Java - 最小值和最大值之间的限制数 [英] Java - limit number between min and max

问题描述

只要数字在限制范围内，我想返回，否则返回限制的最大值或最小值.我可以结合使用 Math.min 和 Math.max.

I want to return the number as long as it falls within a limit, else return the maximum or minimum value of the limit. I can do this with a combination of Math.min and Math.max.

public int limit(int value) {
    return Math.max(0, Math.min(value, 10));
}

我想知道是否存在我忽略的现有 limit 或 range 函数.
如果第三方库很常见(例如:Commons 或 Guava)，欢迎使用它们

I'm wondering if there's an existing limit or range function I'm overlooking.
3rd party libraries welcome if they are pretty common (eg: Commons or Guava)

推荐答案

从第 21 版开始，Guava包括 Ints.constrainToRange() (以及其他原语的等效方法).来自发行说明:

As of version 21, Guava includes Ints.constrainToRange() (and equivalent methods for the other primitives). From the release notes:

添加了 constrainToRange([type] value, [type] min, [type] max) 方法，将给定值限制在 min 定义的封闭范围内，并且max 值.如果它在范围内，则返回值本身，如果低于范围，则返回 min，如果高于范围，则返回 max.

added constrainToRange([type] value, [type] min, [type] max) methods which constrain the given value to the closed range defined by the min and max values. They return the value itself if it's within the range, the min if it's below the range and the max if it's above the range.

由@dimo414 复制自https://stackoverflow.com/a/42968254/122441.

Copied from https://stackoverflow.com/a/42968254/122441 by @dimo414.

不幸的是，这个版本是 2017 年 7 月的最新版本，并且在某些项目中(请参阅 https://stackoverflow.com/a/40691831/122441) Guava 已经破坏了向后兼容性，这需要我暂时保留在版本 19 上.我也很震惊，Commons Lang 和 Commons Math 都没有它！:(

Unfortunately this version is quite recent as of July 2017, and in some projects (see https://stackoverflow.com/a/40691831/122441) Guava had broken backwards compatibility that required me to stay on version 19 for now. I'm also shocked that neither Commons Lang nor Commons Math has it! :(

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