有一段奇怪的历史bit_width。

最终将被称为bit_width开始生命的函数log2，作为添加整数二次函数的提议的一部分。log2指定为在传递 0 时生成 UB。

因为这就是对数的工作原理。

但随后，事情发生了变化。该函数后来变成了log2p1，并且由于未指定的原因被赋予了更广泛的契约（C++ 术语中的“广泛契约”意味着更多的东西被认为是有效的输入）。具体来说，0 是有效输入，并产生 0 值。

这不是对数的工作方式，但无论如何。

随着 C++20 接近标准化，发现了名称冲突 (PDF)。该名称log2p1恰好对应于 IEEE-754 算法的名称，但它完全不同。此外，具有类似输入和结果的其他语言中的函数使用类似bit_length. 所以改名为bit_width.

而且由于它不再假装做对数，0 处的行为可以是我们想要的任何东西。

实际上，Python 函数int.bit_length具有完全相同的行为。前导零不被视为位长度的一部分，并且由于 的​​值0包含所有前导零......

• The function formerly known as `log2p1` actually does work exactly like a logarithm; the issue is that the formula it evaluates is not `floor(log(n) + 1)` but `ceil(log(n + 1))`. Why the committee did not describe it using the second formula, which works for all `n`, but instead used the first one which has an awkward discontinuity at `n=0` and therefore needs to be described more wordily, is anyone's guess (this is the *C++* standard committee we're talking about, after all).
• @psmears: "*including having effects way outside the return value of this function*" That's what happens when you take the logarithm of 0 as well. It is a thing you cannot do, and if you do it, then your math is broken and ceases to be actual math.
• It's perfectly valid (and useful) in mathematics to [extend the real number line](https://en.wikipedia.org/wiki/Extended_real_number_line) to include plus/minus infinity, and to define log 0 to be minus infinity. Or to treat log as a [partial function](https://en.wikipedia.org/wiki/Partial_function) on the reals. This answer currently says "that is _not_ how logarithms work" about a behavior that is 100% conformant with the part you describe as "that _is_ how logarithms work", which seems a bit inconsistent 🙂
• @psmears so what's tan 90

因为在数学上它是有道理的：

bit_width(x) = log2(round_up_to_nearest_integer_power_of_2(x + 1))
bit_width(0) = log2(round_up_to_nearest_integer_power_of_2(0 + 1))
             = log2(1)
             = 0

详细说明评论中的内容：

假设“位宽”意味着“存储（非负整数）数所需的最少位数”。直观地说，我们至少需要log2(n)四舍五入的位，所以它是一个接近 的公式ceil(log2(n))，所以 255 需要ceil(log2(255)) = ceil(7.99..) = 8位，但这不适用于 2 的幂，所以我们可以添加一个 1 的模糊因子n来得到ceil(log2(n+1))。这恰好在数学上等同1+floor(log2(n))于正 n，但log2(0)没有定义或定义为无用的东西，如地板版本中的负无穷大。

如果我们使用 0 的上限公式，我们会得到结果。您还可以看到我没有写出前导零，正如 Nicol Bolas 指出的那样，0 都是前导零。