为什么我会在函数中获得“非穷举模式”？ [英] Why do I get "non-exhaustive patterns in function"?

问题描述

在以下代码中，Haskell抱怨

函数prime中的非穷举模式
 
 
 
  prime :: Int  - > [int] 
 prime x = prime'[​​2..x]其中
 prime'（p：ps）= p：prime'[​​x | x< -ps，mod x p> 0&& prime''x [2..div x 2]] 
 prime''_ [] = True 
 prime''n（x：xs）
 | mod n x == 0 = False 
 |否则= prime''n xs 
 
 prime'[​​] = [] 
  
我找不到我的错误。有人可以解释为什么会发生这种情况，它是什么意思？
 
解决方案
缩进。最后一行定义另一个函数，称为 prime'。因此， prime \prime'（ prime 的where子句中的定义）没有匹配的模式为空列表。
 
 
 
另外，您的缩进遍布整个地方。你仍然混合标签和空格？  
In the following code Haskell complains about  
Non-exhaustive patterns in function prime'


prime :: Int -> [Int]
prime x = prime' [2..x] where
  prime' (p:ps)= p : prime' [x | x <- ps, mod x p > 0 && prime'' x [2..div x 2]]     
  prime'' _ [] = True
  prime'' n (x:xs)
    | mod n x == 0 = False
    |    otherwise = prime'' n xs

prime' []=[]
I can't find my mistake. Could someone explain why this happens, and what it means?
解决方案
Indentation. The last line defines another function called prime'. Therefore, prime\prime' (the definition in the where clause of prime) doesn't have a matching pattern for the empty list.

Also, you're indentation is all over the place. Do you still mix tabs and spaces?

                        
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