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!
Option<T>
has a From<T>
implementation that lets you write Option::from($file_path).map(|path| path.to_string())
to accept both cases in the same expression.
https://doc.rust-lang.org/std/option/enum.Option.html#impl-From<T>-for-Option<T>
Why not add a new()
function that does the same? Macro seems unneccessary
You might be okay with this:
macro_rules! span {
($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,
}
};
}
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>