How can you convert a variable into denary in Python? [duplicate]

Question

I have had problems with converting strings (like inputs) into denary, as a string is a necessary for using hexadecimal (as it can't be stored as an integer because of the characters a-f being included). Here are two attempts that don't work:

>>> hex_number = 'ff' #(255 in denary)
>>> ascii(0xhex_number)
SyntaxError: invalid hexadecimal literal
>>> ascii('0x'+hex_number)
"'0xff'"

I have also seen that you can use format() somehow, but that didn't work either.

>>> format(hex_number, '16') #This was what the demo said to do.
'ff              '

How can I get it so that hex_number (hexadecimal) is converted to denary (aka decimal), or any other n-base number systems?

int('ff', 16)

khelwood
– khelwood

2020-10-11 18:28:55 +00:00
Commented Oct 11, 2020 at 18:28 — khelwood
– khelwood, Commented Oct 11, 2020 at 18:28

spagh-eddie · Accepted Answer · 2020-10-11 18:30:47Z

2

Turn the string into an int then back into a string

>>> n = int('ff', 16); n
255
>>> f'{n:d}'
'255'

edited Oct 11, 2020 at 18:30

answered Oct 11, 2020 at 18:28

spagh-eddie

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2 Comments

waevarc Over a year ago

Oh. I didn't realise that you could adapt int() for n-base numerics. Does that make ascii() completely useless then?

spagh-eddie Over a year ago

I'm sure that ascii is useful for text (e.g. ümlaut or \xfcmlaut), but I have never personally had a need for it. For numbers, do not use ascii. Everything can be done through the "master" integer representation of a number, and convert to/from string representations on input/output.

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