Question

如果小于2，则删除列表的前n个元素。错误说“因使用'print'而没有（showa0）的实例......”我从不理解Haskell错误按摩

func [] _  _ = []
func (x:xs) counter n 
       |  x > 2 && counter < n = x :func xs counter limit 
       | otherwise = func xs (counter+1) limit 

main = do 
 print $  func [3,1,4,2,1] 0 2
 -- expectet output is [3,4,1]

Answer 1

这应该删除编译错误：

func [] _  _ = []
func (x:xs) counter n 
       |  x > 2 && counter < n = x :(func xs counter n )
       | otherwise = func xs (counter+1) n

main = do 
 print $  func [3,1,4,2,1] 0 2

您已定义新变量limit而非n。您也可以浏览take和drop等库函数。

如果在Haskell中小于2，则删除列表中的前n个元素

1 个答案: