Editing Function (section)

===Composition / folding example===
Haskell supports a [[Function composition]] operator: 
<haskell>
(.) :: (b -> c) -> (a ->b) -> (a->c)
(f . g) x = f (g x) 
</haskell>
So, for example, <hask>((insert 1) . (insert 2) . (insert 3)) mySet</hask> is the same as <hask>insert 1 (insert 2 (insert 3 mySet))</hask>. We're almost there -- what we need now is a function that can automatically put the composition operator between every element of <hask>map insert myList</hask>. Such code is included in Haskell, and it's known as ''folding.''

Several variants of the ''fold'' function are defined, but the basic concept is the same: given a function and a list, "collapse" the list by applying the function "between" its elements. This is easiest to see with simple binary operators, but it is not limited to them. For example, <hask>foldr1 (+) [1,1,2,3,5]</hask> eventually creates the expression <hask>1+1+2+3+5</hask>, and thus returns 12. In the set example, <hask>foldr1 (.) (map insert myList)</hask> gives us what we want, the successive insertion of every element of ''myList.'' What is the type of this expression? It is <hask>Set -> Set</hask>, meaning it will take a set and return a set -- in this case, the set it returns will have every element of ''myList'' inserted. To complete the example, 
<haskell>
newSet = (foldr1 (.) (map insert myList)) mySet 
</haskell>
will define ''newSet'' as ''mySet'' with the elements of ''myList'' inserted.