haskell - 将小数值舍入为 Int

Question

我正在尝试学习 Haskell，但我陷入了数字转换的困境，有人可以解释一下为什么 Haskell 编译器对这段代码感到疯狂吗:

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

它打印错误消息:

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多种组合(添加积分、分数、 double 等)。有些东西告诉我，文字 4.236068 与问题有关，但无法弄清楚。

Answer 1

Haskell 不会自动为您转换内容，因此 x * y 仅当 x 和 y 具有相同类型时才有效(您可以例如，不要将 Int 乘以 Double)。

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

请注意，我们可以使用 Prelude 函数 iterate 更自然地表达 phimagic:

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