How to parse an optional token?

[ccorcos]

suppose I have optional arguments: Approve() or Approve(a, b). How would I specify that the arguments are optional?

[ccorcos]

I'm hoping theres something we can do like p.maybe(moreArgs) which returns none and backtracks if theres nothing...

[ccorcos]

here's my solution that i figured out. seems messy though

def optional(parser):
    @p.generate
    def _optional():
        r = yield p.times(parser, 0, 1)
        if len(r) > 0:
            result(r[0])
        else:
            result(None)
    return _optional

[sighingnow]

We can create a dummy parser which consumes nothing and returns None, then use the try_choice combinator:

import parsec as p

def optional(parser):
    @p.generate
    def dummy():
        '''Return None without doing anything.'''
        yield p.string('')
        return None
    return parser ^ dummy    ## (^) means try_choice

s = p.string('xx')

print(optional(s).parse('xx')) // xx
print(optional(s).parse('xy')) // None

Thanks for your proposal @ccorcos and I will add the optional combinator into the library later (maybe with better implementation).

[sighingnow]

If you have better solution, feel free to comment here.

[ccorcos]

That looks fine to me. I'm not terribly familiar with how the internals work but this looks fine.

[sighingnow]

Close as fixed.