TMS570LC4357: How to generate values for setting up the MAC Hash registers

Hi,

I looked at the LWIP demo code, we have 

void EMACFrameSelect(uint32 emacBase, uint64 hashTable)
{
HWREG(emacBase + EMAC_MACHASH1) = (uint32)(hashTable & 0xFFFFFFFFU);
HWREG(emacBase + EMAC_MACHASH2) = (uint32)(hashTable >> 32U);
}

But this function is never been called. The example is not multicast. Are you trying to do multicast in your case?

From reading the 570LC4357 TRM, 32.5.37 MAC Hash Address Register 1 (MACHASH1) 

You have a MAC address this is 48-bit, then you can calculate the hash_function with those bits:

Hash_fun(0)=DA(0) XOR DA(6) XOR DA(12) XOR DA(18) XOR DA(24) XOR DA(30) XOR DA(36) XOR DA(42);
Hash_fun(1)=DA(1) XOR DA(7) XOR DA(13) XOR DA(19) XOR DA(25) XOR DA(31) XOR DA(37) XOR DA(43);
Hash_fun(2)=DA(2) XOR DA(8) XOR DA(14) XOR DA(20) XOR DA(26) XOR DA(32) XOR DA(38) XOR DA(44);
Hash_fun(3)=DA(3) XOR DA(9) XOR DA(15) XOR DA(21) XOR DA(27) XOR DA(33) XOR DA(39) XOR DA(45);
Hash_fun(4)=DA(4) XOR DA(10) XOR DA(16) XOR DA(22) XOR DA(28) XOR DA(34) XOR DA(40) XOR DA(46);
Hash_fun(5)=DA(5) XOR DA(11) XOR DA(17) XOR DA(23) XOR DA(29) XOR DA(35) XOR DA(41) XOR DA(47);

Your output is a 6-bit Hash_fun(5 ... 0), totaling 2^6=64 index range. 

Then you have  MACHASH2 (bit32-63) and MACHASH1 (bit31-0) programmed, each bit can be 0 or 1. If indexing Hash_fun(5...0) into this table and that bit position is 0, then the MAC is rejected. If it is 1, it is accepted. 

As you want to accept two MAC addresses, you need to compute the Hash_fun(5...0) for both addresses and make that index position = 1 in the hash table. 

Regards, Eric

Hi Eric,

Yup, I am trying to use Multicast filtering. We have a bunch of sources that are multicasting, and in order to limit the number of packets we need to process I am attempting to use this H/W filter.

Based on the above approach you mentioned, does this mean, that I can only allow only packets from 64 unique Multicast addresses ? 

i.e MAC Address: 01:00:5E:00:00:00 and MAC address  01:00:5E:00:00:41 would hash to the same bit i.e bit 0xF (15). 

As  for  01:00:5E:00:00:00 -> we get 

Hash_fun(0) = 0 (MSB)

Hash_fun(1) = 0

Hash_fun(2) = 1

Hash_fun(3) = 1

Hash_fun(4) = 1

Hash_fun(5) = 1 (LSB)

And also for 01:00:5E:00:00:41 -> we get 

Hash_fun(0) = 0  (MSB)

Hash_fun(1) = 0

Hash_fun(2) = 1

Hash_fun(3) = 1

Hash_fun(4) = 1

Hash_fun(5) = 1    (LSB)

Which would mean, if i set bit 15 in the MAC hash registers, packets from both address would come through ?