可能的重复：
在java中使用双精度数保持精度 https://stackoverflow.com/questions/322749/retain-precision-with-doubles-in-java
Java 程序中奇怪的浮点行为 https://stackoverflow.com/questions/327544/strange-floating-point-behaviour-in-a-java-program
我正在制作一个直方图类，我遇到了一个奇怪的问题。
这是该类的基础知识，还有更多方法，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");
}
当我绘制频率时就会出现问题。我已经创建了 2 个实例。
新直方图(0,1,10);
新直方图(0, 10, 10);
这就是他们输出的内容。
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非常长的小数限制，但在第二个示例中却没有？
