sscanf ?
Bruce Edge
bedge at troikanetworks.com
Tue Nov 20 16:22:20 EST 2001
In <3BFAC45B.9D246137 at alcyone.com>, Erik Max Francis wrote:
> Bruce Edge wrote:
>
>> Additionally, after parsing with re, I still need to convert to
>> integers,
>> which is a PITA when you consider the "%x" case. The format string
>> "knows"
>> that this is in hex, using re, I get back say a string which I still
>> need
>> to dtermine the number base and convert.
>
> You can use the int-with-arg form for converting strings to integers
> other than in base ten.
>
Ahh, OK, yes, that does it. Thanks.
>> I'm not arguing :)
>> It's just that in this one tiny case, scanf would have been the
>> perfect
>> tool.
>
> It wouldn't be hard to write a simple wrapping script that takes a
> printf format string, converts it to a regular expression, does a match
> against a string, then pulls out the arguments and converts them to
> types as appropriate. But one might just as well be using regular
> expressions directly in the first place.
>
Here's what I did to convert the format string to a regex.
It's not pretty, and many types will break it, but it serves the
purpose, for now:
def fmtstr2regex( str ):
regex = ""
while len(str):
if str[0] == '%':
x = re.match( "%(?P<len>\d*)(?P<type>\w)(?P<rest>.*)$", str )
if not x:
exc = CommandException()
exc.reason = "Invalid print format specifier %s" % str
raise exc
length = int( x.group("len") )
type = x.group("type")
# Some types need to be changed from printf to regexp world
if type == 'x':
type = '['+string.hexdigits+']'
else:
type = "\\%s" % type
regex += "(%s" % type
if length:
regex += "{%d,%d})" % ( length, length )
else:
regex += "+)"
str = x.group('rest')
else:
regex += "\%s" % str[0]
str = str[1:]
return regex
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