[Haskell-beginners] understanding curried function calls

[Dimitri DeFigueiredo]

Hi Brent,

Is this how we go about solving these then?
Keep adding parameters on the right until we have enough of them that we 
are able to apply the function definition.

In other words, in
ex1 = doTwice doTwice

the first doTwice needs another parameter, so you added 'y'.
ex1 y = (doTwice doTwice) y = doTwice doTwice y

Then we can apply the definition.
We got stuck again, so you added another parameter 'z'.
Then again applied the definition of the left most function.
And so on.

Is there a similarly systematic approach for the foldl implemented by 
foldr example?
I'm looking for systematic approaches that I can then practice.


Thanks!


Dimitri


Em 20/08/14 14:30, Brent Yorgey escreveu:
> On Wed, Aug 20, 2014 at 02:19:16AM -0600, Dimitri DeFigueiredo wrote:
>> doTwice :: (a -> a) -> a -> a
>> doTwice f x = f (f x)
>>
>> what does this do?
>>
>> ex1 :: (a -> a) -> a -> a
>> ex1 = doTwice doTwice
>>
>> At least, it is clear that there is a parameter to doTwice missing.
>> So, I wanted to do:
>>
>> ex1 y = (doTwice doTwice) y
>>
>> but this gets me nowhere as I don't know how to apply the definition
>> of doTwice inside
>> the parenthesis without naming the arguments.
> Note that function application associates to the left, so
>
>    (doTwice doTwice) y = doTwice doTwice y
>
> So, we have
>
> ex1 y = doTwice doTwice y
>        = doTwice (doTwice y)   -- definition of doTwice
>
> Now we are stuck again; we can add another arbitrary parameter.
>
> ex1 y z = doTwice (doTwice y) z
>          = (doTwice y) ((doTwice y) z)
>          = doTwice y (doTwice y z)   -- remove unnecessary parentheses
>          = y (y (doTwice y z))
>          = y (y (y (y z)))
>
> Does that help?
>
>> What is the systematic way to evaluate these expressions? I actually
>> got really
>> stumped when I considered.
>>
>> ex2 :: (a -> a) -> a -> a
>> ex2 = doTwice doTwice doTwice doTwice
>>
>> I assume this is not the same as
>>
>> ex2 = (doTwice doTwice doTwice) doTwice
> These ARE exactly the same.  It's always the case that
>
>    f w x y z ... = (((f w) x) y) z ...
>
> -Brent
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