Hello rsFalse,

Is it true that ($match){1}+ equals to simply $match?

Yes, and the + does nothing here, as you can see with use warnings:

Output:

22:21 >perl 1345_SoPW.pl
Useless use of greediness modifier '+' in regex; marked by <-- HERE in
+ m/\A ((a|b){2}){1}+ <-- HERE  \z/ at 1345_SoPW.pl line 27, <DATA> li
+ne 2.
FAIL
FAIL
FAIL
SUCCESS
SUCCESS

22:21 >
[download]

Hope that helps,

Is it true that ($match){1}+ equals to simply $match? Yes, and the + does nothing here, as you can see with use warnings:

That's only true because you the OP included a quantifier in the embedded regex; and you get 3 fails because your anchors guarantee the regex used couldn't succeed (which doesn't seem like a useful test?)

This shows that {n}+ is a valid quantifier:

use strict;
use warnings;

chomp(my $match = <DATA>);
chomp(   $_     = <DATA>);

for my $regex (
                  "($match){1}+",
                  "($match){1}",
                  "($match)",
                  "($match)+",
                  "($match)+?",
              )
{
    print qw(FAIL SUCCESS)[ !! m/$regex/ ], "\n"
}

__DATA__
(?:a|b)
bbab
[download]

C:\test>junk37
SUCCESS
SUCCESS
SUCCESS
SUCCESS
SUCCESS
[download]

---

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In the absence of evidence, opinion is indistinguishable from prejudice.

I'm with torvalds on this Agile (and TDD) debunked I told'em LLVM was the way to go. But did they listen!

Hello BrowserUk,

I still get the warning when the regex isn’t embedded:

23:15 >perl -wE "my $r = qr{(?:a|b)}; say $r;"
(?^u:(?:a|b))

23:17 >perl -wE "my $r = qr{(?:a|b){1}}; say $r;"
(?^u:(?:a|b){1})

23:17 >perl -wE "my $r = qr{(?:a|b){1}+}; say $r;"
Useless use of greediness modifier '+' in regex; marked by <-- HERE in
+ m/(?:a|b){1}+ <-- HERE / at -e line 1.
(?^u:(?:a|b){1}+)

23:17 >perl -v

This is perl 5, version 22, subversion 0 (v5.22.0) built for MSWin32-x
+64-multi-thread
...
[download]

I get the same warning with Strawberry Perl v5.20.2, but not with v5.18.2. Is the warning erroneous?

Athanasius <°(((>< contra mundum

Iustus alius egestas vitae, eros Piratica,

Could you please show a use case where you want to avoid backtracking into a once matched group (thats what atomic/possessive is about)

answering your question

> Is it true that ($match){1}+ equals to simply $match?

No!

  DB<125> $_ = "a"x5
 => "aaaaa"

  DB<126> $match = "(a+)"
 => "(a+)"

  DB<128> $_ =~ /${match}a/   
 => "aaaa"

  DB<129> $_ =~ /${match}{1}a/   
 => "aaaa"

  DB<130> $_ =~ /${match}{1}+a/   

  DB<131> $_ =~ /(?>${match})a/   

  DB<132>
[download]

It seems to show that perlre is wrong with the redundant anotation.¹

edit

Answering your root question, IMHO using + as a second quantifier like in a++ only makes sense if you already need to use a first quantifier in the first level of interpolation, like in

($pat1)++($pat2)*+

i.e. + and * are not part of a sub-pattern $patN.

Otherwise I'd always prefer

(?>$pat1)(?>$pat2)

using something like ($pat){1}+ seems quite confusing for me.

¹) if "redundant" is supposed to mean line 131 and 130 are equivalent, then redundant is a misleading wording

I used a bigger regex, which contained some such groups. And for avoiding unexpected results, I wanted to use atomics/possessives. And I wanted to capture these groups. So instead of writing /(?> ($pat1) ) (?> ($pat2) )/x I decided to use possessive quantifiers to increase readability :) (my eyes dislike too many parentheses)
I don't have access to code, where I used it, but friend wrote to me, that with perl 5.20 he gained something wrong (warning/error).

> but friend wrote to me, that with perl 5.20 he gained something wrong (warning/error).

Couldn't you make this clear from the beginning?

> don't have access to code,

Then please reproduce it next time.

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)
Je suis Charlie!}