Tukukkk

I have a enum where I want to generalize an function pointer. As soon as I add a reference inside the function pointer definition. it fails to compile because it cannot print it with Debug:

fn div1(t: i64, b: i64) -> i64 {

    t / b

}



fn div2(t: i64, b: &i64) -> i64 {

    t / b

}



#[derive(Debug)]

enum Enum {

    FnTest1(fn(i64, i64) -> i64),

    FnTest2(fn(i64, &i64) -> i64),

}



fn main() {

    println!("{:?}", Enum::FnTest1(div1));

    println!("{:?}", Enum::FnTest2(div2));

}

The error I get is this

error[E0277]: `for<'r> fn(i64, &'r i64) -> i64` doesn't implement `std::fmt::Debug`

  --> src/main.rs:12:13

   |

12 |     FnTest2(fn(i64, &i64) -> i64),

   |             ^^^^^^^^^^^^^^^^^^^^ `for<'r> fn(i64, &'r i64) -> i64` cannot be formatted using `{:?}` because it doesn't implement `std::fmt::Debug`

   |

   = help: the trait `std::fmt::Debug` is not implemented for `for<'r> fn(i64, &'r i64) -> i64`

   = note: required because of the requirements on the impl of `std::fmt::Debug` for `&for<'r> fn(i64, &'r i64) -> i64`

   = note: required for the cast to the object type `dyn std::fmt::Debug`

It only shows an error for FnTest2 which has a reference argument while FnTest1 works fine.

Is this a bug in Rust or is there a solution or an alternative method to this issue?

I am running Rust nightly (rustup says: rustc 1.30.0-nightly (ae7fe84e8 2018-09-26)).

Is this a bug in Rust

No, but it is a limitation:

• Cannot derive(Debug) for a struct with a function with a reference parameter (#45048)


• #[derive] Debug, PartialEq, Hash, etc. for any function pointers, regardless of type signature (#54508)

is there a solution or an alternative method

Yes, you must implement Debug for the type Enum yourself.