Re: [Haskell-beginners] Why is type "Integer -> Integer" and not "(Num a) => a -> a"?
This is just a theory, but in my (limited) experience, GHCi is willing to guess at the type of values, where at functions (explicitly or implicitly typed) it can't guess. In your original function, it multiplied a number (Num a) => a by an Int, therefor it must be Int -> Int (because you can't multiply a Double by an Int, don't be crazy. In the new function, though, it can infer the type of the 2, and therefor it can infer it to whatever numerical type you send it. I've run into this problem a few times, but for whatever reason I can't get myself to fall into any of the usual ways I walk blindly into these sorts of problems this time.
On Thu, Nov 12, 2009 at 5:40 PM, Nathan M. Holden
This is just a theory, but in my (limited) experience, GHCi is willing to guess at the type of values, where at functions (explicitly or implicitly typed) it can't guess.
In your original function, it multiplied a number (Num a) => a by an Int, therefor it must be Int -> Int (because you can't multiply a Double by an Int, don't be crazy.
That is not the problem : as the rest of your message acknowledge 2 is not an Int, it is of type (Num a) => a. The issue rest squarely with the fact that the binding f1 don't have parameter expressed on the left side of the "=", thus the monomorphism restriction kicks in (as alluded, probably, by your first paragraph) and GHC has to use the defaulting rules to find a monomorphic type for f1. -- Jedaï
participants (2)
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Chaddaï Fouché -
Nathan M. Holden