Python十六进制比较 [英] Python hexadecimal comparison

问题描述

我遇到了一个问题，希望有人能帮助我找出答案！

I got a problem I was hoping someone could help me figure out!

我有一个字符串，十六进制数= '0x00000000'，这意味着:

I have a string with a hexadecimal number = '0x00000000' which means:

0x01000000 = apple  
0x00010000 = orange  
0x00000100 = banana   

所有这些组合都是可能的.即 0x01010000 = apple&橙色

All combinations with those are possible. i.e., 0x01010000 = apple & orange

如何从字符串中确定它是什么水果?我制作了包含所有组合的字典，然后将其与之进行比较，就可以了！但是我想知道一种更好的方法.

How can I from my string determine what fruit it is? I made a dictionary with all the combinations and then comparing to that, and it works! But I am wondering about a nicer way of doing it.

推荐答案

通过使用内置的 int()函数并指定基数，将字符串转换为整数:

Convert your string to an integer, by using the int() built-in function and specifying a base:

>>> int('0x01010000',16)
16842752

现在，您有了一个代表位集的标准整数.使用& ， | 和任何其他按位运算符来测试各个位.

Now, you have a standard integer representing a bitset. use &, | and any other bitwise operator to test individual bits.

>>> value  = int('0x01010000',16)
>>> apple  = 0x01000000
>>> orange = 0x00010000
>>> banana = 0x00000100
>>> bool(value & apple) # tests if apple is part of the value
True
>>> value |= banana     # adds the banana flag to the value
>>> value &= ~orange    # removes the orange flag from the value

现在，如果您需要转换回字符串:

Now, if you need to convert back to your string:

>>> hex(value)
'0x1000100'

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