问题描述

我有一个非常烦人的问题，即 Java 中的浮点数或双精度数很长.基本上这个想法是，如果我执行:

I have a very annoying problem with long sums of floats or doubles in Java. Basically the idea is that if I execute:

for ( float value = 0.0f; value < 1.0f; value += 0.1f )
    System.out.println( value );

我得到的是:

0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.70000005
0.8000001
0.9000001

我知道有浮动精度误差的累积，但是，如何摆脱这个?我尝试使用 doubles 将错误减半，但结果还是一样.

I understand that there is an accumulation of the floating precision error, however, how to get rid of this? I tried using doubles to half the error, but the result is still the same.

有什么想法吗?

推荐答案

没有将 0.1 精确表示为 float 或 double.由于这种表示错误，结果与您的预期略有不同.

There is a no exact representation of 0.1 as a float or double. Because of this representation error the results are slightly different from what you expected.

您可以使用的几种方法:

A couple of approaches you can use:

• 当使用 double 类型时，只显示你需要的数字.在检查相等性时，无论哪种方式都允许有一个小的容差.
• 或者使用允许您存储要精确表示的数字的类型，例如 BigDecimal 可以精确表示 0.1.

• When using the double type, only display as many digits as you need. When checking for equality allow for a small tolerance either way.
• Alternatively use a type that allows you to store the numbers you are trying to represent exactly, for example BigDecimal can represent 0.1 exactly.

BigDecimal 的示例代码:

BigDecimal step = new BigDecimal("0.1");
for (BigDecimal value = BigDecimal.ZERO;
     value.compareTo(BigDecimal.ONE) < 0;
     value = value.add(step)) {
    System.out.println(value);
}

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