Question

我的职能：

type Block = [Maybe Int] 

checkBlock' :: Block -> Block
checkBlock' (r:rs)
    | r == [] = []
    | isJust r == True = r:checkBlock' rs
    | isNothing r == True = checkBlock' rs

checkBlock :: Block -> Block
checkBlock (r:rs)
    | r == [] = []
    | isNothing r == True = r:checkBlock rs
    | isJust r == True    = checkBlock rs

我想首先检查Just Ints和Nothings的列表，并仅返回Just值。而第二个功能只返回Nothings。

他们在没有r == [] = []的基础的情况下编译正常，但是我得到了错误：

Sudoku.hs:104:12:
    Couldn't match expected type `Maybe Int' with actual type `[t0]'
    In the second argument of `(==)', namely `[]'
    In the expression: r == []
    In a stmt of a pattern guard for
                   an equation for checkBlock':
      r == []
Failed, modules loaded: none.

我可以放在那里而不是[]进行编译？我有很多想法。

Answer 1

你应该在这里使用模式匹配，而不是守卫：

checkBlock :: Block -> Block
checkBlock [] = []
checkBlock (r:rs)
    | isNothing r == True = r:checkBlock rs
    | isJust r == True    = checkBlock rs

或者，最好是：

checkBlock :: Block -> Block
checkBlock [] = []
checkBlock (Nothing : rs) = Nothing : checkBlock rs
checkBlock (Just _ : rs) = checkBlock rs

Answer 2

您无法将r与空列表进行比较，您只能比较整个列表。

作为替代方案，你可以写

checkBlock :: Block -> Block
checkBlock rs
    | rs == [] = []
    | isNothing (head rs) == True = (head rs):checkBlock (tail rs)
    | isJust (head rs) == True    = checkBlock (tail rs)

其次，警卫已经使用Bool，因此== True是不必要的。顺便说一下，尝试使用null rs

的rs == []

checkBlock :: Block -> Block
checkBlock rs
    | null rs              = []
    | isNothing (head rs)  = (head rs):checkBlock (tail rs)
    | isJust (head rs)     = checkBlock (tail rs)

但这看起来有点难看，让我们使用模式匹配：

checkBlock :: Block -> Block
checkBlock [] = []
checkBlock (r:rs)
    | isNothing r  = r:checkBlock rs
    | isJust r     = checkBlock rs

有问题的基本案例Haskell

2 个答案: