haskell - 如何使用 Haskell 读取以指数形式写入的整数？

Question

读取以十进制形式书写的整数非常简单:

Prelude> read "1000000000" :: Int
1000000000

但是如何读取以指数形式写入的整数？

Prelude> read "10e+9" :: Int
*** Exception: Prelude.read: no parse

Prelude中是否有函数这样做，还是我们需要解析表达式？

感谢您的任何答复。

Answer 1

这是一个解析器

readI xs = let (m,e) = break (=='e') xs in 
     read m * 10 ^ case e of
       "" -> 1
       ('e':'+':p) -> read p
       ('e':p) -> read p

给予

Main> readI "3e5"
300000
Main> readI "3e+500"
300000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Main> readI "3e+500" :: Int
0
Main> readI "3e+500" :: Integer
300000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

并且

Main> readI "32e-5" 
Program error: Prelude.^: negative exponent

我们可以尝试让它处理给出整数答案的负指数，但这对于读取函数来说太过分了。