haskell - Infinite lazy lists of digits

Question

So I'm trying to do some number theory work, and I was using Mathematica but thought that Haskell would be more suited to dealing with infinite lists (as AFAIK Mathematica doesn't have lazy evaluation). What I want to do is have Haskell store all the digits of 1/x in an infinite lazy list. So far my searching has not turned up a way to split a ratio into its digits that returns a list of digits rather than an actual floating point number.

Answer 1

我们也可以将其实现为一个简单的流生产者：

divDigits :: Int -> Int -> [Int]
divDigits x y = x `div` y : divDigits (10 * (x `mod` y)) y

实际上有这种使用惰性列表的“无限”精度数字表示的库，请参阅Haskell Wiki。

Answer 2

非常感谢 Sam Yonnou，他提供的链接有正确
的公式使用的公式：
x/y 的第 n 位是 (10^(n-1)*x mod y)/y = floor(10 * (10^(n-1)*x mod y) / y) mod 10

结束代码如下所示：

nDigRat :: Int -> Int -> Int -> Int
nDigRat num denom n = floor (fromIntegral (10*(10^(n-1)*num `rem` denom)) / 
                             fromIntegral denom) 
                      `rem` 10

decExpansionRecipRat :: Int -> [Int]
decExpansionRecipRat n = map (nDigRat 1 n) [1..]