反转 Python 整数的位

2022-01-14 00:00:00 python integer bit-manipulation

问题描述

给定一个十进制整数(例如 65)，如何反转 Python 中的底层位?即以下操作:

Given a decimal integer (eg. 65), how does one reverse the underlying bits in Python? i.e.. the following operation:

65 → 01000001 → 10000010 → 130

看来这个任务可以分解为三个步骤:

It seems that this task can be broken down into three steps:

  1. 将十进制整数转换为二进制表示
  2. 反转位
  3. 转换回十进制

第 2 步和第 3 步看起来非常简单(请参阅 this 和 this SO 问题与步骤#2) 相关，但我被困在步骤#1.第 1 步的问题是检索完整的十进制表示并填充零(即 65 = 01000001，而不是 1000001).

Steps #2 and 3 seem pretty straightforward (see this and this SO question related to step #2), but I'm stuck on step #1. The issue with step #1 is retrieving the full decimal representation with filling zeros (ie. 65 = 01000001, not 1000001).

我四处寻找，但似乎找不到任何东西.

I've searched around, but I can't seem to find anything.

解决方案

int('{:08b}'.format(n)[::-1], 2)

您可以指定任何填充长度来代替 8.如果您想要真正花哨，

You can specify any filling length in place of the 8. If you want to get really fancy,

b = '{:0{width}b}'.format(n, width=width)
int(b[::-1], 2)

让您以编程方式指定宽度.

lets you specify the width programmatically.

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