I'm sure someone else on this list will explain it much more eloquently than I, but for now, here goes: What you are effectively forcing haskell to do is to return a result that hasn't been defined. You can't force a result of type `Int' where `+' has not been defined to return such a result for the sum of a Float(s) and integers(s). Of course, in languages such as C you can use casting for the purpose, but we are talking about very different, in fact, entirely different programming paradigms. hth, Matthew At 00:17 14/05/2014, you wrote:
On 05/12/2014 05:44 PM, Venu Chakravorty wrote:
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?
Just to follow up on the part Brandon didn't explain, no, (Num a, Num b) => a -> b -> a would not work either: there is not enough information there to do anything sensible. That is, we don't know how to turn any arbitrary Num into any other arbitrary Num.
What if it was (4 :: Int) + (7.5 :: Double). The type signature would match but the result is fractional and we promised to return Int.
This is the reason why it's Num a => a -> a -> a; we only know how to add things of the same type together. If we want more, we need to have more information.
Please let me know what I am missing.
Regards, Venu Chakravorty.
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