Front page | perl.perl5.porters |
Postings from January 2012
On 2012-01-27 Father Chrysostomos via RT wrote: > On Fri Jan 27 07:33:32 2012, demerphq wrote: > > /* turn .* into ^.* with an implied $*=1 */ > > > > I have to admit I have not checked to see what the heck $*=1 means. > > $* doesn’t do anything anymore, unless you are using Classic::Perl. > > $* = 1 puts /m on every match in 5.8, bugs aside. > > What the comment means exactly by implied $*=1 I don’t know. Is it > referring to /^/ meaning /^/m in split? But that couldn’t be right. It means that a regexp that starts with .* is implicitly anchored because if it doesn't match at offset 0, it won't match at offsets 1, 2, 3 ... either. /m is implied because (since .* won't cross newlines) there can be multiple possible match locations if the string contains \n. Which means you have to check every embedded \n for a match. (Conversely, if /s is active, leading .* should generate an implicit ^ with /m off (a.k.a. \A).) AFAICS this optimization is valid except when the target string ends with a newline. In that case .* could (and should) match, but /^/m won't. That is, "\n" =~ /^/mg only matches once. So ... I guess the regex code should behave differently if the /^/m is implicit and \n is the last character in the target string? (And maybe there's a missed optimization opportunity here because I don't see why this special case shouldn't trigger for [^\n]* at the beginning of a pattern.)Thread Previous | Thread Next