Reading hex to int from a binary string

[Luc]

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.

 

[Diez B. Roggisch]

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):
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:
415


HTH, Diez

 

[Luc]

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.

 

[Dennis Lee Bieber]

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.

Yeesh... This is what struct is designed for...
ˆand more
8805, a0d

You may need to adjust for expected endian mode...

 

[Luc]

        Yeesh... This is what   struct  is designed for...


 ˆand more




8805,  a0d

        You may need to adjust for expected endian mode...

3338

Nice, thanks!

 

[Diez B. Roggisch]

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

 

[Jack Norton]

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.

Hi,

Check out the ord() function.

Example:
x = '\x34'
print ord(x)

output: 52

Also, if you, lets say read(4), and end up with `x = '\x05\x41\x24\x00'
you can use x to address each character. So if you
print x[1]
output: 'A'

That should be enough to get you started in the right direction.

-Jack

 

[Luc]

Luc schrieb:





(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.

Thanks.

 

[Diez B. Roggisch]

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