[Haskell-cafe] Would you mind explain such a code ?

[zaxis]

thanks for your quick answer!
 
As I understand  foldr (\x g -> g . (`f`x)) id xs  will return a function
such as (`f` 3).(`f` 2).(`f` 1) . You have already made it clear !  However, 
why  does the "step" function below has three parameters ? I think foldr
will call step using two parameters, the 1st is list element and the 2nd is
a funtion whose initial value is id).

myFoldl f z xs = foldr step id xs z
    where step x g a = g (f a x)


staafmeister wrote:
> 
> 
> 
> zaxis wrote:
>> 
>> myFoldl :: (a -> b -> a) -> a -> [b] -> a
>> 
>> myFoldl f z xs = foldr step id xs z
>>     where step x g a = g (f a x)
>> 
>> I know myFoldl implements foldl using foldr. However i really donot know
>> how it can do it ?
>> 
>> Please shed a light one me, thanks!
>> 
> 
> Hi,
> 
> Nice example! Well this is indeed an abstract piece of code. But basically
> foldl f z xs starts with z and keeps applying (`f`x) to it
> so for example foldl f z [1,2,3] = ((`f`3).(`f`2).(`f`1)) z
> 
> Because functions are first-class in haskell, we can also perform a foldl
> where instead of calculating the intermediate values we calculate the
> total function, i.e. ((`f`3).(`f`2).(`f`1)) and apply it to z.
> 
> When the list is empty z goes to z, so the start function must be id.
> So we can write
> (`f`3).(`f`2).(`f`1) = foldr (\x g -> g . (`f`x)) id xs
> 
> This is almost in your form.
> 
> Hope this helps,
> Gerben
> 

-- 
View this message in context: http://www.nabble.com/Would-you-mind-explain-such-a-code---tp25377949p25378882.html
Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.