From @Abigail

Created by @Abigail

  $ perl -wle '/(?{1})?/'

results in

  Quantifier unexpected on zero-length expression in regex;
  marked by <-- HERE in m/(?{1})? <-- HERE / at -e line 1.

While this is a useful (fatal) error for expressions without side-effects,
it's a nuisance for expressions with side-effects. And doing a (?{ })
without side-effects is kind of useless...

The Perl regex machine is a nice framework to do backtracking with. And
then it would be useful to be able to do​:

  (?{ CODE })?

There is a workaround of course​:

  (?​:(?{ CODE })|)

but that's more verbose.

Flags:
    category=core
    severity=medium

Site configuration information for perl v5.8.0:
    
Configured by abigail at Fri Jun 21 18:20:52 CEST 2002.

Summary of my perl5 (revision 5.0 version 8 subversion 0 patch 17339) configurat
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Locally applied patches:
    DEVEL17339


@INC for perl v5.8.0:
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    /home/abigail/Sybase
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From @tamias

On Thu, Jul 11, 2002 at 09​:50​:28AM -0700, Abigail wrote​:

$ perl \-wle '/\(?\{1\}\)?/'

results in

Quantifier unexpected on zero\-length expression in regex;
marked by \<\-\- HERE in m/\(?\{1\}\)? \<\-\- HERE / at \-e line 1\.

While this is a useful (fatal) error for expressions without side-effects,
it's a nuisance for expressions with side-effects. And doing a (?{ })
without side-effects is kind of useless...

The Perl regex machine is a nice framework to do backtracking with. And
then it would be useful to be able to do​:

\(?\{ CODE \}\)?

There is a workaround of course​:

\(?​:\(?\{ CODE \}\)|\)

But (?{ }) always succeeds. The only way for (?{ }) to be backtracked is
for the surrounding regex to backtrack.

Ronald

From @vanstyn

Abigail <abigail@​foad.org> wrote​:
: $ perl -wle '/(?{1})?/'
:
:results in
:
: Quantifier unexpected on zero-length expression in regex;
: marked by <-- HERE in m/(?{1})? <-- HERE / at -e line 1.
:
:
:While this is a useful (fatal) error for expressions without side-effects,
:it's a nuisance for expressions with side-effects. And doing a (?{ })
:without side-effects is kind of useless...

Presumably the only relevant side-effects are the ones that occur under
backtracking, ie rollback of local()ised variables. If I understand you
correctly you want to be able to say​:
  local $count = 0;
  "foobar" =~ m{
  (?{ local $count = 1 })?
  foo
  (??{ $count ? 'baz' : 'bar' })
  };
... and have it match 'foobar' the second time through. Is that right?

Do you have a more concrete example that cannot easily be expressed
using the currently supported syntax?

What would you want m/(?{1})*/ to do? The same as C< 1 while 1 >?

:Flags​:
: category=core
: severity=medium

As far as I am aware we consciously chose to impose this restriction,
so I'd class this as a wishlist item rather than a bug. If the
restriction is underdocumented (which is quite likely), then
the docs need fixing.

Hugo

From @Abigail

On Thu, Jul 11, 2002 at 01​:10​:27PM -0400, Ronald J Kimball wrote​:

On Thu, Jul 11, 2002 at 09​:50​:28AM -0700, Abigail wrote​:

$ perl \-wle '/\(?\{1\}\)?/'

results in

Quantifier unexpected on zero\-length expression in regex;
marked by \<\-\- HERE in m/\(?\{1\}\)? \<\-\- HERE / at \-e line 1\.

While this is a useful (fatal) error for expressions without side-effects,
it's a nuisance for expressions with side-effects. And doing a (?{ })
without side-effects is kind of useless...

The Perl regex machine is a nice framework to do backtracking with. And
then it would be useful to be able to do​:

\(?\{ CODE \}\)?

There is a workaround of course​:

\(?​:\(?\{ CODE \}\)|\)

But (?{ }) always succeeds. The only way for (?{ }) to be backtracked is
for the surrounding regex to backtrack.

Well, yes. Consider the following code to generate all combinations
of a 4 element set​:

"" =~ /(?{ @​x = qw !A B C D!; @​y = (0) x 4; })
  (?​:(?{ local $y [0] = 1 })|)
  (?​:(?{ local $y [1] = 1 })|)
  (?​:(?{ local $y [2] = 1 })|)
  (?​:(?{ local $y [3] = 1 })|)
  (?(?{ print "@​x[grep {$y [$_]} 0 .. 3]\n" }) fail )/x;

I'd find it more logical to be able to write it as​:

"" =~ /(?{ @​x = qw !A B C D!; @​y = (0) x 4; })
  (?{ local $y [0] = 1 })?
  (?{ local $y [1] = 1 })?
  (?{ local $y [2] = 1 })?
  (?{ local $y [3] = 1 })?
  (?(?{ print "@​x[grep {$y [$_]} 0 .. 3]\n" }) fail )/x;

 
as '?' indicates "first try matching, then try without".
 
Abigail

From @Abigail

On Thu, Jul 11, 2002 at 06​:28​:02PM +0100, Hugo van der Sanden wrote​:

Abigail <abigail@​foad.org> wrote​:
: $ perl -wle '/(?{1})?/'
:
:results in
:
: Quantifier unexpected on zero-length expression in regex;
: marked by <-- HERE in m/(?{1})? <-- HERE / at -e line 1.
:
:
:While this is a useful (fatal) error for expressions without side-effects,
:it's a nuisance for expressions with side-effects. And doing a (?{ })
:without side-effects is kind of useless...

Presumably the only relevant side-effects are the ones that occur under
backtracking, ie rollback of local()ised variables. If I understand you
correctly you want to be able to say​:
local $count = 0;
"foobar" =~ m{
(?{ local $count = 1 })?
foo
(??{ $count ? 'baz' : 'bar' })
};
.. and have it match 'foobar' the second time through. Is that right?

Yes.

Do you have a more concrete example that cannot easily be expressed
using the currently supported syntax?

Well, as I said, there's a workaround, so it's always expressable in
the currently supported syntax. But expressing (?{ CODE }){100,200}
as a sequence of | clauses is a bit awkward.

What would you want m/(?{1})*/ to do? The same as C< 1 while 1 >?

Yes.

Abigail

From @vanstyn

Abigail <abigail@​foad.org> wrote​:
:On Thu, Jul 11, 2002 at 06​:28​:02PM +0100, Hugo van der Sanden wrote​:
:> Do you have a more concrete example that cannot easily be expressed
:> using the currently supported syntax?
:
:Well, as I said, there's a workaround, so it's always expressable in
:the currently supported syntax. But expressing (?{ CODE }){100,200}
:as a sequence of | clauses is a bit awkward.

Hence 'cannot _easily_ be expressed'.

:> What would you want m/(?{1})*/ to do? The same as C< 1 while 1 >?
:
:Yes.

I think this is in large part what the current behaviour aims to
guard against.

I could imagine supporting this in 5.10 by way of a new switch, but
I do not think it would be wise to change the default behaviour.
I would also be tempted to allow any such switch also to turn on
the currently disabled backtracking through assertions, which would
allow patterns such as /(?=.*(.))/ to backtrack and find alternate
ways of matching the assertion.

Hugo

From @iabyn

removing this ticket from the (?{}) metabug, as its a wishlist rather
than a bug