Regular expression: Is this a bug or feature?

Discussion in 'Ruby' started by Martin Kahlert, Jan 23, 2007.

  1. Hi ruby experts!

    Is this intended behaviour?

    irb(main):001:0> s1='a=1'
    => "a=1"
    irb(main):002:0> s2='b=1'
    => "b=1"
    irb(main):003:0> s1 =~ /a|b=(.)/
    => 0 <------ expression matches
    irb(main):004:0> $1
    => nil <------ but where is argument?
    irb(main):005:0> s2 =~ /a|b=(.)/
    => 0 <------ expression matches
    irb(main):006:0> $1
    => "1" <------ this has been expected
    irb(main):007:0> s1 =~ /(a|b)=(.)/
    => 0 <------ expression matches
    irb(main):012:0> $2
    => "1" <------ this has been expected

    Tested on ruby 1.8.2 (2004-12-22) [i686-linux]

    Thanks for your help in advance
    Martin Kahlert, Jan 23, 2007
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  2. Martin Kahlert

    George Ogata Guest

    Call this part 1:
    Part 2:
    Part 3:
    I'm not sure why you think it might be a bug. The '|' operator just
    binds very loosely, so you have to group the "a|b" in parens. Note
    that in part 1, The bit that matches is the left side of the '|',
    namely 'a' (no parens), so there are no captures. In part 2, the
    right side ('b=(.)') matches, so there's 1 capture. In part 3, it
    matches the whole thing ('(a|b)=(.)'), so there are 2 captures.

    Does this make sense?

    Note that if you only want 1 capture, you can also use the shy
    grouping operator (?:...), so:

    s1 =~ /(?:a|b)=(.)/

    [$1, $2] #=> ["1", nil]
    George Ogata, Jan 23, 2007
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  3. Martin Kahlert

    Per Velschow Guest

    I assume this is the one you need explanation for? I think you simply
    misinterpret the regexp. /a|b=(.)/ is a union between the two regexp /a/
    and /b=(.)/. So in this case it matches only the first one which has no
    bindings. The regexp you are probably looking for would be
    /(?:a|b)=(.)/. Try that.
    Per Velschow, Jan 23, 2007

  4. I always assumed 'a|b anything' means '(a|b) anything'.

    Thanks for this clarification!

    Martin Kahlert, Jan 24, 2007
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