舍入分數值INT

Question

我試着去學習哈斯克爾，但我被困在數轉換，能不能有人解釋爲什麼Haskell的編譯器在這個代碼要瘋了：

phimagic :: Int -> [Int] 
phimagic x = x : (phimagic (round (x * 4.236068))) 

它打印錯誤信息：

problem2.hs:25:33: 
    No instance for (RealFrac Int) arising from a use of `round' 
    Possible fix: add an instance declaration for (RealFrac Int) 
    In the first argument of `phimagic', namely 
     `(round (x * 4.236068))' 
    In the second argument of `(:)', namely 
     `(phimagic (round (x * 4.236068)))' 
    In the expression: x : (phimagic (round (x * 4.236068))) 


problem2.hs:25:44: 
    No instance for (Fractional Int) 
     arising from the literal `4.236068' 
    Possible fix: add an instance declaration for (Fractional Int) 
    In the second argument of `(*)', namely `4.236068' 
    In the first argument of `round', namely `(x * 4.236068)' 
    In the first argument of `phimagic', namely 
     `(round (x * 4.236068))' 

我已經嘗試了方法簽名（添加積分，分數，雙等等）的一些組合。有些東西告訴我，4.236068與問題有關，但無法解決問題。

Answer 1

哈斯克爾不會自動轉換你的東西，所以x * y只有工作，如果x和y具有相同的類型（你不能Double乘Int，例如）。

phimagic :: Int -> [Int] 
phimagic x = x : phimagic (round (fromIntegral x * 4.236068)) 

報告中，我們可以使用前奏功能iterate更自然地表達phimagic：

phimagic = iterate $ round . (4.236068 *) . fromIntegral