Haskell 检查列表列表是否有空列表

Question

I have tried to create a function that determines if a list of lists has an empty list. However so far I don't seem to find any luck. So far I have tried using: hasEmpty (x:xs) = if null x then True else False However this only return true if the first list is empty. I also tried: hasEmpty (x:xs) = if null x || null xs then True else False hasEmpty (x:xs) = if null x || null xs then True else False But it produces the same result. I have also tried using any and elem but I couldn't get it working. I am truly stumped on this. Any help would be much appreciated

Answer 1

The type of any is any:: Foldable t => (a -> Bool) -> ta -> Bool (use :t any ) to get this.

There are two arguments:

1. The first argument is a function which takes a value and returns a boolean, such as null
2. The second argument is a foldable, such as a list

Hence we can simply use any null on a list.

lst = [[1],[],[3]]
lst2 = [[1],[3],[2]]

any null lst -- returns True
any null lst2 -- returns False

Answer 2

Recursion always has a base case. When you're dealing with lists, it's an empty list. If we try running an anyNull function on an empty list, it should return false.

anyNull :: [a] -> Bool
anyNull [] = False

But we also need to match the non-empty list and converge toward the base case. This is done by recursively calling the function on the list's tail. Fortunately pattern matching makes it easy to tell if the first element is an empty list, and then to handle the case where it is not empty.

anyNull :: [a] -> Bool
anyNull [] = False
anyNull ([]:_) = True 
anyNull (_:xs) = anyNull xs