On Sun, Sep 30, 2012 at 10:08:11PM +0200, Alberto Alassio wrote:
Ok, to me it wasn't so easy to understand count < steps. so it counts till the steps end, right?
Yes. As long as "count < steps", which means "count is lower than steps" evaluates to true the loop keeps repeating and each time the loop repeats in increases "count". Once this is no longer true the clause returns false, the loop stops and we reach the end of the function.
And in this patch count refers to steps but it is also the one who makes the freq increases by inc, I mean, for each count we have, then the freq increases. Is it correct?
Yes, totally correct.
Thank you Kas, it always helps!
No problem. The more I look at it the more I suspect this example is doing unusual things on purpose, demonstrating a few things at the same time. It is also demonstrating "function overloading" in a way that I think is designed to make that principle understandable by looking at this code. Note that two functions named "chirp" are defined here.... Yours, Kas.