[英]Haskell: How is Num held?
In the following example: 在以下示例中:
> let sum::Num a=> a->a->a; sum a b = a +b
> let partialSum = sum 1
> partialSum 2.0
3.0
In step let partialSum = sum 1
it would appear that a is interpreted as Integer but I suppose Haskell delays this until it can figure out the type (and consequently which typeclass instance to use) until the whole expression is built. 在步骤中,
let partialSum = sum 1
,似乎a被解释为Integer,但我认为Haskell将其延迟,直到它可以计算出类型(以及因此使用哪个类型类实例),直到构建整个表达式为止。 We can interpret this because the ultimate result is Fractional (3.0) 我们可以解释这个,因为最终的结果是分数(3.0)
But at this point let partialSum = sum 1
GHCI has to hold 1
in memory, I was wondering what does it hold it as? 但是在这一点上
let partialSum = sum 1
GHCI必须在内存中保持1
,我想知道它是什么呢?
The answer can be broken down in two components: 答案可以分为两个部分:
Typeclass constraints get elaborated as records and they're automatically passed around whenever needed. 类型类约束被详细描述为记录,并且它们会在需要时自动传递。
A numeric literal n
is translated to fromInteger n
数字文字
n
被转换为fromInteger n
So 所以
sum :: Num a => a -> a -> a
sum a b = a + b
would become: 会成为:
sum' :: RecordNum a -> a -> a -> a
sum' dict a b = (plus dict) a b
where 哪里
data RecordNum a = RecordNum
{ plus :: a -> a -> a
; mult :: a -> a -> a
(...)
; fromInteger :: Integer -> a
}
and then 接着
partialSum :: Num a => a -> a
partialSum = sum 1
becomes 变
partialSum' :: RecordNum a -> a -> a
partialSum' dict = sum' dict (fromInteger dict 1)
But at this point
let partialSum = sum 1
GHCi has to hold 1 in memory但是在这一点上,
let partialSum = sum 1
GHCi必须在内存中保持1
Sure, but remember that Haskell number literals are overloaded. 当然,但请记住,Haskell数字文字已经过载。 In this case,
1
gets stored as fromInteger (1 :: Integer) :: Num a => a
. 在这种情况下,
1
被存储为fromInteger (1 :: Integer) :: Num a => a
。 partialSum
holds onto this until it knows what a
should be. partialSum
持有到这个直到它知道什么a
应该的。 GHCi holds onto only 1
the Integer
. GHCi仅保留
1
Integer
。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.