* quantifier in Perl 6

Question

This seems to be something very basic that I don't understand here.

Why doesn't "babc" match / a * / ?

> "abc" ~~ / a /
｢a｣
> "abc" ~~ / a * /
｢a｣
> "babc" ~~ / a * /
｢｣                    # WHY?
> "babc" ~~ / a + /
｢a｣

Answer 1

Because * quantifier makes the preceding atom match zero or more times.

｢｣ is first match of / a * / in any string. For example:

say "xabc" ~~ / a * . /; # OUTPUT: ｢x｣

it's same:

say "xabc" ~~ / (a+)? . /;

If you set the pattern more precise, you will get another result:

say "xabc" ~~ / x a * /; # OUTPUT: ｢xa｣
say "xabc" ~~ / a * b /; # OUTPUT: ｢ab｣

Answer 2

The answers here are correct, I'll just try to present them in a more coherent form:

Matching always starts from the left

The regex engine always starts at the left of the strings, and prefers left-most matches over longer matches

* matches empty strings

The regex a* matches can match the strings '', 'a', 'aa' etc. It will always prefer the longest match it finds, but it can't find a match longer than the empty string, it'll just match the empty string.

Putting it together

In 'abc' ~~ /a*/, the regex engine starts at position 0, the a* matches as many a's as it can, and thus matches the first character.

In 'babc' ~~ /a*/, the regex engine starts at position 0, and the a* can match only zero characters. It does so successfully. Since the overall match succeeds, there is no reason to try again at position 1.