Is it that since both the formal args are defined as "a" they have to be exactly the same instances?
Exactly.
Had "+" been defined something like: (+) :: (Num a, Num b) => a -> b -> a my second addition would have worked?
Yes except that you cannot write this function given the definition of Num
http://hackage.haskell.org/package/base-4.7.0.0/docs/Prelude.html#t:Num
If you want your second example to work, you have to explicitly convert the
floating-point value into an Int (using floor, ceiling or round for
instance).
-Sylvain
2014-05-12 17:44 GMT+02:00 Venu Chakravorty
Hello everyone, I am just starting with Haskell so please bear with me.
Here's my question:
Consider the below definition / output:
Prelude> :t (+) (+) :: (Num a) => a -> a -> a
What I understand from the above is that "+" is a function that takes two args which are types of anything that IS-AN instance of "Num" (Int, Integer, Float, Double) and returns an instance of "Num". Hence this works fine: Prelude> 4.3 + 2 6.3
But I can't understand why this doesn't work: Prelude> 4.3 + 4 :: Int
<interactive>:1:0: No instance for (Fractional Int) arising from the literal `4.3' at <interactive>:1:0-2 Possible fix: add an instance declaration for (Fractional Int) In the first argument of `(+)', namely `4.3' In the expression: 4.3 + 4 :: Int In the definition of `it': it = 4.3 + 4 :: Int
I expected that the second addition would work as both "Float" and "Int" are instances of "Num". Is it that since both the formal args are defined as "a" they have to be exactly the same instances? Had "+" been defined something like: (+) :: (Num a, Num b) => a -> b -> a my second addition would have worked?
Please let me know what I am missing.
Regards, Venu Chakravorty.
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