Thanks for your help, your suggestion works.

But I still don’t quite understand. In the following line:

caluDecimal l = (foldr1 (\x acc -> acc / 10.0 + x) (map digitToInt l)),

After applying digitToInt, the type of ‘x’ in the expression above is Int indeed, but Haskell consider the ’10.0’ to be a Int, is it?

----------------------------------------------------------------------------------------------------------------------------

Hi,

I am doing an exercise in Haskell, which is converting a string like ?$123.312? to double value. Below is my code:

module Main where

import Data.Char

caluInt l = foldl1 (\acc x -> acc * 10 + x) (map digitToInt l)

caluDecimal l = (foldr1 (\x acc -> acc / 10.0 + x) (map digitToInt l))

convert(x:xs) =

let num = [e | e <- xs, e /= ',']

intPart = takeWhile (/='.') num

decimalPart = tail(dropWhile (/='.') num)

in (caluInt intPart) + (caluDecimal decimalPart)

And I got an error in this line: caluDecimal l = (foldr1 (\x acc -> acc / 10.0 + x) (map digitToInt l)),

which says:

No instance for (Fractional Int) arising from a use of `/'

Possible fix: add an instance declaration for (Fractional Int)

In the first argument of `(+)', namely `acc / 10.0'

In the expression: acc / 10.0 + x

In the first argument of `foldr1', namely

`(\ x acc -> acc / 10.0 + x)'

Why Haskell insists that 10.0 is a Int? How can I explicitly tell Haskell I want a Fractional?

Cui Liqiang

Why Haskell insists that 10.0 is a Int? How can I explicitly tell Haskell
I want a Fractional?

Because digitToInt means exactly what it says. If you want it to become

something other than Int, apply fromIntegral to its result.