在javascript中获取数字的最低有效位的最有效方法是什么? [英] What's the most efficient way of getting position of least significant bit of a number in javascript?

问题描述

我有一些数字，我需要获取它们的低位在0位应该移位多少.

I got some numbers and I need to get how much they should be shifted for their lower bit to be at position 0.

例如:
0x40000000 => 30，因为0x40000000 >> 30 = 1
768 = 512 + 256 => 8

ex:
0x40000000 => 30 because 0x40000000 >> 30 = 1
768 = 512+256 => 8

这有效

if (Math.log2(x) == 31)
  return 31;
if (Math.log2(x) > 31)
  x = x & 0x7FFFFFFF;
return Math.log2(x & -x)

在javascript中，有没有更有效或更优雅的方法(内置?)?

Is there any more efficient or elegant way (builtin ?) to do this in javascript ?

推荐答案

您无法通过内置函数立即获得该结果，但是可以避免使用Math.log2.有一个鲜为人知的函数Math.clz32，该函数在其32位二进制表示形式中对数字的前导零进行计数.像这样使用它:

You cannot get that result immediately with a builtin function, but you can avoid using Math.log2. There is a little known function Math.clz32, which counts the number of leading zeroes of a number in its 32-bit binary representation. Use it like this:

function countTrailingZeroes(n) {
    n |= 0; // Turn to 32 bit range
    return n ? 31 - Math.clz32(n & -n) : 0;
}

console.log(countTrailingZeroes(0b11100)); // 2

三元表达式在那里捕获值 n = 0，就像简并的情况:它没有1位.

The ternary expression is there to catch the value n=0, which is like a degenerate case: it has no 1-bit.

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