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, Jan 07 '14 at 10:50

score 6 · Accepted Answer · answered Jan 07 '14 at 09:49

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 07 '14 at 09:49

anubhava

but there is no empty char in `abc`. you mean it also can match a position like `^ $ \b` ? – looping Jan 07 '14 at 09:57
Yes it definitely can. In this case it is just matching `^` – anubhava Jan 07 '14 at 09:59
can you please show me some articles about `.*` match a position? it's a little weird. – looping Jan 07 '14 at 10:02
Better to go through this post: http://www.regular-expressions.info/zerolength.html – anubhava Jan 07 '14 at 10:07
@looping: `/b*a/.exec("abc")` – Karoly Horvath Jan 07 '14 at 10:08
1

@looping Think of it in this way. /a(b*)c/ should obviously match "ac", what does the capturing group (b*) capture? – Taemyr Jan 07 '14 at 10:35

score 0 · Answer 2 · answered Jan 07 '14 at 09:49

0

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

answered Jan 07 '14 at 09:49

guessimtoolate

2 Answers2