Treating idents as string in Rust macros

Question

I have a configuration struct with some top level properties which I would like to put in sections. I order to provide deprecation warnings I made the following macro

macro_rules! deprecated {
($config:expr, $old:ident, $section:ident, $new:ident) => {
    if $config.$old.is_some() {
        println!(
            "$old configuration option is deprecated. Rename it to $new and put it under the section [$section]",
        );

        &$config.$old
    } else {
        if let Some(section) = &$config.$section {
            &section.$new
        } else {
            &None
        }
    }
};

}

This doesn't seem to work as expected as the macro parameters aren't substituted inside a string. How can I change the macro to achieve the desired effect?

Shouldn't it be `println!("{} configuration option is deprecated. Rename it to {} and put it under the section [{}]", $old, $new, $section);` — Alexey S. Larionov, Jun 06 '20 at 19:17
@AlexLarionov that won't work as well. I'm not looking to print the value of the struct section in `$config.$old`, I'm just trying to print `$old` as a literal. — reish, Jun 06 '20 at 19:27

score 8 · Accepted Answer · answered Jun 06 '20 at 19:46

The stringify! macro can turn an element into a string literal, and the concat! macro can concatenate several literals into a single string literal:

println!(
    concat!(
        stringify!($old),
        " configuration option is deprecated. Rename it to ",
        stringify!($new),
        " and put it under the section [",
        stringify!($section),
        "]",
    )
);

Permalink to the playground

Treating idents as string in Rust macros

1 Answers1