How to specify "unnamed" uint arg, e.g. git log -10

[vallentin]

I've been looking around the API and the examples, trying to figure out if it's possible to replicate an arg like -<uint>.

Similarly to git log -<uint>, e.g. git log -10 for commit limiting. Just to clarify, this is not a positional arg.

Preferably using the derive API, otherwise using the combinatoric API.

[pacak]

any + ParseAny::anywhere should do the trick. You can find a complete example here

[vallentin]

Thanks for the literally instant response! I don't even know how I overlooked that...

I can get it to work using combinatoric:

#[derive(Bpaf, Clone, Debug)]
#[bpaf(options)]
pub struct Args {
    #[bpaf(external(limit))]
    n: usize,
    foo: bool,
}

fn limit() -> impl Parser<usize> {
    bpaf::any::<isize, _, _>("LIMIT", isize_to_usize)
        .help("LIMIT")
        .anywhere()
}

fn isize_to_usize(n: isize) -> Option<usize> {
    n.abs().try_into().ok()
}

However, I'm confused how to specify anywhere() using derive? (Such that it can be anywhere and isn't a right-most positional)

#[derive(Bpaf, Clone, Debug)]
#[bpaf(options)]
pub struct Args {
    // anywhere?
    #[bpaf(any::<isize>("LIMIT", isize_to_usize))]
    n: usize,
    foo: bool,
}

fn isize_to_usize(n: isize) -> Option<usize> {
    n.abs().try_into().ok()
}

[pacak]

Should be this I think: #[bpaf(any::<isize>("LIMIT", isize_to_usize), anywhere)]

It is easier to think about items in #[bpaf(XXX)] annotations as chained calls combined with ..

For example this: #[bpaf(external(backing), fallback(false), debug_fallback)] is really this: backing().fallback(false).debug_fallback().

[vallentin]

That sadly gives a compile error:

error: Unexpected attribute in field annotation
  --> src\main.rs:14:51
   |
14 |     #[bpaf(any::<isize>("LIMIT", isize_to_usize), anywhere)]
   |                                                   ^^^^^^^^

[pacak]

(came back from a run)

My bad, turns out anywhere is not supported at all at the moment by derive macro. Will make a fix shortly.

[vallentin]

I will use the combinatoric approach for now then. Thanks! :)