用于Haskell中类型表达式的Lambda？ [英] Lambda for type expressions in Haskell?

问题描述

Haskell或特定的编译器是否有类型级的lambda表达式（如果这甚至是一个术语）？

详细说，假设我有一个参数化类型 Foo ab 并且希望 Foo _ b 是Functor的一个实例。是否有任何机制可以让我做类似于

pre $实例Functor（\ a - > Foo ab）其中
...

？

>解决方案

从TypeCompose：

  newtype Flip（〜>）ba = Flip {unFlip :: a〜> b} 
  

http://hackage.haskell.org/packages/archive/TypeCompose/0.6.3/doc/html/ Control-Compose.html＃t：Flip

另外，如果在两个参数中有一个Functor，你可以使它成为双重函数：

http://hackage.haskell.org/packages/archive/category-extras/0.44.4/doc/html/Control-Bifunctor.html

（或者在以后的类别中，extras是更通用的版本： http://hackage.haskell.org/packages/archive/category-extras/0.53.5/doc/html/Control-Functor.html#t： Bifunctor ）

Does Haskell, or a specific compiler, have anything like type-level lambdas (if that's even a term)?

To elaborate, say I have a parametrized type Foo a b and want Foo _ b to be an instance of, say, Functor. Is there any mechanism that would let me do something akin to

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

?

解决方案

From TypeCompose:

newtype Flip (~>) b a = Flip { unFlip :: a ~> b }

http://hackage.haskell.org/packages/archive/TypeCompose/0.6.3/doc/html/Control-Compose.html#t:Flip

Also, if something is a Functor in two arguments, you can make it a bifunctor:

http://hackage.haskell.org/packages/archive/category-extras/0.44.4/doc/html/Control-Bifunctor.html

(or, in a later category-extras, a more general version: http://hackage.haskell.org/packages/archive/category-extras/0.53.5/doc/html/Control-Functor.html#t:Bifunctor)

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