Re: [Haskell-cafe] Facing issue: "No instance for (Fractional a0) arising from a use of ‘it’ The type variable ‘a0’ is ambiguous"

Thanks Jack it works. On Tue, Sep 13, 2016 at 12:01 AM, Jack Henahan wrote:

The inferred type of findLoot is

findLoot :: (Integral a, Fractional a) => a -> [[a]] -> a

It's impossible to satisfy those constraints.

Rewriting like this does what you want, but there's likely a cleaner way to express this algorithm.

findLoot :: (Integral a, Fractional b) => a -> [[a]] -> b findLoot val [] = 0.0 findLoot val ((x:y:[]):xs) | val == 0 = 0.0 | val < y = ((fromIntegral val) * (fromIntegral x)) / fromIntegral y | val > y = fromIntegral ((div val y)*x) + (findLoot (mod val y) xs)

Balraj Singh writes:

...

Hi,

I am writing a sample code below and getting an error. Please help to resolve it:

Code:

import Data.List

findLoot val [] = 0.0

findLoot val ((x:y:[]):xs) | val == 0 = 0.0

| val < y = ((fromIntegral val) * (fromIntegral x)) / fromIntegral y

| val > y = ((div val y)*x) + (findLoot (mod val y) xs)

Error:

No instance for (Fractional a0) arising from a use of ‘it’

The type variable ‘a0’ is ambiguous

Note: there are several potential instances:

instance HasResolution a => Fractional (Fixed a)

-- Defined in ‘Data.Fixed’

instance Integral a => Fractional (Ratio a)

-- Defined in ‘GHC.Real’

instance Fractional Double -- Defined in ‘GHC.Float’

...plus one other

In the first argument of ‘print’, namely ‘it’

In a stmt of an interactive GHCi command: print it

Sample Input and Output:

Input 1:- val = 50 ((x:y:[]):xs) = [[120, 30], [100,50],[60, 20]] Output 1:- 180.0000

Input 2:- val = 10 ((x:y:[]):xs) = [[500,30]] Output 2:- 166.6667

-- Jack