2009/3/20 Brandon S. Allbery KF8NH

On 2009 Mar 20, at 18:01, Sean Bartell wrote:

...

For a type "a" to be Fractional requires there to be: (/) :: a -> a -> a You can't divide an Int by another Int and (in general) get a third Int. You would probably want something like a "Fractionable" typeclass, with (/) :: a -> a -> b which would result in a Rational, but Haskell doesn't have this.

...but there is (%) :: (Integral a) => a -> a -> Ratio a

Thanks, I knew about % but didn't remember about it when I was working on this sample :) So that being said, consider the following: Prelude Data.Ratio> let x = 5::Int Prelude Data.Ratio> :t x x :: Int Prelude Data.Ratio> :t (x%3) (x%3) :: Ratio Int Prelude Data.Ratio> let y = truncate (x%3) Prelude Data.Ratio> :t y y :: Integer Why does y now have the type Integer instead of Int?

On 2009 Mar 20, at 20:14, Brandon S. Allbery KF8NH wrote:

On 2009 Mar 20, at 20:07, Zachary Turner wrote:

...

2009/3/20 Brandon S. Allbery KF8NH ...but there is (%) :: (Integral a) => a -> a -> Ratio a

Prelude Data.Ratio> let y = truncate (x%3) Prelude Data.Ratio> :t y y :: Integer

Why does y now have the type Integer instead of Int?

Because of the monomorphism restriction and defaulting. Actually use it somewhere and it will take the appropriate (Integral) type.

uh, clarification: because y has no arguments, the monomorphism restriction comes into play; if you define y as above in a program it will also end up defaulting to Integer. The expression "truncate (x %3)", on the other hand, is of type "Integral a => a" (as long as it's not placed into a context that triggers monomorphism). -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery@kf8nh.com system administrator [openafs,heimdal,too many hats] allbery@ece.cmu.edu electrical and computer engineering, carnegie mellon university KF8NH