Can Rust constant expressions use traits like Default?

Question

This code gives an error:

#[derive(Default)]
struct A {
    b: Option<()>,
    c: Option<()>,
}

const a: A = A {
    b: None,
    ..Default::default()
};

error[E0015]: calls in constants are limited to constant functions, tuple structs and tuple variants
 --> src/lib.rs:9:7
  |
9 |     ..Default::default()
  |       ^^^^^^^^^^^^^^^^^^

In this small example it's not a big problem, but if I have a struct composed by multiple structs that implement the Default trait, not being able to use it becomes at minimum an inconvenience.

While I could write this, it wouldn't have the flexibility that Default provides:

impl A {
    const fn new(b: Option<()>) -> Self {
        A { b, c: None }
    }
}

const a: A = A::new(None);

Is there any way to avoid doing that?

Answer 1

The ..Default::default() syntax is not restricted to Default::default(), so you can write a const fn default-like function and use that inside of a constant:

struct A {
    b: Option<()>,
    c: Option<()>,
}

impl A {
    const fn new() -> A {
        A {
            b: None,
            c: None,
        }
    }
}

impl Default for A {
    fn default() -> A {
        // implementing using new() instead of #[derive]
        // to avoid diverging implementations
        A::new()
    }
}

const a: A = A {
    b: None,
    ..A::new()
};

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Answer 2

No, it is not possible to use traits in a constant context. This is still being discussed in RFC #2632 — Calling methods on generic parameters of const fns.

See also:

• Can I use const with overloading operators in Rust?