# Reading hex to int from a binary string

Diez B. Roggisch deets at nospam.web.de
Fri Oct 9 10:45:12 CEST 2009

```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

```