CODEWITHSUNDEEP
haskellpattern-matching
mish
mish

Reputation: 1065

Hakell surplus binding in pattern

I have the following snippet out of an Interpreter:

type Ident = String
type Value = Int
type State = Ident -> Value

iniState :: State

iniState = \ident -> error "internal error initial state"


updateS :: State -> (Ident, Value) -> State

updateS s (ident, val) ident' | ident' == ident = val
                              | otherwise = s ident'

where does the ident' in the pattern of updateS come from?

Upvotes: 0

Views: 103

Answers (2)

Heatsink
Heatsink

Reputation: 7761

updateS is a function that takes three parameters. If you replace State with its definition in the type signature, you get

updateS :: (Ident -> Value) -> (Ident, Value) -> Ident -> Value

So ident' is the third parameter of the function, which has type Ident.

Looking at the bigger picture, updateS turns a State and an (Ident, Value) pair into a new State. We can reason about what kind of State is returned. If we evaluate updateS s (ident, val), we get a State that acts like s in most situations, except that if it's is called with an argument equal to ident, it returns val.

Upvotes: 3

huon
huon

Reputation: 102306

The third argument is from the Ident argument to the function in the type State.

Using type defines a synonym for another type (quite literally a synonym). In other words, the following type signatures are all the same:

State -> (Ident, Value) -> State 
(Ident -> Value) -> (Ident, Value) -> (Ident -> Value)
(Ident -> Value) -> (Ident, Value) -> Ident -> Value

(You could also replace Ident with String and Value with Int to do a full expansion.)

The last one makes it clear that updateS could take 3 parameters: a function Ident -> Value, a pair (Ident, Value) and a single Ident.

Upvotes: 3

Related Questions