将 monad 限制为类型类 [英] Restricting a monad to a type class

问题描述

在 Haskell 中，有没有办法限制 monad M a 以便 a 满足类型类约束?

我正在翻译 概率建模示例来自F# 到 Haskell.但是，在 Haskell 中，我省略了 support 因为它会将 data Distribution a 更改为 data (Ord a) =>分发一个.通过此更改，我收到以下错误:

...probabilisticModeling.hs:42:13:无法从上下文中推导出 (Ord a) ()源于使用总是"在 ...probabilisticModeling.hs:42:13-18可能的修复:将 (Ord a) 添加到 `return' 的类型签名的上下文中在表达中:总是在`return'的定义中:return = always在`Monad Distribution'的实例声明中

确实，always/return 的类型是:(Ord a) =>->分发一个.有没有办法让我拥有一个 monad Distribution，但是在这个 monad 上强制约束 (Ord a) ?我试过了:

instance Monad Distribution where(>>=) = 绑定返回 :: (Ord a) =>->分布 a = 总是

但我收到错误:

...probabilisticModeling2.hs:48:4:实例声明中不允许使用模式绑定(简单变量除外)返回 :: (Ord a) =>->分布 a = 总是失败，加载的模块:无.

那么有没有办法让一个monad M a，但是用Ord a这样的约束来限制a?

谢谢.

解决方案

我的理解是你根本不能，因为 monad 旨在泛化所有类型，而不是一些受限的类型子集，例如 (Ord a).

您可以简单地限制使用该 monadic 类型的函数，而不是限制 monadic 类型 M a，例如，

foo :: Ord a => Int -> M a

实际上，最好保持类型尽可能通用，使用类型类仅用于限制功能.

等

In Haskell, is there a way to restrict a monad M a so that a satisfy a type class constraint?

I am translating the probabilistic modeling example from F# to Haskell. However, in Haskell, I omitted support because it would change data Distribution a to data (Ord a) => Distribution a. With this change, I get the following error:

...probabilisticModeling.hs:42:13:
    Could not deduce (Ord a) from the context ()
      arising from a use of `always'
                   at ...probabilisticModeling.hs:42:13-18
    Possible fix:
      add (Ord a) to the context of the type signature for `return'
    In the expression: always
    In the definition of `return': return = always
    In the instance declaration for `Monad Distribution'

Indeed, the type of always/return is: (Ord a) => a -> Distribution a. Is there a way I can have a monad Distribution, but force the constraint (Ord a) on this monad? I tried:

instance Monad Distribution where
    (>>=) = bind
    return :: (Ord a) => a -> Distribution a = always

But I get the error:

...probabilisticModeling2.hs:48:4:
    Pattern bindings (except simple variables) not allowed in instance declarations
      return :: (Ord a) => a -> Distribution a = always
Failed, modules loaded: none.

So it there a way to have a monad M a, but restrict the a with a constraint such as Ord a?

Thanks.

解决方案

My understanding of this is that you simply cannot, because a monad is meant to be generalized over all types, not some restricted subset of types such as (Ord a).

Instead of restricting the monadic type M a, you can simply restrict functions which use that monadic type, e.g.,

foo :: Ord a => Int -> M a

In fact, it is preferable to keep types as general as possible and use type classes only to restrict functions.

etc.

这篇关于将 monad 限制为类型类的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！