I got it! The msg “ No instance for (Fractional Int) arising from a use of `/‘ “ actually has no business with the operands order at all ! -- Cui Liqiang On Tuesday, October 28, 2014 at 8:00 PM, Cui Liqiang wrote:
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.