在表达式中使用gu ..

Question

有时我会写这样的代码

solveLogic :: Int -> Int -> Int
solveLogic a b =
    let 
        x = 1
        brainiac
            | a >= x     = 1
            | a == b     = 333
            | otherwise  = 5
    in
        brainiac

每次我都强烈要求编写这些东西而不需要"brainiac"功能,如下所示:

solveLogic :: Int -> Int -> Int
solveLogic a b =
    let 
        x = 1
    in
        | a >= x     = 1
        | a == b     = 333
        | otherwise  = 5

哪个代码更像是"Haskellish".有没有办法做到这一点？

Answer 1

是的,使用where条款:

solveLogic a b
        | a >= x     = 1
        | a == b     = 333
        | otherwise  = 5
    where
      x = 1

Answer 2

当我想要守卫作为表达时,我会使用这个有点丑陋的黑客

case () of
_ | a >= x     -> 1
  | a == b     -> 333
  | otherwise  -> 5