How to compare a hex byte with its literal (visual) representation in Python?-CodePudding

I want to compare two entities, one being an int as a single byte and the other a str which is the ASCII code of the visual representation (visual reading) of that byte (not its ASCII value).

For example: I have byte 0x5a which I want to compare with a pure text that says '5a' (or '5A', case is unimportant) as when I speak or I read this in a book. I don't need to compare the byte versus the 'Z' ASCII character, which in my case would be a different thing.

How can I do that? Could be I am in a middle of a logical trap and I don't know how to search for a proper question or similar case.

CodePudding user response：

There are functions that allow you to transform numbers into their string representation, in certain basis. In your case, hex should do the trick. For example:

>>> hex(0x5a)
'0x5a'
>>> hex(0x5a)[2:] # get rid of `0x` if you don't want it
'5a'

CodePudding user response：

You can use hex() to turn the integer into a hex string, and then you slice off the first two characters using string slicing to remove the leading 0x:

lhs = 90
rhs = "5a"

print(hex(lhs)[2:] == rhs)

This outputs:

True