Haskell中的類型問題

Question

我正在寫一個函數來查找實數的整數因數。當我運行的代碼，我得到以下錯誤：

Clean solutions.hs:70:33: 
    Could not deduce (Integral a) arising from a use of `truncate' 
    from the context (RealFrac a) 
     bound by the type signature for divisors :: RealFrac a => a -> [a] 
     at Clean solutions.hs:70:1-74 
    Possible fix: 
     add (Integral a) to the context of 
     the type signature for divisors :: RealFrac a => a -> [a] 
    In the first argument of `(==)', namely `(truncate (n/x))' 
    In the expression: (truncate (n/x)) == (n/x) 
    In the first argument of `filter', namely 
     `(\ x -> (truncate (n/x)) == (n/x))' 

Clean solutions.hs:70:59: 
    Could not deduce (Enum a) 
     arising from the arithmetic sequence `2.0, 3.0 .. n/2' 
    from the context (RealFrac a) 
     bound by the type signature for divisors :: RealFrac a => a -> [a] 
     at Clean solutions.hs:70:1-74 
    Possible fix: 
     add (Enum a) to the context of 
     the type signature for divisors :: RealFrac a => a -> [a] 
    In the second argument of `filter', namely `[2.0, 3.0 .. n/2]' 
    In the second argument of `(:)', namely 
     `filter (\ x -> (truncate (n/x)) == (n/x)) [2.0, 3.0 .. n/2]' 
    In the expression: 
     1 
     : filter (\ x -> (truncate (n/x)) == (n/x)) [2.0, 3.0 .. n/2] 

我發現很難理解我在做什麼錯的，儘管花費一兩個小時在Haskell刷牙的類型。我的代碼如下：

divisors :: RealFrac a => a -> [a] 
divisors n = 1 : filter (\x -> (truncate (n/x)) == (n/x)) [2.0, 3.0.. n/2] 

謝謝， 山姆。

Answer 1

怎麼樣：

divisors :: (RealFrac a, Enum a) => a -> [a] 
divisors n = filter (\x -> n/x == fromIntegral (truncate (n/x))) [1.0..(n/2)]