sqrt和floor的组合模糊变量

Question

试图实现列出某些数字范围内的所有素数的函数，我知道在检查因素时我不必检查该数字的sqrt以外的因素。

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] 

我已经浏览了答案，我尝试了各种方法，如利用类型

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

检查，但有没有运气。

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Answer 1

看起来你看着你写的代码，并计算出后的类型。一般来说，Haskell开发是相反的：首先找出类型，然后实现函数。 factors应该有什么类型？试图编译代码，我们得到以下错误，当

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

现在：

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) 

嗯，你只能比化整数，所以什么类型的，所以这似乎是明智的

它抱怨说您指定了Integral a，但它需要Floating a为sqrt。我们可以通过usinf fromIntegral做到这一点：

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