将小数值舍入为Int

Question

我试图学习Haskell,但我被卡在数字转换上,有人可以解释为什么Haskell编译器会对这段代码感到厌倦:

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

它会输出错误消息:

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(添加了Integral,Fractional,Double等).有人告诉我,文字4.236068与问题有关,但无法弄明白.

Answer 1

哈斯克尔不会自动转换你的东西,所以x * y只能当x和y具有相同的类型(你不能乘Int用Double,例如).

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

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