How to read an integer written in exponential form with Haskell?

Question

To read an integer written in decimal form is quite simple :

Prelude> read "1000000000" :: Int
1000000000

But how to read an integer written in exponetial form ?

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

Is there a function in the Prelude to do that, or do we need to parse the expression?

Thanks for any reply.

Answer 1

Here's a parser

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

Giving

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

And also

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

We could try to make it cope with negative exponents that give integer answers but that would be overkill for a read function.

Answer 2

Depending on the exact format of the string, you could just read it into a floating point type:

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

then convert to an integral type -- I'd recommend Integer instead of Int:

> floor (read "10e+9" :: Double) :: Integer
10000000000