CODEWITHSUNDEEP
haskelltype-inferencetypeclass
J Cooper
J Cooper

Reputation: 17051

How does Haskell know which typeclass instance you mean?

This question arose while reading the new chapter in the excellent Learn You a Haskell about applicative functors.

The Applicative typeclass has, as part of the definition for the Maybe instance:

pure = Just

If I just go to GHCi and import Control.Applicative, and do:

pure (3+)

I don't get Just anything (makes sense). But if I use it in part of the expression:

pure (3+) <*> Just 4

I get Just 7. I guess it also isn't surprising, but I'm missing something integral about how typeclasses work, I think, that there is no ambiguity with the call to pure here.

If my confusion makes sense, can anyone explain what's going on in detail?

Upvotes: 4

Views: 806

Answers (4)

JP Alioto
JP Alioto

Reputation: 45117

Here is an interesting SO thread on type inference. Not Haskell specific, but lots of good links and stuff to read about type inference in functional languages.

Upvotes: 1

Norman Ramsey
Norman Ramsey

Reputation: 202465

It's worth looking at the type the compiler infers for pure (3+):

Prelude Control.Applicative> :t pure (3+)
pure (3+) :: (Num a, Applicative f) => f (a -> a)

The type of this term is overloaded, and the decision about the numeric class and the applicative class is delayed until later. But you can force a particular type with an annotation, for example:

*Showfun Control.Applicative> pure (3+) :: Maybe (Double -> Double)
Just <function>

(This works because Showfun has an instance declaration that prints a function value as <function>.)

It's just a question of when the compiler has accumulated enough information to make a decision.

Upvotes: 3

bdonlan
bdonlan

Reputation: 231063

To expand a bit on newacct's answer, if there isn't enough information to infer the actual type, the compiler may (in some cases) attempt to select a default type, limited to those that would satisfy the type constraints in question. In this case, the type inferred is IO (n -> n) for some difficult-to-determine instance of Num => n. GHCi then evaluates it and throws away the return value, with no visible effect.

Upvotes: 2

newacct
newacct

Reputation: 122419

It's just type inference. The (<*>) operator requires both arguments to use the same Applicative instance. The right side is a Maybe, so the left side has to be a Maybe also. So that's how it figures out which instance is used here. You can look at the type of any expression in the interpreter by typing :t expression, and maybe if you just go through each subexpression and look at the type that was inferred, you will get a better picture of what's going on.

Upvotes: 7

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