Haskell - 在'where'中定义一个带有警卫的功能

Question

我刚刚开始自学Haskell.这段代码应该进行素数分解:

divides :: Integer -> Integer -> Bool
divides small big = (big `mod` small == 0)

lowestDivisor :: Integer -> Integer
lowestDivisor n = lowestDivisorHelper 2 n
    where lowestDivisorHelper m n
        | (m `divides` n) = m  -- these should belong to lowestDivisorHelper
        | otherwise = lowestDivisorHelper (m+1) n

primeFactors :: Integer -> [Integer]
primeFactors 1 = []
primeFactors n
    | n < 1 = error "Must be positive"
    | otherwise = let m = lowestDivisor n
                  in m:primeFactors (n/m)

我在评论行上得到一个解析错误.我认为我的问题可能是lowestDivisorHelper有警卫,但编译器不知道警卫是属于lowestDivisorHelper还是lowestDivisor.我该如何解决这个问题？

我应该补充一点,我不想在顶层定义辅助函数,以隐藏实现细节.导入文件不应该使用辅助函数.

Answer 1

lowestDivisor :: Integer -> Integer
lowestDivisor n = lowestDivisorHelper 2 n where 
  lowestDivisorHelper m n
        | (m `divides` n) = m  -- these should belong to lowestDivisorHelper
        | otherwise = lowestDivisorHelper (m+1) n

您需要使用辅助函数启动一个新语句,以便通过比较使警卫充分缩进.(你也忘记了一个论点,n.)这也有效:

lowestDivisor :: Integer -> Integer
lowestDivisor n = lowestDivisorHelper 2 n 
    where 
  lowestDivisorHelper m n
        | (m `divides` n) = m  -- these should belong to lowestDivisorHelper
        | otherwise = lowestDivisorHelper (m+1) n

但这不是:

lowestDivisor :: Integer -> Integer
lowestDivisor n = lowestDivisorHelper 2 n 
  where lowestDivisorHelper m n
        | (m `divides` n) = m  -- these should belong to lowestDivisorHelper
        | otherwise = lowestDivisorHelper (m+1) n

关键点是|必须比函数名称更靠右边.

通常,只要它在右边,开始一个新行继续前一个行.警卫必须从功能名称继续.