alt operator : alt(A, B) != alt(B, A) ? about parsy HOT 3 CLOSED

BTW, you can make your code more succinct like this:

level_normal = regex(r'[0-4]?[0-9]')
level_specials = regex(r'(66|77|88)')
level_num = (level_specials | level_normal).desc(parse_level_docstring)
# OR:
level_num = alt(level_specials, level_normal).desc(parse_level_docstring)

(That is, for your case you don't need the generate calls).

from parsy.

Yes, this is correct, and the documented behaviour - http://parsy.readthedocs.io/en/latest/ref/methods_and_combinators.html#parsy.alt - the parsers are done in order, and it moves to the next only if the first fails, therefore alt(A, B) != alt(B, A)

In your case, level_normal succeeds at parsing (some of) 88:

>>> re.search(r'[0-4]?[0-9]', '88')
<_sre.SRE_Match object; span=(0, 1), match='8'>

Or:

>>> level_num.parse_partial('88')
('8', '8')

This means that level_specials is not tried in the first case, and so parsing of the whole thing fails with "EOF expected" if you just call parse.

There is also a section in the tutorial about alternatives and backtracking - http://parsy.readthedocs.io/en/latest/tutorial.html#alternatives-and-backtracking

I will see if I can make the docs for alt a bit clearer on this front.

from parsy.

OK, that is clearly documented and explained on http://parsy.readthedocs.io/en/latest/tutorial.html#alternatives-and-backtracking

As I am combining parsers the way I would build a regexp, it's still a bit counter-intuitive to me.

Thank you @spookylukey

from parsy.