P is for Practical | |
PerlMonks |
comment on |
( [id://3333]=superdoc: print w/replies, xml ) | Need Help?? |
The proof for the (x?|y) case follows immediately from regex union being commutative. Regexp union is not commutative when one of the alternates is a leading substring of another: then order becomes important - (E|x) will always match E in preference to x. It is the presence of the outer anchors in the original pattern that disambiguates and thus makes it commutative. Hugo In reply to Re^2: Why machine-generated solutions will never cease to amaze me
by hv
|
|