How to ignore generic argument for `#[derive(Debug)]`? - rust

Question

Here's a minimal example representing the type of problem I'm running into.
use core::fmt::Debug;
pub trait Config {
type A: Debug;
}
#[derive(Debug)]
pub struct Info<T: Config> {
pub field: T::A,
}
pub struct Conf;
impl Config for Conf {
type A = i128;
}
fn main() {
let s = Info::<Conf> {
field: 123
};
dbg!(s);
}
The framework that I'm using (Substrate) uses this concept of a Config trait that aggregates all generic types for a module (pallet).
The problem is that trying to #[derive(Debug)] for a struct that only uses the associated types of the object T implementing Config still requires that T implements Debug itself.
error[E0277]: `Conf` doesn't implement `Debug`
--> src/main.rs:22:5
|
22 | dbg!(s);
| ^^^^^^^ `Conf` cannot be formatted using `{:?}`
|
= help: the trait `Debug` is not implemented for `Conf`
= note: add `#[derive(Debug)]` to `Conf` or manually `impl Debug for Conf`
Moreover, I don't have control of the implementation of the Conf object. Regardless, I'm not trying to print anything about the Conf object itself.
Is there a way to make #[derive(Debug)] for Info ignore T?

Answer 1

Unfortunately, not as of today. You have to impl Debug manually:
impl<T: Config> fmt::Debug for Info<T> {
fn fmt(&self, f: &mut fmt::Formatter<'_>) -> fmt::Result {
f.debug_struct("Info").field("field", &self.field).finish()
}
}
There is an intent to make it possible to create what is called "perfect derive": derive bounds based on what needed and not the generic parameters. See for example this lang team design meeting proposal. But for now there is nothing.