idiomatic way to express (optional) delimiters?

[femtomc]

Hi again!

I want to write a parser where the delimiters are optional. I tried the following:

        let lapp__ = expr
            .clone()
            .then_ignore(just(Token::Space))
            .then_ignore(just(Token::RParen))
            .then(expr.clone());

        let lapp_ = choice((
            just(Token::LParen)
                .ignore_then(lapp__)
                .then_ignore(just(Token::RParen)),
            lapp__,
        ))
        .map(|(e1, e2)| LApp(e1, e2));

the issue here is that lapp_ is moved in choice -- which makes me think that there is a better idiom than what I'm trying to do here.

[zesterer]

My usual approach looks something like the following:

let inner = ... ;

let outer = inner.clone()
    .delimited_by(just(Token::LParen), just(Token::RParen))
    .or(inner);

Cloning a parser is usually relatively cheap and, most importantly, happens during parser construction and not during parsing itself so you need not worry about the performance implications.