Python Regex greedy boundaries

Question

I have a

import re

s = """
/* comments */
vector<CWaypoint> Vparam;     // comment
int cValue=2049;                // comment
double param=0.01;              // comment
"""

exp = re.compile(r"(?:\s|\b)(.+?)=(.+?)\b;")
print(exp.findall(s))

My expected output is

[(cValue,2049), (param,0.01)]

but why am I getting the data type before the variable name like below

[('int cValue', '2049'), ('double param', '0.01')]

Why isn't boundaries working even if they are non greedy

Regex scans from left to right, so they match. You can change your regex from (.+?)= to (\w+)= to get what you want. (Though I don't think parsing code with regex is a good idea). — nhahtdh
– nhahtdh, Commented Jun 18, 2015 at 12:21
There is the option of using (\S+), which matches non-space characters (again, this is very poor approximation, and you should consult the documentation for the exact character set). — nhahtdh
– nhahtdh, Commented Jun 18, 2015 at 12:24
Whats with "Left to right" ? I mean how does . makes so much of a difference? — Bharadwaj
– Bharadwaj, Commented Jun 18, 2015 at 12:26
Regex engine scans from left to right, so it will find the left most match first, and since \b.+?=.+?\b; matches int cValue=2049, it is returned as a match. \S+ or \w+ basically forces the content before = to contain no space, so you only get the variable name. — nhahtdh
– nhahtdh, Commented Jun 18, 2015 at 12:29

Avinash Raj · Accepted Answer · 2015-06-18 12:29:50Z

1

Note that the \s will also match newline characters and .+ will match any char including spaces.

I suggest you to use [^\s=]+ before =

exp = re.compile(r"([^\s=]+)=(.+?);")

answered Jun 18, 2015 at 12:29

Avinash Raj

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