[Haskell-beginners] OOP exercise with closures

[Lawrence Bottorff]

Thanks! Since I studied lambda calc a bit I understand the steps

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

But this is so bizarre! To have a "constructor" that creates an instance,
which is then really a holder of a lambda calculation is a mind-bender.

On Tue, Dec 29, 2020 at 4:07 PM Francesco Ariis <fa-ml at ariis.it> wrote:

> 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
>
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