regex question: $1 vs. \1

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

Replies are listed 'Best First'.
Re: regex question... by Zaxo (Archbishop) on Dec 23, 2001 at 12:39 UTC
Yes. the regex centered ops `s///` and `m//` take `\1` in the regex itself for the first fully matched pattern in parentheses. `$1` means the same in the second argument or until the next regex. After Compline, Zaxo	[reply]
Re: Re: regex question... by count0 (Friar) on Dec 23, 2001 at 12:43 UTC
Additionally: Although you can use \(digit) in the right side of a s/// operation, you shouldn't. This was included (i would guess) just to appease the sed junkies out there ;) Generally the only time \(digit) should be used is on the LHS of a s///, or within a m//. Example: `/(foo)\1/; # is effectively the same as /(foo){2}/;` [download]	[reply] [d/l]
Re: regex question: $1 vs. \1 by robin (Chaplain) on Dec 23, 2001 at 18:32 UTC
`\1` and `$1` mean different things, most of the time. In ordinary code, `$1` refers to the text that was matched by the first pair of parentheses in the last regular expression match or substitution. On the other hand, `\1` is a reference to the constant scalar `1`: `[robin@robin robin]$ perl -le 'print \1' SCALAR(0xd8b8)` [download] In a regular expression, `$1` refers to the text matched by the first pair of parentheses in the previous regular expression, whereas `\1` refers to the text matched by the first pair of parentheses in the current regular expression. So this program: `shift() =~ /(.*)/; shift() =~ /($1)x\1/ && print "yes!\n"` [download] will print `yes!` if you pass it a pair of strings like `foo fooxfoobar` In a substitution string (i.e. the `yyy` part of a `s/xxx/yyy/` command), `\1` and `$1` both refer to the same thing - the text matched by the first pair of parentheses in the pattern. It's usually considered poor style to use the `\1` version here.	[reply] [d/l] [select]
Re: regex question: $1 vs. \1 by tachyon (Chancellor) on Dec 23, 2001 at 16:02 UTC
Also $1 is not initialised on the LHS of the regex whereas \1 is... `use warnings; $_ = 'oooooo'; s/(o)$1/foo/; s/(o)\1/foo/; s/(o)/\1/; s/(o)/$1/; __END__ Use of uninitialized value in concatenation (.) at script line 3. \1 better written as $1 at script line 5.` [download] cheers tachyon s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print	[reply] [d/l]
Re: regex question: $1 vs. \1 by mrbbking (Hermit) on Dec 24, 2001 at 05:55 UTC
I happened across this today, in the third edition of the Camel Book. On page 152, in the discussion of the s/// operator... In the replacement string, $1 returns what the first ... pair of parentheses captured. (You could use also \1 as you would in the pattern, but that usage is deprecated in the replacement. ...)	[reply]