Haskell 将分数值舍入为Int

我试图学习Haskell，但我在数字转换方面遇到了困难，有人能解释一下为什么Haskell编译器会对这段代码发火：

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

它将打印错误消息：

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))'

---

我已经在方法签名上尝试了许多组合（添加整数、分数、双精度等）。有人告诉我，文字4.236068与问题有关，但无法解决问题。

Haskell不会自动为您转换内容，因此

x*y

仅在

x

和

y

具有相同类型（例如，您不能将

Int

乘以

Double

）时才有效

注意：我们可以使用Prelude函数

iterate

更自然地表达

phimagic

：

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

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