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")