添加和减去双打会产生奇怪的结果 [英] Adding and subtracting doubles are giving strange results

问题描述

因此，当我使用Doubles在Java中添加或减去时，它会给我带来奇怪的结果。以下是一些：

So when I add or subtract in Java with Doubles, it giving me strange results. Here are some:

如果我添加 0.0 + 5.1 ，它会给我 5.1 。这是正确的。

If I add 0.0 + 5.1, it gives me 5.1. That's correct.

如果我添加 5.1 + 0.1 ，它会给我 5.199999999999 （重复 9 的次数可能已关闭）。那是错的。

If I add 5.1 + 0.1, it gives me 5.199999999999 (The number of repeating 9s may be off). That's wrong.

如果我减去 4.8 - 0.4 ，它会给我 4.39999999999995 （同样，重复的 9 可能会关闭）。那是错的。

If I subtract 4.8 - 0.4, it gives me 4.39999999999995 (Again, the repeating 9s may be off). That's wrong.

起初我认为这只是添加带小数值的双打的问题，但我错了。以下工作正常：

At first I thought this was only the problem with adding doubles with decimal values, but I was wrong. The following worked fine:

5.1 + 0.2 = 5.3
5.1 - 0.3 = 4.8

现在，添加的第一个数字是保存为变量的double，尽管第二个变量从的JTextField 。例如：

Now, the first number added is a double saved as a variable, though the second variable grabs the text from a JTextField. For example:

//doubleNum = 5.1 RIGHT HERE
//The textfield has only a "0.1" in it.
doubleNum += Double.parseDouble(textField.getText());
//doubleNum = 5.199999999999999

推荐答案

在Java中， double 值为 IEEE浮点数。除非它们是2的幂（或2的幂的总和，例如1/8 + 1/4 = 3/8），否则即使它们具有高精度，也不能精确地表示它们。某些浮点运算会使这些浮点数中出现的舍入误差复杂化。在上面描述的情况下，浮点错误已经变得非常重要，足以显示在输出中。

In Java, double values are IEEE floating point numbers. Unless they are a power of 2 (or sums of powers of 2, e.g. 1/8 + 1/4 = 3/8), they cannot be represented exactly, even if they have high precision. Some floating point operations will compound the round-off error present in these floating point numbers. In cases you've described above, the floating-point errors have become significant enough to show up in the output.

数字的来源无关紧要是，它是从 JTextField 解析字符串还是指定 double 字面值 - 问题是在浮动中继承 - 点数表示。

It doesn't matter what the source of the number is, whether it's parsing a string from a JTextField or specifying a double literal -- the problem is inherit in floating-point representation.

解决方法：

• 如果你认识你的话ll只有这么多的小数点，然后使用整数
  算术，然后转换为小数：

• If you know you'll only have so many decimal points, then use integer arithmetic, then convert to a decimal:

(double) (51 + 1) / 10
(double) (48 - 4) / 10

使用 BigDecimal

如果必须使用 double ，则可以使用： //en.wikipedia.org/wiki/Kahan_summation_algorithm\">Kahan求和算法。

If you must use double, you can cut down on floating-point errors with the Kahan Summation Algorithm.

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