Java 四舍五入到最接近的 0.5
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Java round to nearest .5
提问by user1513171
How is this possible in Java?
这在 Java 中怎么可能？
I have a float and I'd like to round it to the nearest .5.
我有一个浮点数，我想把它四舍五入到最接近的 0.5。
For example:
例如：
1.1 should round to 1.0
1.1 应该四舍五入到 1.0
1.3 should round to 1.5
1.3 应该四舍五入到 1.5
2.5 should round to 2.5
2.5 应该四舍五入到 2.5
3.223920 should round to 3.0
3.223920 应该四舍五入到 3.0
EDIT: Also, I don't just want the string representation, I want an actual float to work with after that.
编辑：另外，我不只是想要字符串表示，我想要一个实际的浮点数。
采纳答案by Michael Yaworski
@SamiKorhonen said this in a comment:
@SamiKorhonen 在评论中说：
Multiply by two, round and finally divide by two
乘以二，圆，最后除以二
So this is that code:
所以这是代码：
public static double roundToHalf(double d) {
return Math.round(d * 2) / 2.0;
}
public static void main(String[] args) {
double d1 = roundToHalf(1.1);
double d2 = roundToHalf(1.3);
double d3 = roundToHalf(2.5);
double d4 = roundToHalf(3.223920);
double d5 = roundToHalf(3);
System.out.println(d1);
System.out.println(d2);
System.out.println(d3);
System.out.println(d4);
System.out.println(d5);
}
Output:
输出：
1.0
1.5
2.5
3.0
3.0
回答by Peter Lawrey
The general solution is
一般的解决办法是
public static double roundToFraction(double x, long fraction) {
return (double) Math.round(x * fraction) / fraction;
}
In your case, you can do
在你的情况下，你可以做
double d = roundToFraction(x, 2);
to round to two decimal places
四舍五入到小数点后两位
double d = roundToFraction(x, 100);
回答by Kostas Chalkias
Although it is not that elegant, you could create your own Round function similarly to the following (it works for positive numbers, it needs some additions to support negatives):
虽然它不是那么优雅，但您可以创建自己的 Round 函数，类似于以下内容（它适用于正数，它需要一些添加来支持负数）：
public static double roundHalf(double number) {
double diff = number - (int)number;
if (diff < 0.25) return (int)number;
else if (diff < 0.75) return (int)number + 0.5;
else return (int)number + 1;
}
