> This is exactly how it is supposed to work! That's why there is a loop > in the atomic increment - you check if you still had the reservation > after the transaction by checking the result from the stwcx, and if not, > retry.
I surely know all the theories you mentioned clearly. But please do look at the case I gave. Correct me if I missed anything. Thanks All the lwarx and stwcx operate on the same address. > Task A Task B > lwarx // Get RESERVATION > ...... > lwarx > stwcx // RESERVATION cleared > > ..... > ..... > lwarx // Get RESERVATION again > stwcx //Note here: RESERVATION is valid, address is the same. So the result is commited, no retry for task A > ..... > stwcx //RESERVATION is cleared, retry atomic op for task B Please be noted that reservation is only identified by reservation bit and address operated on. So different lwarx's on the same address, may be considered as the same reservation.
