Java regex to find a method's arguments

Question

I'm trying to find a regex to split a call to a method

boo(a,b), c

so it would return

boo(a,b)

and

c

after splitting by it. Is it possible doing it with regex? Thanks.

Answer 1

Regular languages cannot express recursion. At the point where you realise that the solution would involve recursion, it is not possible to express this in a "pure" regex engine (in practice, most regex engines have some sort of extension to allow limited violation of this rule).

Since you're writing in a Java context - why not use the nice programming language to perform this fairly simple manipulation of a string, rather than trying to shoehorn a regex into the solution? :)

Answer 2

I'm not sure of what you really regarding your question want but with the given case :

"boo(a,b), c".split("(?<=\\)),") //look for a ',' preceeded by ')'

gives you an array containing ["boo(a,b)","c"] .

EDIT :

For this case : boo(a(a,b),b),c , use :

"boo(a(a,b),b),c".split("(?<=\\)),(?!(\\w\\p{Punct}?)+\\))")

(?!(\\\\w\\\\p{Punct}?)+\\\\)) means you don't accept match if it's followed by (a word and one or zero punctuation) n times , and a right parenthesis .

The result is an array containing ["boo(a(a,b),b)","c"] .