问题
I'm just starting out at teaching myself Haskell. This code is supposed to do prime factorisation:
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)
I get a parse error on the commented line. I think my problem might be that lowestDivisorHelper
has guards, but the compiler doesn't know whether the guards belong to lowestDivisorHelper
or lowestDivisor
. How do I get around this?
I should add that I didn't want to define the helper function at the top level in order to hide the implementation detail. Importing the file shouldn't take the helper function with it.
回答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
You need to start a new statement with your helper function for the guards to be sufficiently indented by comparison.
(And you also forgot an argument, n
.)
This would also work:
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
but this doesn't:
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
The key point is that the |
has to be further to the right than the function name.
In general, starting a new line continues the previous one as long as it is further to the right. The guards have to continue on from the function name.
来源:https://stackoverflow.com/questions/13327374/haskell-defining-a-function-with-guards-inside-a-where