|Re: (Vieira) " IFT " command problem . . .|
Message #4 Posted by Vieira, Luiz C. (Brazil) on 24 Jan 2002, 8:02 a.m.,
in response to message #3 by Tal
Let´s consider that when the program is running, the IFT structure will always place the object to be evaluated (in this case the program, that can be small or big) in the stack (normaly, Level 1). Then, after testing the flag (in Level 2), this object will be evaluated or not. The IF..THEN...(ELSE...)END structure will execute (evaluate) each command inside the THEN clause at a time, needing no extra time (some nano-seconds?) to load them all in the stack prior to their evaluation (execution).
To be honest, I cannot ´guess´ which is the fastest. The sequence to generate the flag - k 16 == i 22 > j 32 < AND i 66 > j 46 == AND OR AND - will be time consuming, too. I am not sure, but for this case it will be the same.
If you need to be sure about it, create a 1000-count loop (say, << 1 1000 START procedure NEXT >>) and test for both procedures. I would do that, you bet.
Sorry not helping much more.