JavaScript函数检查对数字进行编码需要多少位? [英] JavaScript function to check how many bits are required to encode a number?

问题描述

我有以下问题.这是示例输入/输出数据:

I have the following problem. Here is sample input/output data:

• i:0，o:1(0)
• i:1，o:1(1)
• i:2，o:2(10)
• i:3，o:2(11)
• i:4，o:3(100)
• i:5，o:3(101)
• i:6，o:3(110)
• i:7，o:3(111)

但是，这只是使用最少的位数对数字进行编码.我想要一些与众不同的东西，很难把头缠在它上面.我希望输出为2的幂.

However, that is just using the smallest number of bits to encode the number. I want something slightly different, having a hard time wrapping my head around it. I want the output to be a power of 2.

• i:0，o:2(00)
• i:1，o:2(01)
• i:2，o:2(10)
• i:3，o:2(11)
• i:4，o:4(0100)
• i:5，o:4(0101)
• i:6，o:4(0110)
• i:7，o:4(0111)
• i:8，o:4(1000)
• ...
• i .... o:8
• ...
• i .... o:16
• ...

• i: 0, o: 2 (00)
• i: 1, o: 2 (01)
• i: 2, o: 2 (10)
• i: 3, o: 2 (11)
• i: 4, o: 4 (0100)
• i: 5, o: 4 (0101)
• i: 6, o: 4 (0110)
• i: 7, o: 4 (0111)
• i: 8, o: 4 (1000)
• ...
• i.... o: 8
• ...
• i.... o: 16
• ...

我该如何编写一个函数，该函数在JavaScript中使用一个整数，并返回给我o以获取该数字?一种有效地执行此操作，并且不将其转换为字符串并使用随机JS帮助器功能的方法:)

How do I write a function that takes an integer in JavaScript, and returns to me o for that number? One that is doing it efficiently and not converting things to strings and using random JS helper functions :)

推荐答案

如果您有点摆弄，可以找到位置MSb "，然后找到下一个2的幂(对于小于2的两个值，有例外情况)

If you're into bit fiddling, you could find the position of the MSB then find the next power of 2 (with an exceptional case for the two values below 2).

我们还可以找到一个封闭的表格:

We can also find a closed form:

const bits = x => Math.floor(1+Math.log2(x))
const npot = x => Math.pow(2, Math.ceil(Math.log2(x)))
const f = x => x < 2 ? 2 : npot(bits(x))

但是，鉴于范围内只有2的幂，因此仅使用条件阶梯会更容易，更快捷:

However, given the small number of powers of two in range, it's much easier and faster to just use a condition ladder:

const f = x => {
  if (x < 2**2) return 2;
  if (x < 2**4) return 4;
  if (x < 2**8) return 8;
  if (x < 2**16) return 16;
  if (x < 2**32) return 32;
  return 64;
}

(使用bigints时，为此使用循环...)

(When working with bigints, use a loop for that…)

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