Peculiar working of '*' in regex expression

Question

While writing a regex pattern for substituting all the continuos '1's and single '1's as 's'. I found this quite confusing ,usage of '+' (used for matching 1 or more ) gave expected result, but '*' gave strange result

>>> l='100'
>>> import re
>>> j=re.compile(r'(1)*')    
>>> m=j.sub('*',l)
>>> m
'*0*0*'

While usage of '+' gave expected result.

>>> l='100'
>>> j=re.compile(r'1+')
>>> m=j.sub('*',l)
>>> m
'*00'

how does '*' in regex gives this, while its behaviour is to match 0 or more.

Answer 1

(1)* means "match 0 or more 1's". So for 100 it matches the 1, the empty string between 0 and 0, and the empty string after the last 0. You then replace the empty strings with '*'. 1+ requires at least one 1 in the match, so it won't match the boundary between characters.

For those readers curious, yes the python output is *0*0* and not **0*0*. Here is a test python script to play with. (Regex101 has the wrong output for this, because it does not use an actual python regex engine. Online Regex testers will usually use PCRE (which is provided in PHP and Apache HTTP Server), and fake the target regex engine. Always test your regex in live code!)

Here you can see in JavaScript the output will be **0*0* (it will match the empty string between 1 and 0 as a new match) This is a prime example of why 'regex flavor' is important. Different engines use slightly different rules. (in this case, if the new match starts at 0 or the character boundary)

console.log("100".replace(/(1)*/g, '*'))