Log File says another reaction as hardware

Bome Products
21

On the First thing you wrote this…
// Column 1 Channel 1
// bits 0-6 is First Row
// bits 7-13 is Second Row
// bits 14-20 is Third Row
// bits 21-27 id Fader

but where are the bits from 28-31?

22

OK, in your channel switcher presst

change this rule

// set channel switch flags FFF FFFF
zz=268435455

 

to this

 

// set channel switch flags FFFF FFFF
zz=-1

 

23

OK, in your channel switcher presst

change this rule

// set channel switch flags FFF FFFF
zz=268435455

 

to this

 

// set channel switch flags FFFF FFFF
zz=-1

 

24

That was for the bitmap for the values of the pots/faders, not channel switch.

g = 1 vv=0-3
h = 2 vv=4-7
i = 3 vv=8-11
j = 4 vv=12-15
k = 5 vv=16-19
l = 6 vv=20-23
m = 7 vv=24-27
n = 8 vv=28-31

25

Seems that it runs correct. Thank you so much for your help.
Hope, no nothing can breaks with this.

Only one thing is a little bit other, reaction from channel switch is a bit lower after changing the switcher variable. But it runs correct, this is the big thing 🙂

kindest regards
Mike

26

Yep, I was only setting invalid flag on 28 of the 32 bits. I probably got confused between the two bitmaps we are using as well.

Steve

27

So the other setting from the big number was the high bitmapping part you have calculated?
and this one with
FFFF FFFF
zz=-1
Is the normal bitmapping?
why is here one F more than on the other part?

28

FFF FFFF is 28 bits of 1
FFFF FFFF is 32 bits 1 and is represented as a signed integer as -1. The high order bit is the negative sign.

29

Nice.
Thank you for your explaining and solve the project.

So is F = 4 bits FF = 8 bits and so on.
this would be a small part of calculating the higher bit mapping.

30

Yes, each F is 4 bits or 1111 in binary.

31

Thank you Steve

32

You are welcome, Mike!

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