[Haskell-beginners] OOP exercise with closures

[Francesco Ariis]

Il 29 dicembre 2020 alle 15:43 Lawrence Bottorff ha scritto:
> Okay, I'm in Lesson 10 of *Get Programming with Haskell * and we're
> creating an OOP-like world with closures. The first step is a cup object
> with one attribute, its ounce size. Here's a "constructor"
> 
> cup :: t1 -> (t1 -> t2) -> t2
> cup flOz = \message -> message flOz
> 
> so this returns upon use
> 
> > myCup = cup 6
> 
> myCup which has "internally" a lambda function
> 
> (\message -> message) 6
> 
> waiting, correct?

Not exactly. `myCup` is a function with takes another function as input

    λ> :t myCup
    myCup :: (Integer -> t2) -> t2

and the body looks like this

    \f -> f 6

`\f -> f 6`  is different from  `(\f -> f) 6`!


> Now a "method"
> 
> getOz aCup = aCup (\foz -> foz)
> 
> creates a closure on the lambda function (\foz -> foz) . So upon calling
> 
> > getOz myCup
> 6
> 
> I'm guessing myCup (\foz -> foz) was evaluated, but I don't understand how
> the 6 that went in as the bound variable in the constructor came out again
> with getOz.

To recap, `myCup` expands to:

    myCup
    cup 6
    \message -> message 6

Now, applying `getOz` to it…

    getOz myCup
    myCup (\foz -> foz)
    (\message -> message 6) (\foz -> foz)
    (\foz -> foz) 6
    6