类型:(数字a，数字a）与（数字a）？ [英] Types: (Num a, Ord a) versus (Int a)?

问题描述

在一个文件中，我试图定义一个这样的函数：

  myReplicate ::（Int a）= > a  - > b  - > [b] 
 myReplicate n x 
 | n< = 0 = [] 
 |否则= x：myReplicate（n-1）x 
  

当我尝试将文件加载到ghci时购买，我得到这个错误：

  ghci>：l 1.hs 
 [1 of 1] Compiling Main（1。 hs，解释）
 
 1.hs：38：17：
'Int'应用于太多类型参数
在'myReplicate'的类型签名中：
 myReplicate :: Int a => a  - > b  - > [b] 
失败，模块加载：无。 
  

ghci告诉我myReplicate的类型应该是：

  ghci>：t myReplicate 
 myReplicate ::（Num a，Ord a）=> a  - > a1  - >如果我将类型声明更改为ghci推荐的内容：
 
 
 
 
 
 
 b 
 
  myReplicate ::（Num a，Ord a）=> a  - > b  - > [b] 
  
 ...然后函数编译和'工作'。然而，它是这样工作的：
  ghci> myReplicate 3.2 1 
 [1,1,1,1 ] 
  
为什么我不能声明myReplicate只将Int作为第一个参数Int下降（？）来自Ord类的事实）？我想我可以把我的第一个后卫改成x< 1，这样myReplicate 3.2 1会生成[1,1,1，]，但为什么我必须打扰浮动？
解决方案
 Int是一个类型，而不是一个类型类型。你想要的是
 
 
  myReplicate :: Int  - > b  - > [b] 
  
或者可能更精确的是
 
 
  myReplicate :: Int  - > Int  - > [int] 
  
这是一个关于类型和类型类的好教程：http://learnyouahaskell.com/types-and-typeclasses  
 
In a file, I'm trying to define a function like this:
myReplicate :: (Int a) => a -> b -> [b]  
myReplicate n x  
    | n <= 0    = []  
    | otherwise = x : myReplicate (n-1) x  
Buy when I try loading the file into ghci, I get this error:
ghci>:l 1.hs 
[1 of 1] Compiling Main             ( 1.hs, interpreted )

1.hs:38:17:
    `Int' is applied to too many type arguments
    In the type signature for `myReplicate':
      myReplicate :: Int a => a -> b -> [b]
Failed, modules loaded: none.
ghci tells me that the type of myReplicate should be:
ghci>:t myReplicate 
myReplicate :: (Num a, Ord a) => a -> a1 -> [a1]
If I change the type declaration to what ghci recommends:
myReplicate :: (Num a, Ord a) => a -> b -> [b]
...then the function compiles and 'works'.  However, it 'works' like this:
ghci>myReplicate 3.2 1
[1,1,1,1]
Why can't I declare that myReplicate only takes an Int as the first argument (also in light of the fact that Int descends(?) from the Ord class)?  I guess I could change my first guard to be x < 1, so that myReplicate 3.2 1 would produce [1, 1, 1,], but why do I have to bother with floats?
解决方案
Int is a type, not a typeclass. What you want is
myReplicate :: Int -> b -> [b]
or probably more precisely 
myReplicate :: Int -> Int -> [Int]
Here is a good tutorial on types and typeclasses: http://learnyouahaskell.com/types-and-typeclasses

                        
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