[Haskell-beginners] Lambda calc version of Haskell lambda function

[Francesco Ariis]

Il 31 dicembre 2020 alle 22:33 Lawrence Bottorff ha scritto:
> Here is a function declaration
> 
> makeAddress :: Int -> String -> String -> (Int, String, String)
> makeAddress number street town = (number,street,town)
> 
> and here is a lambda function version
> 
> makeAddressLambda =  (\number -> (\street -> (\town -> (number, street,
> town))))

You can lose most of the parentheses:

    makeAddressLambda =  \number -> \street -> \town -> (number, street, town)

> How would this lambda version look in lambda calculus? Like this?
> 
> \number.\street.\town.(number street town)

Yes.

> then
> 
> (\number.\street.\town.(number street town) (123 "Sunny St." "Fergus")

Wait! You have lost a ‘)’ on the lambdas (remember λ goes as far right as
possible). Also

    (123, "Sunny St.", "Fergus")

makes it look like it is a single argument, which leads to a wrong result:

    (λnumber. λstreet. λtown. number street town) (123, "Sunny St.", "Fergus")
    λstreet. λtown. (123, "Sunny St.", "Fergus") street town
    -- woops!

Instead, keeping in mind Haskell follows a beta-nu-mu strategy, we have
a series of 3 βs:

    (λnumber. λstreet. λtown. (number, street, town)) 123 "Sunny St." "Fergus"
    (λstreet. λtown. (123, street, town) "Sunny St." "Fergus"
    (λtown. (123, "Sunny St.", town) "Fergus"
    (123, "Sunny St.", "Fergus")