带警卫的功能:使用“ where”时的语法错误

use*_*035 2 syntax haskell

MWE:

import Control.Monad.State.Lazy

fibStep :: State (Integer, Integer) ()
fibStep = state $ \(a, b) -> ((), (b, a + b))

execStateN :: Int -> State s a -> s -> s
execStateN n m s
  | n == 1 = execState m s
  | n > 1 = let s' = execState m s in
              execStateN (n - 1) m s'
  -- | n > 1 = execStateN (n - 1) m s' where s' = execState m s
  | otherwise = error "undefined behaviour"
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它可以工作,但是一旦我取消注释where变体并对其进行注释let,就会出现语法错误:

错误:解析输入“ |”上的错误

我检查了缩进,它们很好。怎么了?

Wil*_*sem 6

where范围限定在所有的卫兵,所以你把它放在后卫,喜欢的结尾:

execStateN :: Int -> State s a -> s -> s
execStateN n m s
  | n == 1 = execState m s
  | n > 1 = execStateN (n - 1) m s'
  | otherwise = error "undefined behaviour"
  where s' = execState m s
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