CODEWITHSUNDEEP
haskellmonads
user6023611
user6023611

Reputation:

What bind operator will be used?

Look at this example: 

ReaderT Integer (ErrorT String (StateT Integer Identity)) Integer

What operator bind (>>=) will be used here ?
For example, if we use do notation, then what bind operator will be chosen by compiler ?

Upvotes: 0

Views: 54

Answers (1)

David Young
David Young

Reputation: 10793

For reference, (>>=) has the type

(>>=) :: Monad m => m a -> (a -> m b) -> m b

and also note that

ReaderT Integer (ErrorT String (StateT Integer Identity)) Integer

is the same as 

(ReaderT Integer (ErrorT String (StateT Integer Identity))) Integer

because type application (like function application) is left associative in Haskell.

So, given

x :: ReaderT Integer (ErrorT String (StateT Integer Identity)) Integer

and an f with the appropriate type, in the expression

x >>= f

the type checker will try to match the type of x with the left argument type of (>>=). This means it will try to unify ReaderT Integer (ErrorT String (StateT Integer Identity)) Integer with m a. The only way that this unifies is by taking a to be Integer and m to be ReaderT Integer (ErrorT String (StateT Integer Identity)). So, using the ~ type equality notation, we end up with

 m ~ (ReaderT Integer (ErrorT String (StateT Integer Identity)))
 a ~ Integer

As a result, it must use the (ReaderT r n) instance of Monad, where r ~ Integer and n ~ ErrorT String (StateT Integer Identity).

This is a result of how type unification works in Haskell in general, not just for (>>=), so this general idea can be used for interpreting how the type checking of other polymorphic functions works.

Upvotes: 5

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