Cru*_*lad 0 syntax haskell function-call
s和a是类型变量。在构造函数中,前两个参数是数据,然后是父级,在图形中的级别以及子级列表。
data Node s a = Root | Node s a (Node s a) Int [Node s a]
createRoot :: (ProblemState s a) => s-> a -> Node s a
createRoot state act= Node (state act Root 0 [])
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我已经将完全相同的5个参数传递给Node构造函数,但是我遇到了错误。
• Couldn't match expected type ‘Node s a’
with actual type ‘a1
-> Node s1 a1 -> Int -> [Node s1 a1] -> Node s1 a1’
• Probable cause: ‘Node’ is applied to too few arguments
In the expression: Node (state act Root 0 [])
In an equation for ‘createRoot’:
createRoot state act = Node (state act Root 0 [])
• Relevant bindings include
act :: a (bound at Search.hs:43:24)
state :: s (bound at Search.hs:43:18)
createRoot :: s -> a -> Node s a (bound at Search.hs:43:1)
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括号用于对表达式进行分组。(length "hello" + 2)是单个值,而不是4。
同样,Node (...)适用Node于单个参数:(state act Root 0 [])。显然这是错误的(并且将需要state一个带有四个参数的函数)。
解决方法是只删除括号:
Node state act Root 0 []
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现在Node应用于五个参数。
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