Haskell type coercion for strings

Question

On this code I'm able to construct a list of Algos using some direct constructors like Lit and Nom, but also from integers. This happens because Algo derives Num.

Is it possible to do something similar for strings?

type Nr = Double
data Algo
  = Nom Nr
  | Lit String
  | Und 
  deriving (Show)

instance Num Algo where
  (+) (Nom a) (Nom b)=Nom(a+b)
  (+) _ _=Und
  (*) (Nom a) (Nom b)=Nom(a*b)
  (*) _ _=Und
  abs (Nom a)=Nom(abs a)
  abs _=Und
  signum (Nom a)=Nom(signum a)
  signum _=Und
  fromInteger a=Nom(fromInteger a)

main=do
  print ([1,2,3,Und,Nom 5,Lit "x"]::[Algo])

running ok with result:

[Nom 1.0,Nom 2.0,Nom 3.0,Und,Nom 5.0,Lit "x"]

desired code:

print (["test",1,2,3,Und,Nom 5,Lit "x"]::[Algo])

will give an error of course... where does the conversion occurs? at parse/compile?

readPrec didn't help either

Answer 1

For "test" to be automatically "converted" to Algo, you would need to define an instance of IsString and then use the GHC extension OverloadedStrings:

instance IsString Algo where
  fromString = Lit

Live demo

So that the following works correctly:

print ["test", 1, 2, 3, Und, Nom 5, Lit "x"]
-- [Lit "test",Nom 1.0,Nom 2.0,Nom 3.0,Und,Nom 5.0,Lit "x"]