Question

编译器告诉我这个函数有非详尽的模式，但是每个场景都被覆盖了，不是吗？

allNeighbors :: Ord v => Graph v -> [v] -> [v] -> [v]
allNeighbors graph (x:xs) neighborsList
                   | length (x:xs) <= 0 = neighborsList
                   | length (x:xs) == 1 = neighborsList ++ (neighbors x graph)
                   | length (x:xs) > 1 = allNeighbors graph xs (neighborsList ++ (neighbors x graph))

Answer 1

未涵盖：allNeighbors a [] b。（试试吧;你应该收到错误。）

length (x:xs) <= 0案例无法访问（第一个元素为x的列表长度不能为0）。

length (x:xs) == 1最好写成[x]形式的模式。

length (x:xs) > 1不必要地效率低下：它强制程序遍历整个列表来计算长度，只是为了查看是否有多个元素。这可以由x : xs@(_ : _)等模式替换。

事实上，整个功能可以这样重写：

allNeighbors :: Ord v => Graph v -> [v] -> [v] -> [v]
allNeighbors graph list neighborsList =
    case list of
        []     -> neighborsList
        [x]    -> neighborsList ++ neighbors x graph
        x : xs -> allNeighbors graph xs (neighborsList ++ neighbors x graph)

Haskell非穷举模式

1 个答案: