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.

Accepted Answer

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
```

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

Solution

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