[英]Forgetting Cofree annotations using a catamorphism
I have an AST that I'm annotating using Cofree
:我有一个 AST,我正在使用
Cofree
注释:
data ExprF a
= Const Int
| Add a
a
| Mul a
a
deriving (Show, Eq, Functor)
I use type Expr = Fix ExprF
to represent untagged ASTs, and type AnnExpr a = Cofree ExprF a
to represent tagged ones.我使用
type Expr = Fix ExprF
来表示未标记的 AST,并使用type AnnExpr a = Cofree ExprF a
来表示标记的。 I've figured out a function to transform tagged ASTs into untagged ones by throwing away all the annotations:我已经找到了一个函数,通过丢弃所有注释将标记的 AST 转换为未标记的 AST:
forget :: Functor f => Cofree f a -> Fix f
forget = Fix . fmap uncofree . unwrap
This looks like it might be some sort of catamorphism (I'm using the definition from Kmett's recursion-schemes package).这看起来可能是某种变形(我使用的是 Kmett 的recursion-schemes包中的定义)。
cata :: (Base t a -> a) -> t -> a
cata f = c where c = f . fmap c . project
I'd think the above rewritten using a catamorphism would look something like this, but I can't figure out what to put for alg
to make it typecheck.我认为上面使用 catamorphism 重写的内容看起来像这样,但我无法弄清楚为
alg
放置什么来使其类型检查。
forget :: Functor f => Cofree f a -> Fix f
forget = cata alg where
alg = ???
Any help figuring out if this really is a cata/anamorphism, and some intuition for why it is/isn't would be greatly appreciated.任何帮助确定这是否真的是 cata/anamorphism,以及为什么它是/不是的一些直觉将不胜感激。
forget :: Functor f => Cofree f a -> Fix f
forget = cata (\(_ :< z) -> Fix z)
-- (Control.Comonad.Trans.Cofree.:<)
-- not to be confused with
-- (Control.Comonad.Cofree.:<)
Looking only at the types, we can show that there is really only one way to implement forget
.只看类型,我们可以证明实际上只有一种方法可以实现
forget
。 Let's start with the type of cata
:让我们从
cata
的类型开始:
cata :: Recursive t => (Base t b -> b) -> t -> b
Here t ~ Cofree fa
and the type instance of Base
for Cofree
gives:这里
t ~ Cofree fa
和Base
for Cofree
的类型实例给出:
type instance Base (Cofree f a) = CofreeF f a
data CoFreeF f a b = a :< f b
-- N.B.: CoFree also defines a (:<) constructor so you have to be
-- careful with imports.
ie, a fancy pair type.即,花哨的配对类型。 Let's replace it with an actual pair type to make things clearer:
让我们用实际的配对类型替换它,以使事情更清楚:
cata :: Functor f => ((a, f b) -> b) -> Cofree f a -> b
Now we're really specializing cata
with a more concrete b
, namely Fix f
:现在我们真的专门用一个更具体的
b
专门化cata
,即Fix f
:
-- expected type of `cata` in `forget`
cata :: Functor f => ((a, f (Fix f)) -> Fix f) -> Cofree f a -> Fix f
forget
is parametric in a
and f
, so the function we give cata
can do nothing with the a
in the pair, and the only sensible way to implement the remaining f (Fix f) -> Fix f
is the Fix
wrapper. forget
在a
和f
是参数化的,所以我们给cata
的函数不能对a
对中的a
做任何事情,实现剩余f (Fix f) -> Fix f
的唯一明智方法是Fix
包装器。
Operationally, Fix
is the identity, so (\\(_ :< z) -> Fix z)
is really (\\(_ :< z) -> z)
which corresponds to the intuition of removing the annotation, ie, the first component of the pair (_ :< z)
.操作上,
Fix
是恒等式,所以(\\(_ :< z) -> Fix z)
真的是(\\(_ :< z) -> z)
这对应于去除注解的直觉,即第一个组件对(_ :< z)
。
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