203125
d/f= 2.999999999999999555910790149937383830547332763671875
(d-f)*(d-f)= 0.03999999999999999389377336456163902767002582550048828125

您会注意到它0.1太大了，0.3也太小了。这意味着当您将它们相加或相乘时，您会得到一个大约正确的数字。但是，如果使用减法或除法，则会累积错误，并且得到的数字与所表示的数字相距太远。

即您可以看到0.1和0.3的结果与0.4相同，而0.3-0.1的结果与0.2不相同

顺便说一句，不使用BigDecimal就可以舍入答案

System.out.printf("d-f= %.2f%n", d - f);
System.out.printf("d/f= %.2f%n", d / f);
System.out.printf("(d-f)*(d-f)= %.2f%n", (d - f) * (d - f));

版画

d-f= 0.20
d/f= 3.00
(d-f)*(d-f)= 0.04

要么

System.out.println("d-f= " +  roundTo6Places(d - f));
System.out.println("d/f= " +  roundTo6Places(d / f));
System.out.println("(d-f)*(d-f)= " +  roundTo6Places((d - f) * (d - f)));

public static double roundTo6Places(double d) {
    return (long)(d * 1e6 + (d > 0 ? 0.5 : -0.5)) / 1e6;
}

版画

System.out.println("d-f= " +  roundTo6Places(d - f));
System.out.println("d/f= " +  roundTo6Places(d / f));
System.out.println("(d-f)*(d-f)= " +  roundTo6Places((d - f) * (d - f)));

舍入消除了舍入错误（仅保留了toString设计要处理的表示错误）

0.1之前和之后可以表示的值可以计算为

double before_f = Double.longBitsToDouble(Double.doubleToLongBits(f) - 1);
System.out.println("The value before 0.1 is " + new BigDecimal(before_f) + " error= " + BigDecimal.valueOf(0.1).subtract(new BigDecimal(before_f)));
System.out.println("The value after 0.1 is  " + new BigDecimal(f) + " error= " + new BigDecimal(f).subtract(BigDecimal.valueOf(0.1)));

版画

The value before 0.1 is 0.09999999999999999167332731531132594682276248931884765625 
    error= 8.32667268468867405317723751068115234375E-18
The value after 0.1 is  0.1000000000000000055511151231257827021181583404541015625 
    error= 5.5511151231257827021181583404541015625E-18