什么是“n”在RankNTypes中 [英] What is "n" in RankNTypes
问题描述
我知道 forall
是如何让我们编写多态函数的。
根据章节,我们通常写的正常函数是Rank 1类型。此功能属于Rank 2类型:
foo ::(a。a - > a) - > (Char,Bool)
foo f =(f'c',f True)
它是这样解释的:
lockquote
一般来说,rank-n类型是一个函数,它至少有一个
rank-( n-1)的参数,但没有更高级别的参数。
实际上排名参数意味着什么?
有人可以给出类似于上述 foo
函数的Rank 3类型的示例。
(等级T)
等级(T→U)=最大(如果等级T = 0则0等级等级T + 1)(等级U)
等级(a)= 0
请注意箭头左侧的增加量。所以:
等级0:Int
等级1:forall a。 a - > Int
等级2:(forall a。a - > Int) - > Int
等级3 :((aa a。a - > Int) - > Int) - > Int
等等。
I understand how forall
enables us to write polymorphic function.
According to this chapter, the normal function which we generally write are Rank 1 types. And this function is of Rank 2 type:
foo :: (forall a. a -> a) -> (Char,Bool)
foo f = (f 'c', f True)
It explains like this:
In general, a rank-n type is a function that has at least one rank-(n-1) argument but no arguments of even higher rank.
What does it actually mean by rank argument ?
Can somebody give an example of Rank 3 type which is similar to the above foo
function.
Rank is defined inductively on the structure of types:
rank (forall a. T) = max 1 (rank T)
rank (T -> U) = max (if rank T = 0 then 0 else rank T + 1) (rank U)
rank (a) = 0
Note how it increases by one on the left-hand side of an arrow. So:
Rank 0: Int
Rank 1: forall a. a -> Int
Rank 2: (forall a. a -> Int) -> Int
Rank 3: ((forall a. a -> Int) -> Int) -> Int
and so on.
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