s/// operator and undef question...

ismail has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: s/// operator and undef question... by merlyn (Sage) on Dec 13, 2000 at 05:30 UTC
It took me a few minutes to see why it's working, and why you can't use it in general. You are likely missing a /g on the end of that regex match. What's happening is that `$is` is getting set to a list of two items, causing the loop to execute twice. The two values are ignored for the next step though. Since the last regex computed is `is`, that makes the substitute inside the loop work as if you had said `$target =~ s/is/are/`. Yup. Always scanning from the left. So, the only reason this works is because the replacement string cannot once again match the search string. But try it with s/is/iis/, and you'll be totally hosed. -- Randal L. Schwartz, Perl hacker	[reply] [d/l] [select]
Re: s/// operator and undef question... by chromatic (Archbishop) on Dec 13, 2000 at 05:25 UTC
I can answer the first for you. m// matches the nothing at any part of the string: `my $target = 'this is a test of the servise'; $target =~ s//are/; print "$target\n";` [download] There's a spot of nothing at the very start of the string. I think you're missing /g at the end of the regex in the loop condition. As for that, my assumption is that there's Deep Magic here, and you shouldn't rely on these results. Another test: `if ($target =~ /test/) { $target =~ s//quiz/; }` [download] Maybe a blank regex condition on the left stays with what was previously matched. Dunno if that's an intended feature or not. Update: I'm upgrading it to potential bug: `my $target = 'this is a test of the servise'; if ($target =~ /test/) { $target =~ s//test2/; } if ($target =~ /nothing/) { $target = s//null/; } else { $target =~ s//foo/; }` [download] The moral of the story is, don't do a substitution inside a match on the same string.	[reply] [d/l] [select]
Re: Re: s/// operator and undef question... by merlyn (Sage) on Dec 13, 2000 at 05:41 UTC
`my $target = 'this is a test of the servise'; if ($target =~ /test/) { $target =~ s//test2/; } if ($target =~ /nothing/) { $target = s//null/; } else { $target =~ s//foo/; }` [download] My guess is that this returns `this is a foo2 of the servise` as documented in perlop: `[....] If the pattern evaluates to the empty string, the last successfully executed regular expression is used instead. [....]` [download] And so my standard admonition about "don't use $1 except in the context of a conditional" would also apply here, since "most recent success" is contextual. -- Randal L. Schwartz, Perl hacker	[reply] [d/l] [select]
s/// tomfoolery explained (again) by japhy (Canon) on Dec 13, 2000 at 06:42 UTC
Ok, regardless of what is happening, merlyn explained the situation. `$string =~ /foo/; $string =~ //; # this repeats the last successful match $string =~ /foo/; $string =~ s//bar/; # again, /foo/ is used in place of //` [download] `japhy` -- Perl and Regex Hacker	[reply] [d/l]
(Ovid) Re: s/// operator and undef question... by Ovid (Cardinal) on Dec 13, 2000 at 05:29 UTC
I could only duplicate your results with adding `/g` to the first regex. I have no frickin' clue how the substitution works. Incidentally, the following is a no-op: `for ($target =~ /is/g) { s//are/; }` [download] But if I take of the /g switch: `for ($target =~ /is/) { s//are/; }` [download] I get the following error message: `Modification of a read-only value attempted at eval.pl line 6.` [download] This is beginning to look like a bug to me. Cheers, Ovid Join the Perlmonks Setiathome Group or just click on the the link and check out our stats.	[reply] [d/l] [select]
Re: s/// operator and undef question... by Fastolfe (Vicar) on Dec 13, 2000 at 05:34 UTC
Your expression `$target =~ /is/` in a list context will return `(1)` for success (see `m//` in perlop). Thus, your loop is equivalent to this: `for my $is (1) { # one iteration with $is=1 $target =~ s//are/; }` [download] I, however, am getting different results than you under 5.6: `thare is a test of the servise` [download] I get the same results by dropping the loop entirely and doing just this: `$target =~ /is/; $target =~ s//are/;` [download] Like you, I find that odd. Somehow Perl is doing the equivalent of `s/is/are/;`.	[reply] [d/l] [select]
Re: s/// operator and undef question... by ismail (Acolyte) on Dec 13, 2000 at 05:37 UTC
I'm sorry, i threw everybody off... the first regex is missing the "/g" from the end, thus: `my $target = 'this is a test of the servise'; for my $is ($target =~ /is/g) { $target =~ s//are/; } print $target\n"; #yields 'thare are a test of the servaree'` [download] sorry about that! Ismail	[reply] [d/l]