#equality #type-level #propositional

refl

Provides a refl encoding which you can use to provide a proof witness that one type is equivalent (identical) to another type. You can use this to encode a subset of what GADTs allow you to in Haskell

6 releases

0.2.1 Feb 24, 2020
0.2.0 Mar 23, 2019
0.1.3 Mar 23, 2019
0.1.2 Feb 4, 2018
0.1.1 Jan 28, 2018

#572 in Rust patterns

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Used in 12 crates (2 directly)

MIT/Apache

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refl

Build Status

Provides a refl encoding which you can use to provide a proof witness that one type is equivalent (identical) to another type. You can use this to encode a subset of what GADTs allow you to in Haskell.

This is encoded as:

use std::mem;
use std::marker::PhantomData;

pub struct Id<S: ?Sized, T: ?Sized>(PhantomData<(*mut S, *mut T)>);

impl<T: ?Sized> Id<T, T> { pub const REFL: Self = Id(PhantomData); }

pub fn refl<T: ?Sized>() -> Id<T, T> { Id::REFL }

impl<S: ?Sized, T: ?Sized> Id<S, T> {
    /// Casts a value of type `S` to `T`.
    ///
    /// This is safe because the `Id` type is always guaranteed to
    /// only be inhabited by `Id<T, T>` types by construction.
    pub fn cast(self, value: S) -> T where S: Sized, T: Sized {
        unsafe {
            // Transmute the value;
            // This is safe since we know by construction that
            // S == T (including lifetime invariance) always holds.
            let cast_value = mem::transmute_copy(&value);
    
            // Forget the value;
            // otherwise the destructor of S would be run.
            mem::forget(value);
    
            cast_value
        }
    }

    // ..
}

In Haskell, the Id<A, B> type corresponds to:

data a :~: b where
    Refl :: a :~: a

However, note that you must do the casting manually with refl.cast(val). Rust will not know that S == T by just pattern matching on Id<S, T> (which you can't do).

Limitations

Please note that Rust has no concept of higher kinded types (HKTs) and so we can not provide the general transformation F<S> -> F<T> given that S == T. With the introduction of generic associated types (GATs), it may be possible to introduce more transformations.

Example - A GADT-encoded expression type

extern crate refl;
use refl::*;

trait Ty { type Repr: Copy + ::std::fmt::Debug; }

#[derive(Debug)]
struct Int;
impl Ty for Int { type Repr = usize; }

#[derive(Debug)]
struct Bool;
impl Ty for Bool { type Repr = bool; }

#[derive(Debug)]
enum Expr<T: Ty> {
    Lit(T::Repr),
    Add(Id<usize, T::Repr>, Box<Expr<Int>>, Box<Expr<Int>>),
    If(Box<Expr<Bool>>, Box<Expr<T>>, Box<Expr<T>>),
}

fn eval<T: Ty>(expr: &Expr<T>) -> T::Repr {
    match *expr {
        Expr::Lit(ref val) =>
            *val,
        Expr::Add(ref refl, ref l, ref r) =>
            refl.cast(eval(&*l) + eval(&*r)),
        Expr::If(ref c, ref i, ref e) =>
            if eval(&*c) { eval(&*i) } else { eval(&*e) },
    }
}

fn main() {
    let expr: Expr<Int> =
        Expr::If(
            Box::new(Expr::Lit(false)),
            Box::new(Expr::Lit(1)),
            Box::new(Expr::Add(
                refl(),
                Box::new(Expr::Lit(2)),
                Box::new(Expr::Lit(3)),
            ))
        );

    assert_eq!(eval(&expr), 5);
}

Contribution

Unless you explicitly state otherwise, any contribution intentionally submitted for inclusion in the work by you, as defined in the Apache-2.0 license, shall be dual licensed as above, without any additional terms or conditions.

No runtime deps