On Oct 8, 11:13 pm, "Diez B. Roggisch" <de...@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. However you pointed me in the right direction and I found that int (binascii.hexlify(a + b, 16)) does the job. Thanks. -- http://mail.python.org/mailman/listinfo/python-list
