Why is following regular expression as greedy?

Question

Why does (.*?) group in /({\/\* )?#if (.*?)( \*\/})?/ not capture process.env.MODE === "std" when using {/* #if process.env.MODE === "std" */} as input?

I am aware removing ? from last group works… that said, it is a requirement.

Is it possible to work around this?

Thanks for helping out @JvdV. Is there a way to write this so that last group is optional (which is a requirement)? — sunknudsen, Aug 26 '22 at 12:06
What about end of string anchor? `/({\/\* )?#if (.*?)( \*\/})?$/` — Wiktor Stribiżew, Aug 26 '22 at 12:08
@WiktorStribiżew Using `$` is promising! Thanks! Are you aware of any other workarounds? — sunknudsen, Aug 26 '22 at 12:11

Wiktor Stribiżew · Accepted Answer · 2022-08-26T12:19:01.330

If you need to have an optional pattern at the end after an optional pattern (like .*?), you can convert the .*? lazy dot pattern into a tempered greedy token:

({\/\* )?#if ((?:(?! \*\/}).)*)( \*\/})?

See the regex demo.

The .*? is changed into a pattern that matches any char, zero or more but as many as possible occurrences, that is not a starting point of the */} char sequence.

Details:

({\/\* )? - an optional Group 1: /* and space
#if - an #if string
((?:(?! \*\/}).)*) - Group 2: any char other than line break chars, zero or more but as many as possible occurrences, that is not a starting point of the */} char sequence
( \*\/})? - an optional Group 3: */} string

Ahmed Gaafer · Answer 2 · 2022-08-26T12:07:18.630

0

remove the optional '?' from the end of the regex which is making the last group optional

it should be

({\/\* )?#if (.*?)( \*\/})

edited Aug 26 '22 at 12:07

answered Aug 26 '22 at 12:04

Ahmed Gaafer

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Thanks for helping out… problem is that last group is optional. – sunknudsen Aug 26 '22 at 12:06

Why is following regular expression as greedy?

2 Answers2