[TheNine90]

Im looking for a way to disable CC messages using CC-ID codes through coding. Similar to how sos is coded out by changing 5th byte from FF and F7 in CC_AKTIVIERUNG_5.

Is there a way to map the codes to the CC_AKTIVIERUNG in the cafd? Or cc messages that are under specific CC_AKTIVIERUNG block?

[iyade46]

Im looking for a way to disable CC messages using CC-ID codes through coding. Similar to how sos is coded out by changing 5th byte from FF and F7 in CC_AKTIVIERUNG_5.

Is there a way to map the codes to the CC_AKTIVIERUNG in the cafd? Or cc messages that are under specific CC_AKTIVIERUNG block?

what fault you wana disable i can find you what CC-ID for that system

[TheNine90]

I have a complete list of the CC-ID. I want to know how it relates to the fdl and how to work it out. Also how to delete individual CC message.

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You can choose any CC-ID and illustrate how figure out it location in the fdl and to disable it.

[Node]

IMHO general idea is that there is 1 bit for each CC-ID and all of them are sequential.

AFAIK SOS failure is CC-ID 299.

Each group has 8 bytes so 64 (= 8 * 8) bits total.

299 / 64 = 4.xxx so it's 5th group (as groups are counted from 1: 4 + 1 = 5).

299 - 64 * 4 = 43 so it's 43rd bit in 5th group.

43 / 8 = 5.xxx so it's 5th byte (as bytes are counted from 0: 5 + 0 = 5).

43 - 5 * 8 = 3 so it's 3rd bit (as bits are also counted from 0)

3rd bit has value of 8 (= 2 ^ 3) so FF - 8 = F7. Well, to be more precise - works when bit WAS SET ONLY. Normally you would do NOT-AND operation which works always. So initial value FF & (~8). ~8 = F7. FF & F7 = F7.

ELEMENTARY!!!

[TheNine90]

Thanks for the explanation.

[elthox]

IMHO general idea is that there is 1 bit for each CC-ID and all of them are sequential.

AFAIK SOS failure is CC-ID 299.

Each group has 8 bytes so 64 (= 8 * 8) bits total.

299 / 64 = 4.xxx so it's 5th group (as groups are counted from 1: 4 + 1 = 5).

299 - 64 * 4 = 43 so it's 43rd bit in 5th group.

43 / 8 = 5.xxx so it's 5th byte (as bytes are counted from 0: 5 + 0 = 5).

43 - 5 * 8 = 3 so it's 3rd bit (as bits are also counted from 0)

3rd bit has value of 8 (= 2 ^ 3) so FF - 8 = F7. Well, to be more precise - works when bit WAS SET ONLY. Normally you would do NOT-AND operation which works always. So initial value FF & (~8). ~8 = F7. FF & F7 = F7.

ELEMENTARY!!!

Assuming cc-id 256, how this math will work?

256 - 64 * 4 = 0
0 / 8 = 0
0 - 0 * 8 = 0

[Node]

IMHO general idea is that there is 1 bit for each CC-ID and all of them are sequential.

AFAIK SOS failure is CC-ID 299.

Each group has 8 bytes so 64 (= 8 * 8) bits total.

299 / 64 = 4.xxx so it's 5th group (as groups are counted from 1: 4 + 1 = 5).

299 - 64 * 4 = 43 so it's 43rd bit in 5th group.

43 / 8 = 5.xxx so it's 5th byte (as bytes are counted from 0: 5 + 0 = 5).

43 - 5 * 8 = 3 so it's 3rd bit (as bits are also counted from 0)

3rd bit has value of 8 (= 2 ^ 3) so FF - 8 = F7. Well, to be more precise - works when bit WAS SET ONLY. Normally you would do NOT-AND operation which works always. So initial value FF & (~8). ~8 = F7. FF & F7 = F7.

ELEMENTARY!!!

Assuming cc-id 256, how this math will work?

256 - 64 * 4 = 0
0 / 8 = 0
0 - 0 * 8 = 0

So it will be 0'th bit in byte 0
0'th bit value is 2^0 = 1, so FF & (~1) = FE (assuming initial value was FF).

[elthox]

IMHO general idea is that there is 1 bit for each CC-ID and all of them are sequential.

AFAIK SOS failure is CC-ID 299.

Each group has 8 bytes so 64 (= 8 * 8) bits total.

299 / 64 = 4.xxx so it's 5th group (as groups are counted from 1: 4 + 1 = 5).

299 - 64 * 4 = 43 so it's 43rd bit in 5th group.

43 / 8 = 5.xxx so it's 5th byte (as bytes are counted from 0: 5 + 0 = 5).

43 - 5 * 8 = 3 so it's 3rd bit (as bits are also counted from 0)

3rd bit has value of 8 (= 2 ^ 3) so FF - 8 = F7. Well, to be more precise - works when bit WAS SET ONLY. Normally you would do NOT-AND operation which works always. So initial value FF & (~8). ~8 = F7. FF & F7 = F7.

ELEMENTARY!!!

Assuming cc-id 256, how this math will work?

256 - 64 * 4 = 0
0 / 8 = 0
0 - 0 * 8 = 0

So it will be 0'th bit in byte 0
0'th bit value is 2^0 = 1, so FF & (~1) = FE (assuming initial value was FF).

Got it. So in another scenario, 250 would have:
7th group
FF FF FF FF FF FF FF FB

[Node]

IMHO general idea is that there is 1 bit for each CC-ID and all of them are sequential.

AFAIK SOS failure is CC-ID 299.

Each group has 8 bytes so 64 (= 8 * 8) bits total.

299 / 64 = 4.xxx so it's 5th group (as groups are counted from 1: 4 + 1 = 5).

299 - 64 * 4 = 43 so it's 43rd bit in 5th group.

43 / 8 = 5.xxx so it's 5th byte (as bytes are counted from 0: 5 + 0 = 5).

43 - 5 * 8 = 3 so it's 3rd bit (as bits are also counted from 0)

3rd bit has value of 8 (= 2 ^ 3) so FF - 8 = F7. Well, to be more precise - works when bit WAS SET ONLY. Normally you would do NOT-AND operation which works always. So initial value FF & (~8). ~8 = F7. FF & F7 = F7.

ELEMENTARY!!!

Assuming cc-id 256, how this math will work?

256 - 64 * 4 = 0
0 / 8 = 0
0 - 0 * 8 = 0

So it will be 0'th bit in byte 0
0'th bit value is 2^0 = 1, so FF & (~1) = FE (assuming initial value was FF).

Got it. So in another scenario, 250 would have:
7th group
FF FF FF FF FF FF FF FB

Yes. (except 7th byte, not group , in the 4th group)

[elthox]

Hi Node,

How would you operate among the below codes? so basically when there are multiple codes for the same message.

CC-ID 253
Parking Brake Failed
CC-ID 239
Parking Brake Failed
CC-ID 240
Parking Brake Failed
CC-ID 241
Parking Brake Failed
CC-ID 242
Parking Brake Failed
CC-ID 243
Parking Brake Failed