Reading hex to int from a binary string

Diez B. Roggisch deets at nospam.web.de
Sat Oct 10 03:05:12 CEST 2009


Luc schrieb:
> On Oct 9, 10:45 am, "Diez B. Roggisch" <de... at nospam.web.de> wrote:
>> Luc schrieb:
>>
>>
>>
>>> On Oct 8, 11:13 pm, "Diez B. Roggisch" <de... at nospam.web.de> wrote:
>>>> Luc schrieb:
>>>>> Hi all,
>>>>> I read data from a binary stream, so I get hex values as characters
>>>>> (in a string) with escaped x, like "\x05\x88", instead of 0x05.
>>>>> I am looking for a clean way to add these two values and turn them
>>>>> into an integer, knowing that calling int() with base 16 throws an
>>>>> invalid literal exception.
>>>>> Any help appreciated, thanks.
>>>> Consider this (in the python interpreter):
>>>>  >>> chr(255)
>>>> '\xff'
>>>>  >>> chr(255) == r"\xff"
>>>> False
>>>>  >>> int(r"ff", 16)
>>>> 255
>>>> In other words: no, you *don't* get hex values. You get bytes from the
>>>> stream "as is", with python resorting to printing these out (in the
>>>> interpreter!!!) as "\xXX". Python does that so that binary data will
>>>> always have a "pretty" output when being inspected on the REPL.
>>>> But they are bytes, and to convert them to an integer, you call "ord" on
>>>> them.
>>>> So assuming your string is read bytewise into two variables a & b, this
>>>> is your desired code:
>>>>  >>> a = "\xff"
>>>>  >>> b = "\xa0"
>>>>  >>> ord(a) + ord(b)
>>>> 415
>>>> HTH, Diez
>>> Sorry I was not clear enough. When I said "add", I meant concatenate
>>> because I want to read 0x0588 as one value and ord() does not allow
>>> that.
>> (ord(a) << 8) + ord(b)
>>
>> Diez
> 
> Yes that too. But I have four bytes fields and single bit fields to
> deal with as well so I'll stick with struct.

For the future: it helps describing the actual problem, not something 
vaguely similar - this will get you better answers, and spare those who 
try to help you the effort to come up with solutions that aren't ones.

Diez



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