Haskell函數在使用數字時有效，但與變量無關

Question

我正在使用ghci，並且遇到了用於獲取數字因子的函數的問題。

我想工作的代碼是：

let factors n = [x | x <- [1..truncate (n/2)], mod n x == 0] 

我努力（在這種情況下，與66）來使用它，它不抱怨，當我按下回車鍵，但只要我得到這個錯誤信息：

Ambiguous type variable 't0' in the constraints: 
    (Integral t0) 
    arising from a use of 'factors' at <interactive>:30:1-10 
    (Num t0) arising from the literal '66' at <interactive>:30:12-13 
    (RealFrac t0) 
    arising from a use of 'factors' at <interactive:30:1-10 
    Probable fix: add a type signature that fixes these type variable(s) 
    In the expression: factors 66 
    In the equation for 'it': it = factors 66 

下面的代碼工作完美：

let factorsOfSixtySix = [x | x <- [1..truncate (66/2)], mod 66 x == 0] 

我是新來的Haskell，並查找類型和ty之後佩奇，我還不確定我的意思。

Answer 1

使用div整數除法來代替：

let factors n = [x | x <- [1.. n `div` 2], mod n x == 0] 

的問題在你的代碼是/需要RealFrac類型n而mod一個Integral之一。在定義時這很好，但是不能選擇適合兩個約束的類型。

另一個選項可能是在使用mod之前截斷n，但更麻煩。畢竟，你不想打電話給factors 6.5，是嗎？ ;-)

let factors n = [x | x <- [1..truncate (n/2)], mod (truncate n) x == 0] 

Answer 2

如果你把這個頂級綁定（慣用的Haskell）類型註釋，你會得到不同的，可能更多有用的錯誤信息。

GHCi> let factors n = [x | x <- [1..truncate (n/2)], mod n x == 0] 
GHCi> :t factors 
factors :: (Integral t, RealFrac t) => t -> [t] 
GHCi> let { factors :: Double -> [Double]; factors n = [x | x <- [1..truncate (n/2)], mod n x == 0]; } 

<interactive>:30:64: 
    No instance for (Integral Double) arising from a use of `truncate' 
    Possible fix: add an instance declaration for (Integral Double) 
    In the expression: truncate (n/2) 
    In the expression: [1 .. truncate (n/2)] 
    In a stmt of a list comprehension: x <- [1 .. truncate (n/2)] 
GHCi> let { factors :: Integer -> [Integer]; factors n = [x | x <- [1..truncate (n/2)], mod n x == 0]; } 

<interactive>:31:66: 
    No instance for (RealFrac Integer) arising from a use of `truncate' 
    Possible fix: add an instance declaration for (RealFrac Integer) 
    In the expression: truncate (n/2) 
    In the expression: [1 .. truncate (n/2)] 
    In a stmt of a list comprehension: x <- [1 .. truncate (n/2)] 

<interactive>:31:77: 
    No instance for (Fractional Integer) arising from a use of `/' 
    Possible fix: add an instance declaration for (Fractional Integer) 
    In the first argument of `truncate', namely `(n/2)' 
    In the expression: truncate (n/2) 
    In the expression: [1 .. truncate (n/2)] 

Answer 3

我是新來的Haskell所以請原諒我的勇氣拿出一個答案在這裏，但最近我已經這樣做了如下;

factors :: Int -> [Int] 
factors n = f' ++ [n `div` x | x <- tail f', x /= exc] 
      where lim = truncate (sqrt (fromIntegral n)) 
        exc = ceiling (sqrt (fromIntegral n)) 
        f' = [x | x <- [1..lim], n `mod` x == 0] 

我相信效率更高。如果你喜歡，你會注意到;

sum (factors 33550336)