pub use / "export" many submodules at once

Question

I need to define many types in one file (->submodule) each and expose them all on the same level of the module. This creates lots of repetitive overhead in the mod.rs:

mod foo;
mod bar;
mod baz;
[...]

pub use self::foo::*;
pub use self::bar::*;
pub use self::baz::*;
[...]

I tried to fix this with a macro, but it does not compile:

macro_rules! expose_submodules {
    ( $( $x:expr ),* ) => {
        $(
            mod $x;
            pub use self::$x::*;
        )*
    };
}

yields

error: expected identifier, found `foo`
 --> src/mod.rs:4:17
  |
4 |             mod $x;
  |                 ^^ expected identifier
  |
 ::: src/mod.rs:10:1
  |
2 | expose_submodules![foo, bar, baz];
  | ----------------------------------------------------------------- in this macro invocation
  |
  = note: this error originates in the macro `expose_submodules` (in Nightly builds, run with -Z macro-backtrace for more info)

(same issue when passing the parameters as strings).

How would I best fix this macro or do the whole task in more idiomatic rust?

Answer 1

To fix the macro you need to use ident fragment instead of expr:

macro_rules! expose_submodules {
    ( $( $x:ident ),* ) => {
        $(
            mod $x;
            pub use self::$x::*;
        )*
    };
}