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Editorial Team
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Asked: 2026-05-25T00:46:35+00:00

‘aaa bbb’.match(/(|’)[^\1]+\1/g) // [‘aaa bbb’] ‘aaa bbb’.match(/(|’)[^]+\1/g) // [‘aaa’, ‘bbb’] Why does [^\1]+ instead

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'"aaa" "bbb"'.match(/("|')[^\1]+\1/g)
// ['"aaa" "bbb"']

'"aaa" "bbb"'.match(/("|')[^"]+\1/g)
// ['"aaa"', '"bbb"']

Why does [^\1]+ instead of [^"]+ make RegExp greedy?

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1 Answer

  1. Editorial Team

    Why does [^\1]+ instead of [^"]+ make RegExp greedy?

    That isn’t what you think it is doing.

    First of all, a + is always a maximal match, the thing you are calling “greedy”. It’s +? which is a minimal match.

    Second and more importantly, back‐referencing doesn’t happen in square‐bracketed character classes. You have accidentally just asked for any character except for Control‑A. That’s because backslash followed by digits means that code point in octal notation, as in \177 for DELETE ᴀᴋᴀ \x7F, or \40 for SPACE ᴀᴋᴀ \x20, or \0 for NULL . So when you wrote \1, you have just used U+0001 or \x01. Don’t do that. 🙂

    You probably mean to use

    (["'])(?:(?!\1).)+\1

    instead. You’ll need /s mode so that dot can match newlines, which I seem to recall Javascript has some screwed‐up–ness with.

    EDIT: According to this, clumsy old Javascript has no way to make dot match linebreaks. What rimnods! And of course because Javascript can’t do Unicode regexes, you can’t use the \p{Any} required by UTS#18’s RL1.2.

    That means you’ll have to use some lame kludge like [\S\s] instead if there’s a chance that you might have linebreaks in your quoted strings.

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