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Allow argument in macro to be Option or T (sopuli.xyz)

submitted 5 months ago* (last edited 5 months ago) by silva@sopuli.xyz to c/rust@programming.dev

3 comments fedilink hide all child comments

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. 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!

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[–] crispy_kilt@feddit.de 1 points 5 months ago

Why not add a new() function that does the same? Macro seems unneccessary

[–] tuna@discuss.tchncs.de 1 points 4 months ago

You might be okay with this:

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

Playground

However, sometimes I don't want to pass in the file path directly but through a variable that is Option<String>.

Essentially I took this to mean str literals will be auto wrapped in Some, but anything else is expected to be Option<String>

[–] Ogeon@programming.dev 1 points 5 months ago

Option<T> has a From<T> implementation that lets you write Option::from($file_path).map(|path| path.to_string()) to accept both cases in the same expression.

https://doc.rust-lang.org/std/option/enum.Option.html#impl-From%3CT%3E-for-Option%3CT%3E