Haskell让/在哪里和如果缩进

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之后我想检查filterLength的数量，我试试：

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

我收到错误：

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

如何使用if和in？

正确缩进缩进

谢谢。

Answer 1

if始终需要then和else分支，因此您可能需要if filterLength == 2 then true else false，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”之后缺少关闭“）”。你也不需要if-then-true表达式。