Hello Antoine! Thank you for the explanation. I think I get closer with it. So my understanding is this: In my example both 'a' and 'b' were of the same typeclass. In fact they were of the same type, too, since in the example the class had only one instance. However, the compiler/type checker could not prove this to be true. But if this was the case then I should be able to convince the compiler via type annotations like if p then (a :: (MyTC a) => a) else (b :: (MyTC a) => a) which does not work ('Inferred type is less polymorphic then expected.') Probably I do not yet understand well the relationship between types and type classes... Regards, Thomas On 21.07.2011 14:04, Antoine Latter wrote: [...]
The difference comes down to two things:
1. The type of 'if'. The haskell 'if ... then ... else ...' is conceptually just a function with the type (Bool -> a -> a -> a). The two branches must be of the same type.
2. In the code:
f (if p then a else b)
If somehow the 'if' could type-check with 'a' and 'b' being of different types, the compiler wouldn't know which type to pick for 'f'. In Haskell, we're not allowed to delay type-checking until run-time - the compiler demands that it can solve everything before running any of it.
I hope I haven't confused things even more!
Antoine