wrapping burst in AHB

[sun_ray]

In the AHB specification it is written that:

"A four-beat wrapping burst of word (4-byte) accesses wraps at
16-byte boundaries. Therefore, if the start address of the transfer is 0x34, then it
consists of four transfers to addresses 0x34, 0x38, 0x3C, and 0x30."

My question about the above quote is as follows:

If 16-byte is the wrap boundary, how in this case the address wraps at 0x3c as in this case the 12 byte of data has been transferred and not the 16 byte. So 0x3C is not the 16-byte boundary from which the address should wrap. Can anyone please clarify?

 

[wangkl]

address will be wrapped from 0x3c to 0x30, and last 4 byte with 0x30 address.

0x34( byte0~byte3)
0x38 (byte4~byte7)
0x3c (byte8~byte11)
0x30 (byte12~byte15)

 

[yx.yang]

Hi, just thinking that: when it's 16bytes address wrap, then just the low 4bits address will change. And all other higher address bits will not change. Then, when you begin at 0x34, and goest to 0x38, 0x3C, what's the next? Of case x030 (higher address bit can't change). 0x40 is wrong, as the higher address bits changes.

Another example: for 32 bytes wrap begin at 0x3C, what you will get?
The rule is: just the low 5bits of address can changes. So, we get: 0x3C, 0x20, 0x24, 0x28, 0x2C, 0x30, 0x34, 0x38.
Wish you get the point.

 

[sun_ray]

- - - Updated - - -

address will be wrapped from 0x3c to 0x30, and last 4 byte with 0x30 address.

0x34( byte0~byte3)
0x38 (byte4~byte7)
0x3c (byte8~byte11)
0x30 (byte12~byte15)

In that case there is another example of WRAP8 in the specification file. There it writes as follows for another WRAP8 transfer:

"Because the burst is an eight-beat burst of word transfers, the address wraps at 32-byte boundaries, and the transfer to address 0x3C is followed by a transfer to address 0x20."

The address sequence in the above case is 0x34, 0x38, 0x3c, 0x20 as shown in the timing diagram in the specification.

How will you explain this?

Regards