Hey,

Is there any way to create a macro that allows a Some<T> or T as input?

It’s for creating a Span struct that I’m using:

struct Span {
    line: usize,
    column: usize,
    file_path: Option<String>,
}

…and I have the following macro:

macro_rules! span {
    ($line:expr, $column:expr) => {
        Span {
            line: $line,
            column: $column
            file_path: None,
        }
    };

    ($line:expr, $column:expr, $file_path: expr) => {
        Span {
            line: $line,
            column: $column
            file_path: Some($file_path.to_string()),
        }
    };
}

…which allows me to do this:

let foo = span!(1, 1);
let bar = span!(1, 1, "file.txt");

However, sometimes I don’t want to pass in the file path directly but through a variable that is Option<String>. To do this, I always have to match the variable:

let file_path = Some("file.txt");

let foo = match file_path {
    Some(file_path) => span!(1, 1, file_path),
    None => span!(1, 1),
}

Is there a way which allows me to directly use span!(1, 1, file_path) where file_path could be "file.txt", Some("file.txt") or None?

Thanks in advance!

      • v9CYKjLeia10dZpz88iU
        1·
        6 months ago

        Two of your macro rules are not used 😉 (expand to see which ones).

        The macro rules are all used. (Macros are matched from top to bottom by the declared match types. The ident/expressions can’t match until after the more text based Option matching.)

        let _foo = Span { line: 1, column: 1, file_path: None };
let _bar = Span { line: 1, column: 1, file_path: "file.txt".upgrade() };
let _baz = Span { line: 1, column: 1, file_path: Some("file.txt".to_string()) };
let _baz = Span { line: 1, column: 1, file_path: None };
let _baz = Span { line: 1, column: 1, file_path: borrowed.upgrade() };
let _baz = Span { line: 1, column: 1, file_path: owned.upgrade() };

        This doesn’t support Option<&str>. If it did, we would lose literal None support 😉

        I didn’t make Option<&str> an option because the struct is for type Option<String>. It does support Option<String> though.

        impl OptionUpgrade for Option<String> {
    fn upgrade(self) -> Option<String> {
        self
    }
}

        It looks like the following, and uses the last match case.

        let opt: Option<String> = Some("text".into());
let opt = span!(1, 1, opt);

        With macro expansion

        let opt: Option<String> = Some("text".into());
let opt = Span { line: 1, column: 1, file_path: opt.upgrade() };

        There’s not anything stopping it from supporting Option<&str> though. This would be the implementation

        impl OptionUpgrade for Option<&str> {
    fn upgrade(self) -> Option<String> {
        self.map(|v| v.into())
    }
}
        • BB_C
          1·
          6 months ago

          The macro rules are all used.

          Oops. I was looking at it wrong.

          I didn’t make Option<&str> an option because the struct is for type Option<String>.

          Re-read the end of OP’s requirements.

          • v9CYKjLeia10dZpz88iU
            1·
            6 months ago

            I made an edit.

            There’s not anything stopping it from supporting Option<&str> though. This would be the implementation

            impl OptionUpgrade for Option<&str> {
    fn upgrade(self) -> Option<String> {
        self.map(|v| v.into())
    }
}

            It’s also possible to just make it generic over Option types

            impl<A: ToString> OptionUpgrade for Option<A> {
    fn upgrade(self) -> Option<String> {
        self.map(|v| v.to_string())
    }
}