why regular expression .*? only match an empty string in javascript?

Question

/.*?/.exec("abc");//output [""]

I think .*? is non-greedy and it should return a

If you want just one character, you can use /./.exec('abc') without any modifiers — Maxim Krizhanovsky
– Maxim Krizhanovsky, Commented Jan 7, 2014 at 10:50

anubhava · Accepted Answer · 2014-01-07 09:49:17Z

6

Well that is expected since .* means 0 or more and by putting ? you make it non-greedy hence it match an empty string.

If you want to match a then you should use:

 /.+?/.exec("abc");

DIfference is + instead of * which means match 1 or more characters using non-greedy quantifier.

answered Jan 7, 2014 at 9:49

anubhava

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6 Comments

but there is no empty char in abc. you mean it also can match a position like ^ $ \b ?

Yes it definitely can. In this case it is just matching ^

can you please show me some articles about .* match a position? it's a little weird.

@looping Think of it in this way. /a(b*)c/ should obviously match "ac", what does the capturing group (b*) capture?

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guessimtoolate · Accepted Answer · 2014-01-07 09:49:03Z

0

By using * instead of e.g. + you allowed for matching empty string as a non-greedy option.

answered Jan 7, 2014 at 9:49

guessimtoolate

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