Alignment Address Calculation in AHB

The 3 examples in Vanhealsing's reply from 2 years ago are all correct and meet all the address alignment requirements of the protocol.

With an 8-beat 4 byte burst you have 32-bytes total to transfer (0x20 in hex), so the "wrap_boundary" the burst will wrap down to in example 3 is 0x20, as the burst will wrap if an address increment were to hit the next higher wrap boundary at 0x40.

If there is something that worries you here, can you expand on your concern ?

The 3 examples in Vanhealsing's reply from 2 years ago are all correct and meet all the address alignment requirements of the protocol.

With an 8-beat 4 byte burst you have 32-bytes total to transfer (0x20 in hex), so the "wrap_boundary" the burst will wrap down to in example 3 is 0x20, as the burst will wrap if an address increment were to hit the next higher wrap boundary at 0x40.

If there is something that worries you here, can you expand on your concern ?