hex -> 16bit signed int (newbie)

[Steven Taschuk]

Quoth zif:
> How can I get 16bit signed int from a hex string?
> Eval('0x...') gives me 16bit unsigned,

... which you could easily convert to signed.

    def hex2uint16(s):
        # assumed s is up to four hex digits
        i = eval('0x' + s)
        if i >= 2**15:
            i -= 2**16
        return i

> struct.pack('i', '\x..\x..') works fine, but
> Python doesn't allow me to concatenate strings
> with '\' (str(92) gives me '\\', which doesn't
> work as well).

To be clear: concatenating strings containing backslashes is fine.
What you want, however, is to construct on the fly a string
containing backslash-escapes, and have those escapes interpreted
the same way they would be in source.  That calls for an eval.

    >>> hexstr = 'AB0E'
    >>> s = ''
    >>> for i in range(0, len(hexstr), 2):
    ...     s += '\\x%s' % hexstr[i:i+2]
    ... 
    >>> s
    '\\xAB\\x0E'
    >>> eval('"' + s + '"')
    '\xab\x0e'

But there's an easier way:

    >>> hexstr = 'AB0E'
    >>> import binascii
    >>> binascii.unhexlify(hexstr)
    '\xab\x0e'

Whereupon you can do your struct.unpack trick if you want.
(Though you should be aware that the meaning of 'i' is
machine-dependent; on my machine it's 32 bits, for example.  If
you want 16-bit unsigned on all machines, you want '>h' or some
such.  See the struct module's documentation.)

-- 
Steven Taschuk                                7\ 7'Z {&~         .
staschuk at telusplanet.net                        Y r          --/hG-
                                            (__/ )_             1^1`