Type class definition with functions depending on an additional type

Question

Still new to Haskell, I have hit a wall with the following: I am trying to define some type classes to generalize a bunch of functions that use gaussian elimination to solve linear systems of equations. Given a linear system M x = k the type a of the elements m(i,j) \elem M can be different from the type b of x and k. To be able to solve the system, a should be an instance of Num and b should have multiplication/addition operators with b, like in the following: class MixedRing b where (.+.) :: b -> b -> b (.*.) :: (Num a) => b -> a -> b (./.) :: (Num a) => b -> a -> b Now, even in the most trivial implementation of these operators, I'll get Could not deduce a ~ Int. a is a rigid type variable errors (Let's forget about ./. which requires Fractional) data Wrap = W { get :: Int } instance MixedRing Wrap where (.+.) w1 w2 = W $ (get w1) + (get w2) (.*.) w s = W $ ((get w) * s) I have read several tutorials on type classes but I can find no pointer to what actually goes wrong. JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

Let us have a look at the type of the implementation that you would have to provide for (.*.) to make Wrap an instance of MixedRing. Substituting Wrap for b in the type of the method yields (.*.) :: Num a => Wrap -> a -> Wrap As Wrap is isomorphic to Int and to not have to think about wrapping and unwrapping with Wrap and get, let us reduce our goal to finding an implementation of (.*.) :: Num a => Int -> a -> Int (You see that this doesn't make the challenge any easier or harder, don't you?) Now, observe that such an implementation will need to be able to operate on all types a that happen to be in the type class Num. (This is what a type variable in such a type denotes: universal quantification.) Note: this is not the same (actually, it's the opposite) of saying that your implementation can itself choose what a to operate on); yet that is what you seem to suggest in your question: that your implementation should be allowed to pick Int as a choice for a. Now, as you want to implement this particular (.*.) in terms of the (*) for values of type Int, we need something of the form n .*. s = n * f s with f :: Num a => a -> Int I cannot think of a function that converts from an arbitary Num-type a to Int in a meaningful way. I'd therefore say that there is no meaningful way to make Int (and, hence, Wrap) an instance of MixedRing; that is, not such that the instance behaves as you would probably expect it to do.