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CSL718 : Superscalar Processors. Speculative Execution 2nd Feb, 2006. Handling Control Dependence. Simple pipeline Branch prediction reduces stalls due to control dependence Wide issue processor Mere branch prediction is not sufficient
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CSL718 : Superscalar Processors Speculative Execution 2nd Feb, 2006 Anshul Kumar, CSE IITD
Handling Control Dependence • Simple pipeline • Branch prediction reduces stalls due to control dependence • Wide issue processor • Mere branch prediction is not sufficient • Instructions in the predicted path need to be fetched and EXECUTED (speculated execution) Anshul Kumar, CSE IITD
What is required for speculation? • Branch prediction to choose which instructions to execute • Execution of instructions before control dependences are resolved • Ability to undo the effects of incorrectly speculated sequence • Preserving of correct behaviour under exceptions Anshul Kumar, CSE IITD
Types of speculation • Hardware based speculation • done with dynamic branch prediction and dynamic scheduling • used in Superscalar processors • Compiler based speculation • done with static branch prediction and static scheduling • used in VLIW processors Anshul Kumar, CSE IITD
i i x x x f x f Extending Tomasulo’s scheme for speculative execution • Introduce re-order buffer (ROB) • Add another stage – “commit” Normal execution • Issue • Execute • Write result Speculative execution • Issue • Execute • Write result • Commit Anshul Kumar, CSE IITD
Extending Tomasulo’s scheme for speculative execution – contd. • Write results into ROB in the “write result” stage • Write results into register file or memory in the “commit” stage • Dependent instructions can read operands from ROB • A speculative instruction commits only if the prediction is determined to be correct • Instructions may complete execution out-of-order, but they commit in-order Anshul Kumar, CSE IITD
Recall Tomasulo’s scheme ...... Anshul Kumar, CSE IITD
Issue • Get next instruction from instruction queue • Check if there is a matching RS which is empty • no: structural hazard, instruction stalls • yes: issue the instruction to that RS • For each operand, check if it is available in RF • yes: put the operand in the RS • no: keep track of FU that will produce it Anshul Kumar, CSE IITD
Execute • If one or more operands not available, wait and monitor CDB • When an operand becomes available, it is placed in RS • When all operands are available, start execution • Choice may need to be made if multiple instructions become ready at the same time Anshul Kumar, CSE IITD
Write result • When result is available • write it on CDB and • from there into RF and relevant RSs • Mark RS as available Anshul Kumar, CSE IITD
More formal description ...... Anshul Kumar, CSE IITD
RS and RF fields Anshul Kumar, CSE IITD
Issue • Get instruction <op, rd, rs, rt> from instruction queue • Wait until r RS[r].busy = no • if (RF[rs].Qi 0) {RS[r].Qj RF[rs].Qi} else {RS[r].Vj RF[rs].val; RS[r].Qj 0} • similarly for rt • RS[r].op op; RS[r].busy yes; RF[rd].Qi r Anshul Kumar, CSE IITD
Execute • Wait until RS[r].Qj = 0 and RS[r].Qk = 0 • Compute result: operation is RS[r].op, operands are RS[r].Vj and RS[r].Vk Anshul Kumar, CSE IITD
Write result • Wait until execution complete at r and CDB available • x if (RF[x].Qi = r) {RF[x].val result; RF[x].Qi 0} • x if (RS[x].Qj = r) {RS[x].Vj result; RS[x].Qj 0} • similarly for Qk / Vk • RS[r].busy no Anshul Kumar, CSE IITD
Tomasulo’s scheme plus ROB...... Anshul Kumar, CSE IITD
Issue • Get next instruction from instruction queue • Check if there is a matching RS which is empty and an empty slot in ROB • no: structural hazard, instruction stalls • yes: issue the instruction to that RS and mark the ROB slot, also put ROB slot number in RS • For each operand, check if it is available in RF or ROB • yes: put the operand in the RS • no: keep track of FU that will produce it Anshul Kumar, CSE IITD
Execute (no change) • If one or more operands not available, wait and monitor CDB • When an operand becomes available, it is placed in RS • When all operands are available, start execution • Choice may need to be made if multiple instructions become ready at the same time Anshul Kumar, CSE IITD
Write result • When result is available • write it on CDB with ROB tag and • from there into ROB RF and relevant RSs • Mark RS as available Anshul Kumar, CSE IITD
Commit (non-branch instruction) • Wait until instruction reaches head of ROB • Update RF • Remove instruction from ROB Anshul Kumar, CSE IITD
Commit (branch instruction) • Wait until instruction reaches head of ROB • If branch is mispredicted, • flush ROB • Restart execution at correct successor of the branch instruction • else • Remove instruction from ROB Anshul Kumar, CSE IITD
More formal description ...... Anshul Kumar, CSE IITD
RS fields Anshul Kumar, CSE IITD
RF fields Anshul Kumar, CSE IITD
ROB fields Anshul Kumar, CSE IITD
Issue • Get instruction <op, rd, rs, rt> from instruction queue • Wait until r RS[r].busy=no and ROB[b].busy=no, where b = ROB tail • if (RF[rs].busy) {h RF[rs].Qi; if (ROB[h].rdy) {RS[r].Vj ROB[h].val; RS[r].Qj 0} else {RS[r].Qj h} } else {RS[r].Vj RF[rs].val; RS[r].Qj 0} • similarly for rt • RS[r].op op; RS[r].busy yes; RS[r].Qi b • RF[rd].Qi b; RF[rd].busy yes; ROB[b].busy yes • ROB[b].inst op; ROB[b].dst rd; ROB[b].rdy no Anshul Kumar, CSE IITD
Execute (no change) • Wait until RS[r].Qj = 0 and RS[r].Qk = 0 • Compute result: operation is RS[r].op, operands are RS[r].Vj and RS[r].Vk Anshul Kumar, CSE IITD
Write result • Wait until execution complete at r and CDB available • b RS[r].Qi; RS[r].busy no • x if (RF[x].Qi = r) {RF[x] result; RF[x].Qi 0} • x if (RS[x].Qj = b) {RS[x].Vj result; RS[x].Qj 0} • similarly for Qk / Vk • ROB[b].rdy yes; ROB[b].val result Anshul Kumar, CSE IITD
Commit (non-branch instruction) • Wait until instruction reaches head of ROB (entry h) and ROB[h].rdy = yes • d ROB[h].dst • RF[d].val ROB[h].val • ROB[h].busy no • if (RF[d].Qi = h) {RF[d].busy no} Anshul Kumar, CSE IITD
Commit (branch instruction) • Wait until instruction reaches head of ROB (entry h) and ROB[h].rdy = yes • If branch is mispredicted, • clear ROB, RF[ ].Qi • fetch branch dest • else • ROB[h].busy no • if (RF[d].Qi = h) {RF[d].busy no} Anshul Kumar, CSE IITD