maths functions in interrupt routime

[Bernard Mentink]

I am having trouble with doing a 'mod' instruction in an interrupt routine. It crashes FF without a reset cause being printed to the terminal.
The code looks like this:

: my_int
  1 esc.seq.step  +!              \ increment step by 1, next of the seq of 6
    esc.seq.step @ 6 mod esc.seq.step !      \ modular 6 division
  ;i

Any reason why this should not work? The inetrrupt is the comparator int.
Do you have to disable comp interrupt on entry to the routine?

[Mikael Nordman]

There was a bug in UM/MOD which is the base for all division and MOD routines. It has now been made interrupt safe.
A fix has been uploaded.
I recommend using a simple IF statement instead of MOD in the interrupt routine. That is much faster.

: my_int
  1 esc.seq.step  +!              \ increment step by 1, next of the seq of 6
    esc.seq.step @ 5 > if 0 esc.seq.step ! then
  ;i

If you have to use MOD I recommend using U/MOD or UM/MOD which are faster than MOD.

[Bernard Mentink]

Many thanks. I had already changed to an IF..THEN
So latest changes in github I presume. EDIT found it :)