myLength1 = foldl (\n _ -> n + 1) 0

and this 

myLength2 = foldr (\_ n -> n + 1) 0

I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to the independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well

 :t foldl
foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b

LB