在Haskell中引用多态性的透明度 [英] Referential transparency with polymorphism in Haskell
问题描述
说我有一个函数:
f :: Int - > (Rational,Integer)
fb =((toRational b)+1,(toInteger b)+1)
我想像这样抽象出(+1):
f :: Int - > (Rational,Integer)
fb =(h(toRational b)
,h(toInteger b))
where h =(+1)
这不会很明显,但如果我指定类型签名,它将起作用:
f :: Int - > (Rational,Integer)
f b =(h(toRational b)
,h(toInteger b))
where h :: Num a => a - > a
h =(+1)
假设我现在想通过传递h作为参数:
f :: Num a => Int - > (a - > a) - > (Rational,Integer)
f b g =(h(toRational b)
,h(toInteger b))
其中h :: Num a => a - > a
h = g
我得到一个错误,即内部a与a不同外部的。
有谁知道如何正确编写这个函数吗?
我想将一个多态函数 g
传递给 f
,并以多态方式使用它。
我现在在多个不同的项目中遇到过这种情况,而且我找不到一个好的解决方案。
}
f :: Int - > (全部数量a => a - > a) - > (Rational,Integer)
f b g =(h(toRational b)
,h(toInteger b))
其中h :: Num a => a - > a
h = g
当然可以变成:
f :: Int - > (全部数量a => a - > a) - > (Rational,Integer)
f b g =(g(toRational b)
,g(toInteger b))
Say I have a function:
f :: Int -> (Rational, Integer)
f b = ((toRational b)+1,(toInteger b)+1)
I want to abstract away the (+1) like so:
f :: Int -> (Rational, Integer)
f b = (h (toRational b)
,h (toInteger b))
where h = (+1)
This wont work obviously, but if I specify the type signature it will work:
f :: Int -> (Rational, Integer)
f b = (h (toRational b)
,h (toInteger b))
where h :: Num a => a -> a
h = (+1)
Say I now want to further abstract the function by passing h as a parameter:
f :: Num a => Int -> (a -> a) -> (Rational, Integer)
f b g = (h (toRational b)
,h (toInteger b))
where h :: Num a => a -> a
h = g
I get an error that the inner a is not the same a as the outer one.
Does anyone know how to write this function correctly?
I want to pass a polymorphic function g
to f
and use it polymorphically.
I have encountered this situation multiple times now in very different projects, and I could not find a good solution.
I found the solution: using the forall quantifier like so:
{-# LANGUAGE RankNTypes #-}
f :: Int -> (forall a. Num a=> a -> a) -> (Rational, Integer)
f b g = (h (toRational b)
,h (toInteger b))
where h :: Num a => a -> a
h = g
Which of course can be turned into:
f :: Int -> (forall a. Num a=>a -> a) -> (Rational, Integer)
f b g = (g (toRational b)
,g (toInteger b))
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