>>>I guess you could simplify the expression above (in VFP) to ceiling(log(N) / log(2)).
>>
>>Hmmm, are we running some sort of contest of getting the shortest code again :) ?
>>
>>Walter,
>
>Not really, but I do like to look for shorter ways of coding different things.
>
>OK, here is one "contest": you have to run three functions one after the other. func2() should only run if func1() is successful, and func3(), again, depends on the successful completion of func2(). Do you see an alternative that is shorter than two nested IFs?
>
>Hilmar.

return func1() and func2() and func3()