How does GHCi pick an instance of the Monad type class to use for polymorphic actions?

Question

I'm new to Haskell so this might be a noob question.

When I do return 10 >>= return GHCi shows 10. When I check the type of return 10 with :t it just says return 10 :: (Monad m, Num a) => m a, and of I do typeOf return 10 I get an error.

But as far as I understand, Haskell must have used a particular instance of >>= to evaluate return 10 >>= return, so which instance did it use and how did it decide which one to use?

Answer 1

This follows the idea that GHCi is sort of like a giant do block of IO. Whenever you type in something that is an expression it first tries to see if the type of the result can be specialized to something of the form IO a. If it can, it executes the IO action and just prints the result. Only otherwise does it print the result of the expression itself.

To force GHCi to go to whatever specific monad you want, you can add a type annotation. Notice how IO gets treated differently (and the same way as the expression would have been treated without any annotation).

ghci> return 10 >>= return :: Maybe Int
Just 10
ghci> return 10 >>= return :: [Int]
[10]
ghci> return 10 >>= return :: IO Int
10

---

As an aside, there is an entirely different problem regarding what instance of Num is chosen, and that one has everything to do with defaulting rules and the monomorphism restriction.