in reply to Re^4: AND and OR on Regular Expressions
in thread AND and OR on Regular Expressions

aren't equivalent in the first place.

Aside from the missing leading ^, what do you mean?

Granted, you need some extra code if you want to know what those captures are, but I don't see how they'll capture something different.

Or are you simply referring to the renumbering of the variables in backrefs?

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Re^6: AND and OR on Regular Expressions
by JavaFan (Canon) on Aug 25, 2009 at 23:31 UTC
    Some examples: 
        $_ = "foo bar foo";
    $pat1 = "(foo)";
    $pat2 = "(bar)";
    /$pat1/ && /$pat2/;          # Sets $1 (twice).
    /^(?=.*?$pat1)(?=.*?$pat2)/; # Sets $1 and $2.

    $pat1 = "(foo)";
    $pat2 = "(baz)";
    /$pat1/ && /$pat2/;          # Sets $1.
    /^(?=.*?$pat1)(?=.*?$pat2)/; # Doesn't set $1 or $2.

    $pat1 = "(foo)";
    $pat2 = "( )\\g{1}";
    /$pat1/ && /$pat2/;          # No match.
    /^(?=.*?$pat1)(?=.*?$pat2)/; # Match!

    [download]
    As for the leading ^ - it only has a speed influence, and only if there's no match. But it can be a significant speed difference if the string is long, so it's better not to omit it.
[reply]
[d/l]

      So "yes".

      You're missing the line that initialises $_ to something "foo" for the second block.

[reply]
[d/l]
[select]
        I'm not missing that line. All matches are against the same $_.
[reply]

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