haskell - Haskell 让 in/where 和 if 缩进

Question

我有一个功能：

isSimpleNumber :: Int -> Bool
isSimpleNumber x = let deriveList = map (\y -> (x `mod` y)) [1 .. x]
                       filterLength = length ( filter (\z -> z == 0) deriveList
                       ....

在 filterLength 我想检查多少 filterLength 之后，我尝试：

isSimpleNumber :: Int -> Bool
isSimpleNumber x = let deriveList = map (\y -> (x `mod` y)) [1 .. x]
                       filterLength = length ( filter (\z -> z == 0) deriveList
                       in if filterLength == 2
                          then true

我得到错误：

    parse error (possibly incorrect indentation)
Failed, modules loaded: none.

如何使用 if 和 in 正确放置缩进？

谢谢你。

Answer 1

一个if总是需要 athen和一个else分支，所以你可能需要if filterLength == 2 then true else false，这相当于filterLength == 2.

Answer 2

这将编译：

isSimpleNumber :: Int -> Bool
isSimpleNumber x = let deriveList = map (\y -> (x `mod` y)) [1 .. x]
                       filterLength = length ( filter (\z -> z == 0) deriveList )
                   in filterLength == 2

main = print $ isSimpleNumber 5

在“deriveList”之后缺少一个结束“）”。您也不需要 if-then-true 表达式。