如何在Java中舍入*向下*整数？
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How to round *down* integers in Java?
提问by Redandwhite
I'd like to round integers down to their nearest 1000 in Java.
我想在 Java 中将整数四舍五入到最接近的 1000。
So for example:
例如：
- 13,623 rounds to 13,000
- 18,999 rounds to 18,000
- etc
- 13,623 轮到 13,000
- 18,999 轮到 18,000 轮
- 等等
采纳答案by abyx
Simply divide by 1000 to lose the digits that are not interesting to you, and multiply by 1000:
简单地除以 1000 去掉你不感兴趣的数字，然后乘以 1000：
i = i/1000 * 1000
Or, you can also try:
或者，您也可以尝试：
i = i - (i % 1000)
回答by Poindexter
You could divide the number by 1000, apply Math.floor, multiply by 1000 and cast back to integer.
您可以将数字除以 1000，应用Math.floor，乘以 1000 并转换回整数。
回答by Alex
int i = Math.floorDiv(-13623, 1000) * 1000
//i => -14000
The above code will always round down (towards negative infinity) assuming the divisor (1000 in the example) is positive.
假设除数（示例中为 1000）为正数，上述代码将始终向下舍入（向负无穷大方向）。
The other answer (i = i/1000 * 1000) rounds down when iis positive, but up when iis negative.
另一个答案 ( i = i/1000 * 1000)i为正时向下舍入，而i为负时向上舍入。
-13623 / 1000 * 1000 == -13000
There is also a version of Math.floorDivfor longs which will work for very large numbers where the Math.floormethod might fail due to the precision of double.
还有一个Math.floorDivfor longs版本，它适用于非常大的数字，其中该Math.floor方法可能由于 的精度而失败double。
There are also Math.floorModmethods to go with the floorDivs which might allow you to shorten it a bit:
还有一些Math.floorMod方法可以与floorDivs搭配使用，这可能会让您将其缩短一点：
int i = -13623;
i -= Math.floorMod(i, 1000);
//i => -14000
