CODEWITHSUNDEEP
rust
Johan Bjäreholt
Johan Bjäreholt

Reputation: 780

Function pointer with a reference argument cannot derive Debug

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)).

Upvotes: 0

Views: 966

Answers (1)

Shepmaster
Shepmaster

Reputation: 431489

Is this a bug in Rust

No, but it is a limitation:

is there a solution or an alternative method

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

Upvotes: 4

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