将monad限制为一个类型类

问题描述

在Haskell中，有没有办法限制monad M a ，这样 a 满足一个类型约束？

我正在翻译，从到。然而，在Haskell中，我省略了 support ，因为它会将数据分配a 更改为 data（ Ord a）=>分配一个。有了这个变化，我得到了以下错误：

  ... probabilisticModeling.hs：42：13：
不是从上下文（）中得出的（$ a 
 $ b）在...中使用`always'
的概率probabilisticModeling.hs：42：13-18 
可能的修正：
在'return'类型签名的上下文中添加（Ord a）
在表达式中：always 
在'return'的定义中：return = always 
在`Monad Distribution'
  

确实，的类型总是 / return 是：（Ord a）=> a - >分配一个。有没有一种方法可以让monad Distribution ，但是在这个monad上强制约束（Ord a）？我尝试过：

pre $ $ $ $ $ $ $实例Monad分布其中
（>> =）=绑定
return： ：（Ord a）=> a - >发行a = always

但是我收到错误：
$ b $

return ::（Ord a ）=> a - >分发a =总是
失败，模块加载：无。

所以有办法让monad M a ，但限制 a 的约束，如 Ord a ？

谢谢。

解决方案

我对此的理解是你根本无法做到，因为monad是为了可以概括为所有类型，而不是某些限制子类型，如（Ord a）。

一元类型 M a ，您可以简单地限制使用该一元类型的函数，例如

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

事实上，最好保持类型尽可能通用，并使用类型类来限制功能。

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.

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