How slow can division really be?

theteaman · Joined: 04 Aug 2006 Posts: 98

Hi

I've got timer0 set up as setup_timer_0(RTCC_DIV_16 | RTCC_8_BIT);

In my interrupt hander for this timer, I have practically nothing, just a few conditional flag setting statements.

When i compile my code I get the good ol' "interrupts have been disabled to prevent re-entrance @DIV3232" warning. But the problem is my interrupt handler is not calling any functions. The offending line is actually a division in another part of the program that is not called, in any way, by the clock interrupt - it's in the main loop.

When I change the division to an addition the warning disappears.

Does anyone know why this is happening?

Thanks

ps. i am dividing int32s.

Ttelmah · Guest

Post the contents of your interrupt handler.
You won't get this message, unless you _are_ calling a division. The 'odds' are that something you don't think uses division, does.

Best Wishes

theteaman · Joined: 04 Aug 2006 Posts: 98

Here it is:

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

See this post which explains how to create an 2nd instance of the math
library code, to avoid getting that compiler warning:
http://www.ccsinfo.com/forum/viewtopic.php?t=25464&start=4

theteaman · Joined: 04 Aug 2006 Posts: 98

Thanks PCM, I will give it a try.

Ttelmah · Guest

It is also worth working out if you can perform this some other way.
The 'modulus' function, is a division. Modulus is the remainder from a division operation, and hence involves a division. Since the compiler refers to div32, it implies at least one of the numbers is a 32bit value, which implies you are talking about something in excess of 500 machine instructions, for this one operation... :(
Now if (for instance), 'MOD_CLK_RES', could be selected to be a binary value (8, 16 etc.), then you can get the same result, using:

if(int_count & (MOD_CLK_RES-1) == 0)

Which takes less than a dozen instructions.
Also, if this is a counter, and you want to know when it is divisible by (say) 60, to give a zero result, it is about 50 times quicker, to just run a second counter. So: