「SF-QC」2 TypeClasses - 黄玄的博客

Considerring printing different types with this common idiom:

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showBool : bool → string
showNat : nat → string
showList : {A : Type} (A → string) → (list A) → string
showPair : {A B : Type} (A → string) → (B → string) → A * B → string

Definition showListOfPairsOfNats := showList (showPair showNat showNat)   (* LOL *)

The designers of Haskell addressed this clunkiness through typeclasses, a mechanism by which the typechecker is instructed to automatically construct “type-driven” functions [Wadler and Blott 1989].

Coq followed Haskell’s lead as well, but

because Coq’s type system is so much richer than that of Haskell, and because typeclasses in Coq are used to automatically construct not only programs but also proofs, Coq’s presentation of typeclasses is quite a bit less “transparent”

Basics

Classes and Instances

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Class Show A : Type := {
  show : A → string
}.

Instance showBool : Show bool := {
  show := fun b:bool ? if b then "true" else "false"
}.

Comparing with Haskell:

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class Show a where
  show :: a -> string

--  you cannot override a `instance` so in reality you need a `newtype` wrapper to do this
instance Show Bool where
  show b = if b then "True" else "Fasle"

The show function is sometimes said to be overloaded, since it can be applied to arguments of many types, with potentially radically different behavior depending on the type of its argument.

Next, we can define functions that use the overloaded function show like this:

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Definition showOne {A : Type} `{Show A} (a : A) : string :=
  "The value is " ++ show a.

Compute (showOne true).
Compute (showOne 42).

Definition showTwo {A B : Type}
           `{Show A} `{Show B} (a : A) (b : B) : string :=
  "First is " ++ show a ++ " and second is " ++ show b.

Compute (showTwo true 42).
Compute (showTwo Red Green).

The parameter `{Show A} is a class constraint, which states that the function showOne is expected to be applied only to types A that belong to the Show class.

Concretely, this constraint should be thought of as an extra parameter to showOne supplying evidence that A is an instance of Show — i.e., it is essentially just a show function for A, which is implicitly invoked by the expression show a.

读时猜测（后来发现接下来有更正确的解释）：show 在 name resolution 到 class Show 时就可以根据其参数的 type（比如 T）infer 出「我们需要一个 Show T 的实现（instance，其实就是个 table）」，在 Haskell/Rust 中这个 table 会在 lower 到 IR 时才 made explicit，而 Coq 这里的语法就已经强调了这里需要 implicitly-and-inferred {} 一个 table，这个 table 的名字其实不重要，只要其 type 是被 A parametrized 的 Show 就好了，类似 ML 的 functor 或者 Java 的 generic interface。

This is Ad-hoc polymorphism.

Missing Constraint

What if we forget the class constrints:

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Error:
Unable to satisfy the following constraints:
In environment:
A : Type
a : A

?Show : "Show A"

Class Eq

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Class Eq A :=
  {
    eqb: A → A → bool;
  }.

Notation "x =? y" := (eqb x y) (at level 70).

Instance eqBool : Eq bool :=
  {
    eqb := fun (b c : bool) ?
       match b, c with
         | true, true ? true
         | true, false ? false
         | false, true ? false
         | false, false ? true
       end
  }.

Instance eqNat : Eq nat :=
  {
    eqb := Nat.eqb
  }.

Why should we need to define a typeclass for boolean equality when Coq’s propositional equality (x = y) is completely generic?
while it makes sense to claim that two values x and y are equal no matter what their type is, it is not possible to write a decidable equality checker for arbitrary types. In particular, equality at types like nat → nat is undecidable.

x = y 返回一个需要去证的 Prop (relational) 而非 executable Fixpoint (functional)
因为 function 的 equality 有时候会 undeciable，所以才需要加 Functional Extensionality Axiom（见 LF-06）

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Instance eqBoolArrowBool: Eq (bool -> bool) :=
  {
    eqb := fun (f1 f2 : bool -> bool) =>
      (f1 true) =? (f2 true) && (f1 false) =? (f2 false)
  }.

Compute (id =? id).      (* ==> true *)
Compute (negb =? negb).  (* ==> true *)
Compute (id =? negb).    (* ==> false *)

这里这个 eqb 的定义也是基于 extensionality 的定义，如果考虑到 effects（divergence、IO）是很容易 break 的（类似 parametricity）

Parameterized Instances: New Typeclasses from Old

Structural recursion

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Instance showPair {A B : Type} `{Show A} `{Show B} : Show (A * B) :=
  {
    show p :=
      let (a,b) := p in
        "(" ++ show a ++ "," ++ show b ++ ")"
  }.
Compute (show (true,42)).

Structural equality

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Instance eqPair {A B : Type} `{Eq A} `{Eq B} : Eq (A * B) :=
  {
    eqb p1 p2 :=
      let (p1a,p1b) := p1 in
      let (p2a,p2b) := p2 in
      andb (p1a =? p2a) (p1b =? p2b)
  }.

Slightly more complicated example: typical list:

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(* the book didn't use any from ListNotation *)
Fixpoint showListAux {A : Type} (s : A → string) (l : list A) : string :=
  match l with
    | nil ? ""
    | cons h nil ? s h
    | cons h t ? append (append (s h) ", ") (showListAux s t)
  end.
Instance showList {A : Type} `{Show A} : Show (list A) :=
  {
    show l := append "[" (append (showListAux show l) "]")
  }.

(* I used them though *)
Fixpoint eqListAux {A : Type} `{Eq A} (l1 l2 : list A) : bool :=
  match l1, l2 with
    | nil, nil => true
    | (h1::t1), (h2::t2) => (h1 =? h2) && (eqListAux t1 t2)
    | _, _ => false
  end.

Instance eqList {A : Type} `{Eq A} : Eq (list A) :=
  {
    eqb l1 l2 := eqListAux l1 l2
  }.

Class Hierarchies

we might want a typeclass Ord for “ordered types” that support both equality and a less-or-equal comparison operator.

A bad way would be declare a new class with two func eq and le.

It’s better to establish dependencies between typeclasses, similar with OOP class inheritence and subtyping (but better!), this gave good code reuses.

We often want to organize typeclasses into hierarchies.

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Class Ord A `{Eq A} : Type :=
  {
    le : A → A → bool
  }.
Check Ord. (* ==>
Ord
     : forall A : Type, Eq A -> Type
*)

class Eq is a “super(type)class” of Ord (not to be confused with OOP superclass)

This is Sub-typeclassing.

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Fixpoint listOrdAux {A : Type} `{Ord A} (l1 l2 : list A) : bool :=
  match l1, l2 with
  | [], _ => true
  | _, [] => false
  | h1::t1, h2::t2 => if (h1 =? h2)
                     then (listOrdAux t1 t2)
                     else (le h1 h2)
  end.

Instance listOrd {A : Type} `{Ord A} : Ord (list A) :=
  {
    le l1 l2 := listOrdAux l1 l2
  }.

(* truthy *)
Compute (le [1] [2]).
Compute (le [1;2] [2;2]).
Compute (le [1;2;3] [2]).

(* falsy *)
Compute (le [1;2;3] [1]).
Compute (le [2] [1;2;3]).

How It works

Implicit Generalization

所以 `{...} 这个 “backtick” notation is called implicit generalization，比 implicit {} 多做了一件自动 generalize 泛化 free varabile 的事情。

that was added to Coq to support typeclasses but that can also be used to good effect elsewhere.

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Definition showOne1 `{Show A} (a : A) : string :=
  "The value is " ++ show a.

Print showOne1.
(* ==>
    showOne1 =
      fun (A : Type) (H : Show A) (a : A) => "The value is " ++ show a
           : forall A : Type, Show A -> A -> string

    Arguments A, H are implicit and maximally inserted
*)

notice that the occurrence of A inside the `{...} is unbound and automatically insert the binding that we wrote explicitly before.

The “implicit and maximally generalized” annotation on the last line means that the automatically inserted bindings are treated (注：printed) as if they had been written with {...}, rather than (...).

The “implicit” part means that the type argument A and the Show witness H are usually expected to be left implicit
whenever we write showOne1, Coq will automatically insert two unification variables as the first two arguments.

This automatic insertion can be disabled by writing @, so a bare occurrence of showOne1 means the same as @showOne1 _ _

这里的 witness H 即 A implements Show 的 evidence，本质就是个 table or record，可以 written more explicitly:

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Definition showOne2 `{_ : Show A} (a : A) : string :=
  "The value is " ++ show a.

Definition showOne3 `{H : Show A} (a : A) : string :=
  "The value is " ++ show a.

甚至

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Definition showOne4 `{Show} a : string :=
  "The value is " ++ show a.

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showOne =
fun (A : Type) (H : Show A) (a : A) => "The value is " ++ show a
     : forall A : Type, Show A -> A -> string

Set Printing Implicit.

showOne =
fun (A : Type) (H : Show A) (a : A) => "The value is " ++ @show A H a     (* <-- 注意这里 *)
     : forall A : Type, Show A -> A -> string

vs. Haskell

顺便，Haskell 的话，Show 是可以直接 inferred from the use of show 得

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Prelude> showOne a = show a
Prelude> :t showOne
showOne :: Show a => a -> String

但是 Coq 不行，会退化上「上一个定义的 instance Show」，还挺奇怪的（

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Definition showOne5 a : string :=  (* not generalized *)
  "The value is " ++ show a.

Free Superclass Instance

``{Ord A} led Coq to fill in both A and H : Eq A because it's the superclass of Ord` (appears as the second argument).

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Definition max1 `{Ord A} (x y : A) :=
  if le x y then y else x.

Set Printing Implicit.
Print max1.
(* ==>
     max1 =
       fun (A : Type) (H : Eq A) (H0 : @Ord A H) (x y : A) =>
         if @le A H H0 x y then y else x

   : forall (A : Type) (H : Eq A),
       @Ord A H -> A -> A -> A
*)
Check Ord.
(* ==> Ord : forall A : Type, Eq A -> Type *)

Ord type 写详细的话可以是：

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Ord : forall (A : Type),大专栏  「SF-QC」2 TypeClasses - 黄玄的博客an class="w">
				


				
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