Why capturing group results in double matches regex

Question

Consider these two scripts:

1st: \" \".match(/(\\s)/)

and

2nd: \" \".match(/\\s/)

Results

1st: [\" \"

Answer 1

First script: The first result is the whole pattern, the second is the capturing group

Second script: the only result is the whole pattern.

Capturing groups are not only to refer later in the pattern, they are displayed in results too.

When you use a capturing group with split, the capturing group is returned with results and since the separator is supposed to slice the string, it is normal that you obtain ["", " ", ""] as result with
" " as input string and /(\s)/ as pattern.

More informations about split.

When you write " ".match(/(\s)/) the result returned is the first match. This result is unique and contains:

• the whole match
• capturing group(s)
• index of the match
• input string

When you write " ".match(/(\s)/g) the result returned is all the matches:

• whole match 1
• whole match 2
• etc.

(in the present case you have only one match)

This behaviour is normal. The match method as two different behaviours (with or without /g). It is a kind of two functions in one. For comparison in PHP (or other languages) which doesn't have the g modifier, you have two different functions: preg_match and preg_match_all