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Java:使用 double 不准确

转载 作者:搜寻专家 更新时间:2023-11-01 01:18:29 25 4
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Possible Duplicate:
Retain precision with Doubles in java
Strange floating-point behaviour in a Java program

我正在制作直方图类,但遇到了一个奇怪的问题。

这是类的基础知识,还有更多的方法,但它们与问题无关。

private int[] counters;
private int numCounters;
private double min, max, width;

public Histogram(double botRange, double topRange, int numCounters) {
counters = new int[numCounters];
this.numCounters = numCounters;
min = botRange;
max = topRange;
width = (max - min) / (double) numCounters;
}

public void plotFrequency() {
for (int i = 0; i < counters.length; i++) {
writeLimit(i * width, (i + 1) * width);
System.out.println(counters[i]);
}
}

private void writeLimit(double start, double end) {
System.out.print(start + " <= x < " + end + "\t\t");
}

问题发生在我绘制频率时。我创建了 2 个实例。 新直方图(0、1、10); 新直方图(0, 10, 10);

这是他们输出的。

Frequecy
0.0 <= x < 0.1 989
0.1 <= x < 0.2 1008
0.2 <= x < 0.30000000000000004 1007
0.30000000000000004 <= x < 0.4 1044
0.4 <= x < 0.5 981
0.5 <= x < 0.6000000000000001 997
0.6000000000000001 <= x < 0.7000000000000001 1005
0.7000000000000001 <= x < 0.8 988
0.8 <= x < 0.9 1003
0.9 <= x < 1.0 978

Frequecy
0.0 <= x < 1.0 990
1.0 <= x < 2.0 967
2.0 <= x < 3.0 1076
3.0 <= x < 4.0 1048
4.0 <= x < 5.0 971
5.0 <= x < 6.0 973
6.0 <= x < 7.0 1002
7.0 <= x < 8.0 988
8.0 <= x < 9.0 1003
9.0 <= x < 10.0 982

所以我的问题是,为什么我在第一个示例中得到了非常长的小数限制,而在第二个示例中却没有?

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