* 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).
That's right. step x g a = g (f a x) is, thanks to currying, another way to write step x g = \a -> g (f a x) This is what we want -- function of two arguments, which returns a new value of accumulator (which in this case is a function itself). And g (f a x) can be rewritten as g ((f a) x) = g (a `f` x) = g ((`f` x) a) = (g . (`f` x)) a, thus step x g = \a -> g (f a x) = \a -> (g . (`f` x)) a = g . (`f` x) (the last step is called 'eta-reduction'). Remember that g is a previous value of accumulator and step x g is its new value, so on each step we compose accumulator with (`f` x) to get the new value. So in the end it will look like you wrote above.
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
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