# Exploring Writing foldl Using foldr

Jan 27, 2016

It is widely known that left folding can be implemented by right folding. But the implementation is not so obvious. Let’s try writing one from scratch. First, we need these two extensions of GHC to make life better.

```
{-# LANGUAGE ScopedTypeVariables, PartialTypeSignatures #-}
```

Let’s just ignore the original implementations in `Prelude`

and try writing them by hand:

```
import Prelude hiding (foldr, foldl)
foldr :: (b -> a -> a) -> a -> [b] -> a
foldr _ r [] = r
foldr f r (x:xs) = f x (foldr f r xs)
foldl :: (a -> b -> a) -> a -> [b] -> a
foldl f r [] = r
foldl f a (x:xs) = foldl f (f a x) xs
```

If we want to write `foldl`

using `foldr`

,
that defination should looks like this:

```
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f i bs = foldr undefined undefined undefined
```

The only thing we can be sure is that we will still be working on list `bs`

.
So let’s refine the above one with this observation:

```
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f i bs =
let fold :: _ -> _ -> [b] -> _
fold = foldr
in fold undefined undefined bs
```

Let’s think about a small example. According to our original implantation,
`foldl f i [b1, b2]`

should be expanded into `f (f i b1) b2`

but the form in `foldr g j [b1, b2]`

is `g b1 (b g2 j)`

.
The most important step is the definition of `g`

and `j`

.
The function `g`

should have type looks like `b -> r -> r`

and
if the first parameter is `b1`

, the second parameter (`r`

) should
be able to use `f i b1`

in itself to re-build the original `foldl`

expansion.
That sounds like a continuation `a -> r`

instead of just `r`

.
Let’s try this idea:

```
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f i bs =
let fold :: (b -> (a -> r) -> (a -> r)) -> (a -> r) -> [b] -> (a -> r)
fold = foldr
g :: b -> (a -> r) -> (a -> r)
g b1 cont = cont (f i b1)
in fold g undefined bs
```

Bad thing happens.
We get two type errors because now `g`

and `fold`

are of type ending with `(a -> r)`

.
Let’s look into them one by one.
For `g`

, we are using a `b`

to turn a continuation into another continuation.
So `g`

actually need a extra parameter `a`

.
And `fold`

return a continuation and need an initial value `a`

to get the final result.
Let’s refine the signature and definition:

```
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f i bs =
let fold :: (b -> (a -> r) -> (a -> r)) -> (a -> r) -> [b] -> (a -> r)
fold = foldr
g :: b -> (a -> r) -> (a -> r)
g b1 cont a = cont (f i b1)
in fold g undefined bs undefined
```

Now our compiler warns us that `a`

is not used.
According to its type, the only place to use it is where `i`

is.
This makes sense because we are using the future result in our computation.
But where `i`

should go? The only place seems to using it to complete our `fold`

.
(I think this is the only tricky part and hope to find better thoughts on this in the future.)
Let’s refine it again:

```
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f i bs =
let fold :: (b -> (a -> r) -> (a -> r)) -> (a -> r) -> [b] -> (a -> r)
fold = foldr
g :: b -> (a -> r) -> (a -> r)
g b1 cont a = cont (f a b1)
in fold g undefined bs i
```

We are almost done and only one `undefined`

is left.
Notice that the final result is of type `r`

but our target type is `a`

.
That just means `r`

is `a`

.
So the only left `undefined`

is of type `forall a. a -> a`

and its only implantation is `id`

according to parametricity.
This also fits our experience with continuation by finishing it with `\a -> a`

.
Our final version is:

```
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f i bs =
let g b1 cont a = cont (f a b1)
in foldr g id bs i
```

It works!

```
λ> foldl (\r x -> r ++ [x]) [0 :: Int] [1 .. 10]
[0,1,2,3,4,5,6,7,8,9,10]
```