Henning Thielemann
I figure it's (constant vs. linear) vs. (linear vs. quadratic), for more involved examples.
I can't see it. If I consider (x++y) but I do not evaluate any element of (x++y) or only the first element, then this will need constant time. If I evaluate the first n elements I need n computation time units. How is (.) on difference lists faster than (++) here?
It's in multiple calls to length if you do ((x++y)++z), the first run over x can be avoided. It basically gets rewritten to (x++y++z) by another level of abstraction. -- (c) this sig last receiving data processing entity. Inspect headers for past copyright information. All rights reserved. Unauthorised copying, hiring, renting, public performance and/or broadcasting of this signature prohibited.