superpollo wrote:
Sebastian Bassi wrote:
On Fri, Jul 24, 2009 at 7:28 PM, superpollo<u...@example.net> wrote:
is there a pythonic and synthetic way (maybe some standard module) to
"pack"
an integer (maybe a *VERY* big one) into a string? like this:
What do you mean to pack? Maybe Pickle is what you want.
import cPickle
variable = 124348654333577698
cPickle.dump(variable,open('filename', 'w'))
To load it:
import cPickle
vaariable = cPickle.load(open('filename'))
It is not "into a string", but instead of a filename, use a string
with stringio
>>> number
252509952
>>> f = cStringIO.StringIO()
>>> cPickle.dump(number , f)
>>> f.getvalue()
'I252509952\n.'
>>>
as you can see, the pickled representation is not what i wanted...
OTOH, using a higher protocol:
>>> cPickle.dump(number , f , 1)
>>> f.getvalue()
'J\x00\xff\x0c\x0f.'
which is very close to '\xf0\xcf\xf0\x00', that i meant originally... so
i i change my spec as; '\x0f\x0c\xff\x00' (i.e. left- intead of right-
zeropadding) i am almost there...
bye
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