sqrt和floor的組合模糊變量

Question

試圖實現列出某些數字範圍內的所有素數的函數，我知道在檢查因素時我不必檢查該數字的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] 

我已經瀏覽了答案，我嘗試了各種方法，如利用類型

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

檢查，但有沒有運氣。

任何幫助表示讚賞！

Answer 1

看起來你看着你寫的代碼，並計算出後的類型。一般來說，Haskell開發是相反的：首先找出類型，然後實現函數。 factors應該有什麼類型？試圖編譯代碼，我們得到以下錯誤，當

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

現在：

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，你只能比化整數，所以什麼類型的，所以這似乎是明智的

它抱怨說您指定了Integral a，但它需要Floating a爲sqrt。我們可以通過usinf fromIntegral做到這一點：

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] 

爲了保持可讀性，

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