get datatype length in python by parsing a string

Question

I have strings like

uint8_t
char[5]
int[3]

How can I write a short function to get the type and length separately in an elegant way

for eg

uint8_t // return 'uint8_t', '1'
char[5] // return 'char', '5'
...

Answer 1

In [1]: import re
In [2]: r = re.compile('([\w_]+)(?:\[(\d+)\])?')
In [3]: m = r.match('char[5]')                                                                                                                                                        
In [4]: m.group(1), m.group(2) or 1
Out[4]: ('char', '5')
In [5]: m = r.match('uint8_t')
In [6]: m.group(1), m.group(2) or 1
Out[6]: ('uint8_t', 1)

Making a function is left as an exercise to the reader.

Answer 2

import re

def parse_type(text):
     match = re.match(r'(.+)\[(\d+)\]', text)
     if match:
         return match.groups()
     return text, 1

print parse_type('uint8_t')
print parse_type('char[5]')
print parse_type('int[3]')

Answer 3

given these tests:

>>> s = "char[5]"
>>> p = s.split("[")
>>> p
['char', '5]']
>>> p[1].strip("]")
'5'
>>> s = "uint8_t"
>>> p = s.split("[")
>>> p
['uint8_t']
>>>

here's a little function that gives you what you want:

def SplitNicely(s):
  p = s.split("[")
  if len(p) == 1:
    size = 1
  else:
    size = int(p[1].strip("]"))

  return p[0], size

more error checking would be useful too

Answer 4

Let's make it a one-liner:

import re

def type_and_size(s):
    return re.split('[][]', s+'[1]', 2)[:2]

type_and_size('char')
['char', '1']

type_and_size('char[5]')
['char', '5']

Obviously you can do:

type, size = type_and_size('char[5]')