Improve fold slides

VL2/content/VL2_fold2.tex

@@ -1,14 +1,14 @@
 \ifger{Um es kurz zu machen, die abstrakte Lösung ist:}{To cut the story short, the abstract solution looks like this:}
 \begin{haskellcode}
-fold :: b -> (a -> b -> b) -> [a] -> b
-fold z f []     = z
-fold z f (x:xs) = x `f` (fold z f xs)
+fold :: (a -> b -> b) -> b -> [a] -> b
+fold f z []     = z
+fold f z (x:xs) = x `f` (fold f z xs)
 \end{haskellcode}
 Whoa! What's going on here?\\
 \ifger{Schauen wir genauer hin...}{Let's see...}
 \begin{itemizep}
-\item \hinline{z} \ifger{ist was die Funktion zurückgibt, wenn die Liste leer ist}{is what we return if the list is empty}
 \item \hinline{f} \ifger{ist unsere Funktion}{is our function} (\ifger{z.b.}{e.g.} \hinline{(*)} \ifger{oder}{or} \hinline{(+)})
+\item \hinline{z} \ifger{ist was die Funktion zurückgibt, wenn die Liste leer ist}{is what we return if the list is empty}
 \item \ifger{das letzte Argument ist die eigentliche Liste, auf der wir arbeiten}{and the last remaining argument is the actual list we are working on}
 \end{itemizep}
 \slidep

VL2/content/VL2_fold3.tex

@@ -2,13 +2,13 @@
 \pause
 \begin{haskellcode}
 sum :: [Int] -> Int
-sum xs = fold 0 (\x y -> x + y) xs
+sum xs = fold (\x y -> x + y) 0 xs
 -- a Haskeller would write
-sum = fold 0 (+)
+sum = fold (+) 0
 
 prod :: [Int] -> Int
-prod xs = fold 1 (\x y -> x * y) xs
+prod xs = fold (\x y -> x * y) 1 xs
 
 length :: [a] -> Int
-length xs = fold 0 (\x y -> 1 + y) xs
+length xs = fold (\x y -> 1 + y) 0 xs
 \end{haskellcode}