我可以为->编写更高阶的类型吗? b-> *? [英] Can I write a higher order type for a -> b -> *?

问题描述

我知道(->) a是类型* -> *的高阶类型，将其应用于类型实参b会得到类型a -> b

I understand that (->) a is a higher order type of kind * -> *, that when applied to a type argument b gives the type a -> b

我可以写一种类型的* -> *应用于c时会给出a -> b -> c吗?

Can I write a type of kind * -> * that when applied to c would give a -> b -> c?

如果没有，为什么不呢?也许使用某些语言扩展名和forall?

If not, why not? Maybe using some language extensions and forall?

这将允许我编写Functor和Applicative(以及其他类)的实例，其中函子结构为"a -> b ->"，如下所示:

This would let me write instances of Functor and Applicative (and other classes) where the functorial structure is "a -> b ->" as in:

(<*>) :: Applicative t => t (c -> d) -> t c -> t d

(<*>) :: (a -> b -> c -> d) -> (a -> b -> c) -> a -> b -> d

作为二进制(咖喱)函数的组合器，这将很有用.

This would be useful as a combinator for binary (curried) functions.

NB.也许与有关类型(*-> *)的类的Functors和Applicatives相关. -> * 但我不确定，因为它越过我的头:-)

NB. Maybe this is related Functors and Applicatives for types of kind (* -> *) -> * but I'm not sure, because it went over my head :-)

推荐答案

不，您不能.您可以想象有许多语言功能可以支持此功能.例如类型级的lambda会很自然:

No, you can't. You could imagine many language features to support this; e.g. type-level lambdas would be a natural one:

instance Functor (\c. a -> b -> c) where ...

不幸的是，类型级别的lambda意味着我们必须在类型推断期间从一阶统一转变为高阶统一，这非常困难. (我想说的是犹豫不决，但我不确定.)

Unfortunately, type-level lambdas would mean we have to move from first-order unification to higher-order unification during type inference, which is notably difficult. (I want to say undecidable off the top of my head, but I'm not certain of this.)

如果通过newtype包装器插入显式的类型检查提示，则可以到达一半.标准的是 Compose :

You can get halfway there if you insert an explicit type-checking hint by way of a newtype wrapper. The standard one is Compose:

a -> b -> c ~= Compose (a ->) (b ->) c
\c. a -> b -> c ~= Compose (a ->) (b ->)

实际上，Compose (a ->) (b ->)的Functor和Applicative实例恰好是您希望\c. a -> b -> c的实例，但是在创建和使用此类型的值时会花费一些语法噪音.

And, indeed, the Functor and Applicative instances for Compose (a ->) (b ->) are exactly the ones you would expect for \c. a -> b -> c, at the cost of a bit of syntactic noise when creating and consuming values of this type.

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