IF write or read reordering depth is 4, does this mean that the transaction coming with ARIDs for ex 3, 2, 1, 0 can be re ordered and give response for 0,1,2,3 in this order ?
what is the case when reordering depth is 1?...like above case example why it completes in order?
if there 10 transactions and depth(read or write reordering) is 4 then what happens?...…. i am thinking like first any 4 four transactions will be reordered and again in pending (6) transactions 4 will be reordered? is it correct?
Yes to your first question.
Second question, if reorder depth is 1 it means the slave cannot reorder transactions.
The reordering depth of a slave is the slave's ability to process multiple transactions (using different IDs) at the same time, so that possibly a later started transaction could actually complete before earlier started transactions.
Not sure about your last question. If there are 10 transactions (presumably all with different IDs) targeting a slave with a reorder depth of 4, it will be actively processing those first 4 received transactions, and as one of those 4 transaction completes, the next received transaction (the 5th) will then replace it in the 4-deep reordering capability. The pending transactions remain ordered until they can take a place in the reordering depth capability of the slave.
In simpler slaves you will just process each received transaction in the order it was received (no reordering), so this feature might only be used in a complex slave, perhaps some sort of high performance memory controller.
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