haskell - 如何在列表上正确进行 Haskell 模式匹配？

Question

我正在尝试完成 this page 底部的 Haskell 练习。我正在研究的是#7。问题是这样的:

定义一个使用分隔符值将列表列表连接在一起的函数。

这是类型签名:

-- file: ch03/Intersperse.hs
intersperse :: a -> [[a]] -> [a]

这是它应该表现的方式:

ghci> :load Intersperse
[1 of 1] Compiling Main             ( Intersperse.hs, interpreted )
Ok, modules loaded: Main.
ghci> intersperse ',' []
""
ghci> intersperse ',' ["foo"]
"foo"
ghci> intersperse ',' ["foo","bar","baz","quux"]
"foo,bar,baz,quux"

这就是我的想法:

intersperse :: a -> [[a]] -> [a]
intersperse a [] = ""
intersperse a [x] = x
intersperse a [x:xs] = x ++ a ++ (intersperse a xs)

但我收到此错误:

intersperse.hs:4:24: error:
    • Couldn't match expected type ‘[a]’ with actual type ‘a’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          intersperse :: forall a. a -> [[a]] -> [a]
        at intersperse.hs:1:16
    • In the first argument of ‘(++)’, namely ‘x’
      In the expression: x ++ a ++ (intersperse a xs)
      In an equation for ‘intersperse’:
          intersperse a [x : xs] = x ++ a ++ (intersperse a xs)
    • Relevant bindings include
        xs :: [a] (bound at intersperse.hs:4:18)
        x :: a (bound at intersperse.hs:4:16)
        a :: a (bound at intersperse.hs:4:13)
        intersperse :: a -> [[a]] -> [a] (bound at intersperse.hs:2:1)

我是 Haskell 的新手，所以我不太理解这个错误。我究竟做错了什么？

编辑:

让它与这个一起工作:

intersperse :: a -> [[a]] -> [a]
intersperse a [] = []
intersperse a [x] = x
intersperse a (x:xs) = x ++ [a] ++ (intersperse a xs)

事实证明，有一些问题。

Answer 1

模式[x:xs]仅与长度为1的列表匹配，其单个元素非空(与x:xs匹配)。例如。它与 [[1,2,3]] 与 x=1 和 xs=[2,3] 匹配。它与 [[1],[2,3]]、[] 或 [[]] 不匹配。

您需要(x:xs)。这可以将 [[1,2],[3],[4,5]] 与 x=[1,2] 和 xs=[[ 3],[4,5]].