5

读取以十进制形式写入的整数非常简单：

Prelude> read "1000000000" :: Int
1000000000

但是如何读取以指数形式写的整数呢？

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

中是否有一个函数Prelude可以做到这一点，还是我们需要解析表达式？

感谢您的回复。

4

2 回答 2

3

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。

于 2012-11-27T11:35:47.130 回答
2

根据字符串的确切格式，您可以将read其转换为浮点类型：

> read "10e+9" :: Double
1.0e10

然后转换为整数类型——我建议Integer不要Int：

> floor (read "10e+9" :: Double) :: Integer
10000000000
于 2012-11-26T15:18:33.137 回答