如果我传递不同类型的"Nothing"值,那么`fun ::(SomeTypeClass a)=>结果可能会改变 - >结果`吗？

Question

几乎是标题,但让我们详细说明:

假设我有一些类型的类SomeTypeClass和不同类型A,B,C它们都有一个实例吧.

让我们进一步假设我有一个功能

fun :: (SomeTypeClass a) => Maybe a -> Result

对于某些类型常数Result.

可以表达

1. fun (Nothing :: Maybe A)
2. fun (Nothing :: Maybe B)
3. fun (Nothing :: Maybe C)

某种程度上产生不同的结果

(可能通过一些编译器扩展来提供不同的实现,以fun取决于参数类型或其他东西.)

Answer 1

当然,有Result= String和SomeTypeClass a= (Monoid a, Show a):

import Data.Foldable
doSomething :: (Show a, Monoid a) => Maybe a -> String
doSomething = show . fold

doSomething (Nothing :: Maybe String) == "\"\""
doSomething (Nothing :: Maybe [Int] == "[]"
doSomething (Nothing :: Maybe (Product Int)) == "Product {getProduct=1}"
doSomething (Nothing :: Maybe (Sum Int)) == "Sum {getSum=0}"

当然不是唯一的例子,但可能是最简单的(也就是说,至少类似于你在实际代码中可能会遇到的东西).