英文:
Conditioned disjunction in Haskell
问题
这是你要翻译的部分：
"编写一个函数，该函数以一个布尔值列表作为输入，其中元素的数量是三的倍数，并返回通过顺序应用“条件析取”操作到列表元素的第k个三元组（第k个三元组是具有编号k、k+1和k+2的三个元素，其中k = 1,3，...，n-2）而获得的值。将操作应用于第k对元素的结果将成为中间列表的第(k+2)个元素。给出三个使用该函数的示例。"
以下是你提供的Haskell代码：
calc::[Bool]->Bool
calc [] = False
calc (p:[]) = False
calc (p:q:[]) = False
calc (p:q:r:[]) = ((not q||p)&&(q||r))
calc (p:q:r:xs) = calc(((not q||p)&&(q||r)):r:xs)
编译器报告以下错误：
"21.hs:3:19: parse error on input ‘=’"
如果需要进一步帮助，请提出具体问题。
英文:
> Write a function that takes as input a list of Boolean values, the number of elements in which is a multiple of three, and returns a value that is obtained by sequentially applying the operation "conditional disjunction" to the k-th triple of list elements (the k-th triple is a triple of elements with numbers k, k+1 and k+2, k = 1,3,...,n-2). The result of applying the operation to the k-th pair of elements becomes the (k+2)th element of the intermediate list. Give three examples of using the function.
I have tried the following, but it doesn't work.
calc::[Bool]->Bool
calc [] = False
calc (p:[]) = False
calc (p:q:[]) = False
calc (p:q:r:[]) = ((not q||p)&&(q||r))
calc (p:q:r:xs) = calc(((not q||p)&&(q||r)):r:xs)
The compiller says the following.
21.hs:3:19: parse error on input ‘=’
答案1
得分: 3
你的缩进有问题 - 你为你的模式编写了单独的函数实现，所以它们每个都应该从第0列开始：
calc::[Bool]->Bool
calc [] = False
calc (p:[]) = False
calc (p:q:[]) = False
calc (p:q:r:[]) = ((not q||p)&& (q||r))
calc (p:q:r:xs) = calc(((not q||p)&& (q||r)):r:xs)
参见Learn you a Haskell: Syntax in functions
英文:
Your indentation is wrong - you've got separate function implementations for your patterns, so each of them should start on column 0:
calc::[Bool]->Bool
calc [] = False
calc (p:[]) = False
calc (p:q:[]) = False
calc (p:q:r:[]) = ((not q||p)&&(q||r))
calc (p:q:r:xs) = calc(((not q||p)&&(q||r)):r:xs)
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