Haskell From First Principles Ch5, Ex 7,8

4
votes

I'm new to Haskell and reading Haskell from first principles.

right now I'm on chapter 5. while solving its exercises, specifically 7,8 I could not understand why I'm not coming up with right answer

so here is the question

u can find questions and solutions here

If the type of kessel is (Ord a, Num b) => a -> b -> a, then the type of kessel 1 2 is:

  1. Integer
  2. Int
  3. a
  4. (Num a, Ord a) => a
  5. Ord a => a
  6. Num a => a

I think its answer is 5. Ord a => a, bcs its one of its possible implementation which I have come up with is to completely ignore parameter b

like this

kessel ::(Ord a, Num b) => a -> b -> a
kessel a b = a
--if u have any other implementation Please share

since Its completely ignoring b it should not effect the type of a but still ghci show its type is

:t kessel 1 2       
kessel 1 2 :: (Ord a, Num a) => a

What I'm missing ? same goes with Q:8

1
haskellghci

1 Answers

3
votes

Num b is a red herring. The Num a constraint is because you passed a numeric literal for the a type, and those require the Num typeclass. Try :t kessel [] 2 to see it go away.