如何将 int 转换为十六进制字符串?

2022-01-14 00:00:00 python string int hex

问题描述

我想将一个整数(即 <= 255)转换为十六进制字符串表示

I want to take an integer (that will be <= 255), to a hex string representation

eg: 我想传入 65 得到 'x41'，或者 255 得到 'xff'.

e.g.: I want to pass in 65 and get out 'x41', or 255 and get 'xff'.

我试过用 struct.pack('c',65) 来做这件事，但是 9 上面的任何东西都会窒息因为它想接收单个字符串.

I've tried doing this with the struct.pack('c',65), but that chokes on anything above 9 since it wants to take in a single character string.

解决方案

您正在寻找 chr 函数.

您似乎混合了整数的十进制表示和整数的十六进制表示，因此您需要什么并不完全清楚.根据您提供的描述，我认为其中一个片段显示了您想要的内容.

You seem to be mixing decimal representations of integers and hex representations of integers, so it's not entirely clear what you need. Based on the description you gave, I think one of these snippets shows what you want.

>>> chr(0x65) == 'x65'
True


>>> hex(65)
'0x41'
>>> chr(65) == 'x41'
True

请注意，这与 包含整数为十六进制的字符串完全不同.如果这是您想要的，请使用 hex 内置.

Note that this is quite different from a string containing an integer as hex. If that is what you want, use the hex builtin.

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