> What about my doubts on why it works so lazily?

Suppose 'f' is the function '(1 :)', and 'g' is some other unspecified other
function of type '[Integer] -> [Integer]' ("difference list").

You may evaluate the concatenation '(f . g) []' like this:

(f . g) []
  = f (g [])
  = 1 : g []

and at this point, you already have partial knowledge of the resulting list.

Note that we use "lazy evaluation", in the sense that we do not evaluate the
argument 'g []' to the function 'f' right away, but proceed by first
substituting the definition of 'f'.

_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe