Question

尝试实现一个函数来列出某些数字范围内的所有素数，我知道当我检查因素时，我不必检查该数字的sqrt之外。

factors n = [x | x <- [1..(floor (sqrt n))], mod n x == 0]
prime n = factors n == [1,n]
listPrimesFromTill n z = [ xs | xs <- [n..z], prime xs == True]

我一直在寻找答案，我尝试了各种方法，比如使用

进行类型检查

factors :: (RealFrac b, Integral c, Floating b) => b -> c

但没有运气。

感谢任何帮助！

Answer 1

看起来你看了你编写的代码，然后想出了类型。一般来说，Haskell开发是另一种方式：首先你弄清楚类型，然后你实现这些功能。 factors应该包含哪种类型？好吧，你只能对整数进行分解，所以这类似的东西，所以这看起来很明智：

factor :: Integral a => a -> [a]

现在尝试编译代码时出现以下错误：

Could not deduce (Floating a) arising from a use of `sqrt` from the context (Integral a)

和

Could not deduce (RealFrac a) arising from a use of `sqrt` from the context (Integral a)

它抱怨您指定了Integral a但Floating a需要sqrt。我们可以通过usinf fromIntegral执行此操作：

sqrt         ::          Floating a => a -> a
fromIntegral :: (Integral a, Num b) => a -> b

factors :: Integral a => a -> [a]       vvvvvvvvvvvvvv
factors n = [x | x <- [1..(floor (sqrt (fromIntegral n)))], mod n x == 0]

为了保持可读性，

factors n = [x | x <- [1..isqrt n], mod n x == 0]
    where isqrt = floor . sqrt . fromIntegral

结合sqrt和floor

1 个答案: