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!

  • @[email protected]
    1•11 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>

      • @[email protected]
        2•1 year ago

        A generic impl is impossible.

        Imagine you want to turn a Into<String> to Some(val.into()) and Option<Into<String>> to val.map(Into::into).

        Now, what if there is a type T where impl From <Option<T>> for String is implemented?
        Then we would have a conflict.

        If you only need this for &str and String, then you can add a wrapper type OptionStringWrapper(Option<String>) and implement From<T> for OptionStringWrapper for all concrete type cases you want to support, and go from there.