剪辑数字的最有效/优雅的方法?

2021-12-17 00:00:00 algorithm math logic c++

给定一个实数 (n)，这个实数可以是的最大值(上)，以及这个实数可以是的最小值(下)，我们如何最有效地裁剪 n，使其保持在下限和上限之间?

Given a real (n), a maximum value this real can be (upper), and a minimum value this real can be (lower), how can we most efficiently clip n, such that it remains between lower and upper?

当然，使用一堆 if 语句可以做到这一点，但这很无聊！更紧凑、优雅/有趣的解决方案怎么样?

Of course, using a bunch of if statements can do this, but that's boring! What about more compact and elegant/fun solutions?

我自己的快速尝试(C/C++):

My own quick attempt (C/C++):

float clip( float n, float lower, float upper )
{
    n = ( n > lower ) * n + !( n > lower ) * lower;
    return ( n < upper ) * n + !( n < upper ) * upper;
}

我确信还有其他更好的方法可以做到这一点，这就是我将它放在那里的原因..！

I'm sure there are other, better ways to do this, that's why I'm putting this out there..!

推荐答案

乏味、陈旧、可读和最短的又如何:

What about boring, old, readable, and shortest yet:

float clip(float n, float lower, float upper) {
  return std::max(lower, std::min(n, upper));
}

?

这个表达式也可以像这样泛化":

This expression could also be 'genericized' like so:

template <typename T>
T clip(const T& n, const T& lower, const T& upper) {
  return std::max(lower, std::min(n, upper));
}

更新

比利・奥尼尔补充说:

请注意，在 Windows 上，您可能必须定义 NOMINMAX，因为它们定义了冲突的最小和最大宏

Note that on windows you might have to define NOMINMAX because they define min and max macros which conflict

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