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I'm pretty sure you can emulate the absent operator with look-ahead expressions. Untested: (?=((?!exp).)*$) That is: a zero-width assertion where at every char

I'm pretty sure you can emulate the absent operator with look-ahead expressions. Untested: (?=((?!exp).)*$) That is: a zero-width assertion where at every character until the end exp doesn't match. It's a nice idea to include this in the standard, and also illustrates how convoluted regular expression syntax has become over the years, mostly because it wasn't designed to be extensible.

runako 10y ago
I haven't completely ingested this post yet, but this seems to be incorrect: > I'm pretty sure you can emulate the absent operator with look-ahead expressions. Untested: (?=((?!exp).)*$) From TFA: "Note that this is not the same as a negative look-behind or look-ahead — we’ll see how shortly."
- perlgeek 10y ago
  Well, it's not the same as a simple positive or negative look-ahead, but that's not what I proposed either.
- falsedan 10y ago
  The article explains that (?~foo) is not the same as (?!foo), since the latter fails the match if it matches, but the former succeeds if it can match anything that's not 'foo' (including 'oo'). The first examples anchor the patterns, so that this behaviour: "foo" ~= /(?~foo)/ # => 1 isn't revealed.
al2o3cr 10y ago
See also the second comment on the original issue: https://bugs.ruby-lang.org/issues/5588 https://bugs.ruby-lang.org/issues/5588
rntz 10y ago
You propose: (?=((?!exp).)*$) I don't believe you're right, but can't be sure, because you haven't given a method of translating an arbitrary regular expression containing a negation (?~E) into one without. For example, how would you translate: (?~exp)a.* which asks for "anything not exp, then an `a`, then anything"? (?=((?!exp).)*$).*a.* is wrong, since it doesn't match "aexp". (?=((?!exp).)*$)a.* is also wrong, for the same reason. (?=((?!exp).)*a)a.* is wrong, because it doesn't match "expba". EDIT: argh, HN formatting rules about asterisks.
- mnarayan01 10y ago
  I believe he's saying you'd want: (?=((?!exp).)*a.*) I'm a little dubious that full equivalence is maintained (particularly w.r.t. to e.g. #match), but my brain isn't up to handling convoluted REs this early. Edit: Another hangup I have is how would you do something like: ((?~E1)|E2)E3 Obviously you can distribute by duplicating E3, but then it's pretty clear what the absent operator buys you.
  - rntz 10y ago
    (?=((?!exp).)*a.*) is also wrong; it doesn't match "expba", but (?~exp)a.* does.
    - mnarayan01 10y ago
      /(?=((?!exp).)*a.*)/ =~ "expba" # => 1 for me at least (using #match is also truthy, but the matched string is not what I'd guess the absent operator would give).
      - rntz 10y ago
        Huh! Now I am super confused.

runako 10y ago

I haven't completely ingested this post yet, but this seems to be incorrect: > I'm pretty sure you can emulate the absent operator with look-ahead expressions. Untested: (?=((?!exp).)*$) From TFA: "Note that this is not the same as a negative look-behind or look-ahead — we’ll see how shortly."

perlgeek 10y ago
Well, it's not the same as a simple positive or negative look-ahead, but that's not what I proposed either.
falsedan 10y ago
The article explains that (?~foo) is not the same as (?!foo), since the latter fails the match if it matches, but the former succeeds if it can match anything that's not 'foo' (including 'oo'). The first examples anchor the patterns, so that this behaviour: "foo" ~= /(?~foo)/ # => 1 isn't revealed.

al2o3cr 10y ago

See also the second comment on the original issue: https://bugs.ruby-lang.org/issues/5588 https://bugs.ruby-lang.org/issues/5588

rntz 10y ago

You propose: (?=((?!exp).)*$) I don't believe you're right, but can't be sure, because you haven't given a method of translating an arbitrary regular expression containing a negation (?~E) into one without. For example, how would you translate: (?~exp)a.* which asks for "anything not exp, then an `a`, then anything"? (?=((?!exp).)*$).*a.* is wrong, since it doesn't match "aexp". (?=((?!exp).)*$)a.* is also wrong, for the same reason. (?=((?!exp).)*a)a.* is wrong, because it doesn't match "expba". EDIT: argh, HN formatting rules about asterisks.

mnarayan01 10y ago
I believe he's saying you'd want: (?=((?!exp).)*a.*) I'm a little dubious that full equivalence is maintained (particularly w.r.t. to e.g. #match), but my brain isn't up to handling convoluted REs this early. Edit: Another hangup I have is how would you do something like: ((?~E1)|E2)E3 Obviously you can distribute by duplicating E3, but then it's pretty clear what the absent operator buys you.
- rntz 10y ago
  (?=((?!exp).)*a.*) is also wrong; it doesn't match "expba", but (?~exp)a.* does.
  - mnarayan01 10y ago
    /(?=((?!exp).)*a.*)/ =~ "expba" # => 1 for me at least (using #match is also truthy, but the matched string is not what I'd guess the absent operator would give).
    - rntz 10y ago
      Huh! Now I am super confused.