CAN ExtID filter for exactly two IDs

[Anno]

Dear Community,

even after reading what feels like 100 posts, I still can't manage to set the filter so that exclusively 2 IDs are let through.

I use the extendet ID, 1 Byte IDs and have managed to let exactly one ID through with the MASK_MODE, as you can see here:

canfilterconfig.FilterActivation = CAN_FILTER_ENABLE;
canfilterconfig.FilterFIFOAssignment = CAN_FILTER_FIFO0;
canfilterconfig.FilterBank = 1;
canfilterconfig.FilterMode = CAN_FILTERMODE_IDMASK;
canfilterconfig.FilterScale = CAN_FILTERSCALE_32BIT;

canfilterconfig.FilterIdHigh = CAN_ModuleAddr<<3;
canfilterconfig.FilterIdLow = 0;

canfilterconfig.FilterMaskIdHigh = 0xF<<3; //Only the Address Bits will be compared
canfilterconfig.FilterMaskIdLow = 0x0000;

canfilterconfig.FilterFIFOAssignment = CAN_RX_FIFO0;
canfilterconfig.SlaveStartFilterBank = 1;

Now I not only want to let the CAN_ModuleAddr through, but also the ID 0xFF.

I would be infinitely grateful for any tips, illustrative material or further help.

Warmest regards
Anno

[SofLit]

Hello @Anno ,

Here you need to configure the filter in ID list mode.

One ID: Mask mode. More than one specific ID:  ID list.

See this thread.

[SofLit]

I didn't test it, but try it:

#define REMOTE_FRAME   0 /* If = 1 the frame should be a remote frame. If = 0 the frame will be a data frame */
#define EXTID          1 /* If = 0 the frame should be a frame with standard ID. If = 1 the frame should be a frame with extended ID */

#define ID1   0x12345678
#define ID2   0x1ABCDEF0

#define FILTER_ID    ((ID1 << 3) | (REMOTE_FRAME<<1) | (EXTID <<2))  // ID1 = 0x12345678 (FxFR1[31:0])
#define FILTER_MASK  ((ID2 << 3) | (REMOTE_FRAME<<1) | (EXTID <<2))  // ID2 = 0x1ABCDEF0 (FxFR2[31:0])
  
  sFilterConfig.FilterMode = CAN_FILTERMODE_IDLIST;
  sFilterConfig.FilterScale = CAN_FILTERSCALE_32BIT;
  sFilterConfig.FilterIdHigh = (FILTER_ID >> 16);          // Filter ID2 LSB   (FxFR2[0:15])
  sFilterConfig.FilterIdLow = (FILTER_ID & 0xFFFF);        // Filter ID1 LSB   (FxFR1[0:15])
  sFilterConfig.FilterMaskIdHigh = (FILTER_MASK >> 16);    // Filter ID2 MSB   (FxFR2[16:31])
  sFilterConfig.FilterMaskIdLow = (FILTER_MASK & 0xFFFF);  // Filter ID1 MSB   (FxFR1[16:31])

 It receives two extended IDs: 0x12345678 and 0x1ABCDEF0

[Anno]

Hey, thank you for your fast reply.

I tested your code and it is not working. Sending messages to 0x12345678 or 0x1ABCDEF0 won't interrupt through the callback function.

I use the other Bits of the Extendet ID as commands, as follow 

            |ID             | |Command                          |
|0|0|0|0|0| |0|0|0|0|0|0|0|0| |0|0|0|0|0|0|0|0| |0|0|0|0|0|0|0|0|

These are variable. Wouldn't I have to put a mask over the list so that only the ID bits are taken into account?

[SofLit · Accepted solution]

Hello,

It works well on an STM32F769 NUCLEO board:

This is my IAR screenshot and I'm receiving the two IDs:

Attached the project with which I got this result.

So please double check your CAN config as well as your filter config. Are you using CAN2?

[Anno]

Thank you, I will have a look.

Yes I am using CAN2 and can not change it.

[SofLit]

Using CAN2 is another story ..

1- You need to enable CAN1 RCC clock (__HAL_RCC_CAN1_CLK_ENABLE()) even if it was not used.

2- Regarding the filtering you need to take care about these two parameters:

.SlaveStartFilterBank and .FilterBank.

Please read the attached text file.

[Anno]

With your explanation and the knowledge that the behavior is different with CAN2, I tried the example you gave above and it works. Many thanks for that!

Only the two extended IDs are really allowed through here. How can I adjust the filter here so that only a certain area of the bits is looked at? Just as I mentioned at the beginning.

No matter how I adjust the filter mask, it doesn't work. Is there something I'm missing?
Do I have to put a submask over it again at the end?

Many thanks for your help and a happy new year

[SofLit]

As your first question was answered this thread is closed, please open a new thread for your latest question.

Thank you for your understanding.