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AHB WRAP address boundaries

myarm myarm
Offline over 6 years ago
Note: This was originally posted on 18th June 2008 at http://forums.arm.com

AMBA spec (v2.0) only shows how the addresses wrap when hsize = 2 (word). Is it because the address boundary remains the same for each WRAP4, WRAP8, and WRAP16 cases? Or, should I re-calculate the wrap boundary based on hsize?

For example, I know from the spec that hsize = 2 and WRAP4 will sequence through addresses like this:
0x4 - 0x8 - 0xC - 0x0

Then, does hsize = 1 (halfword) and WRAP4, mean the sequence should be
0x4 - 0x6 - 0x8 - 0xA    (using 4 word boundary)
or
0x4 - 0x6 - 0x0 - 0x2     (using 4 halfword boundary)?

Thanks!
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  • 0 Offline over 6 years ago
    Note: This was originally posted on 2nd December 2008 at http://forums.arm.com

    As Hemamth case 1 : If Start addr is 0x34, WRAP4, HSize is 2
    Total no of bytes is 16
    Therefore 4 bit alignment is there

    So 4 consecutive addr are :

    0011 0100 - beat1 - 0x34
    0011 1000 - beat2 - 0x38
    0011 1100 - beat3 - 0x3c
    +0000 0100
    -------------------
    0100 0000 ---> But here 6th bit goes high, And bit alignment is 4??? -- on the basis of what we can say that its crossing the boundry?? Please clear me for the same... Thanks

    we have to aligning it  to 0011 0000 - 0x30
    0011 0000 - beat4 - 0x30
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  • 0 Offline over 6 years ago
    Note: This was originally posted on 2nd December 2008 at http://forums.arm.com

    As Hemamth case 1 : If Start addr is 0x34, WRAP4, HSize is 2
    Total no of bytes is 16
    Therefore 4 bit alignment is there

    So 4 consecutive addr are :

    0011 0100 - beat1 - 0x34
    0011 1000 - beat2 - 0x38
    0011 1100 - beat3 - 0x3c
    +0000 0100
    -------------------
    0100 0000 ---> But here 6th bit goes high, And bit alignment is 4??? -- on the basis of what we can say that its crossing the boundry?? Please clear me for the same... Thanks

    we have to aligning it  to 0011 0000 - 0x30
    0011 0000 - beat4 - 0x30
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