My understanding is that issueLat is the minimum number of cycles you
have to wait before scheduling an instruction of the same type on the
FU. Specifically, if opLat is 6 and issueLat is 1, you have a pipelined
unit with a latency of 6 cycles but you have a throughput of 1 op/cycle
once the pipe is full.
The delay between dispatch and issue indeed depends on how soon the
operands become ready wrt to contention (number of FUs, issue width).
The scheduler schedules oldest instruction first without regard for the
instruction type. Furthermore, there is not speculative scheduling,
meaning that if you increase issueToExecute to more than one, you cannot
schedule depending instructions back to back (correct me if I'm wrong).
Hope it helps.
Arthur.
Le 17/07/2013 03:03, Jianghao a écrit :
Take the following from config.ini file, my understanding is here we
defined 2 function units, both of them can execute IntMult and IntDiv
type of instructions, and opLat defines how many ticks need to execute
that instruction. Please correct me if my understanding is wrong.
My confusion comes from the meaning of issueLat. Does it define the
latence from dispatch to issue? From my understanding about O3, after
instruction decoding/ renaming, if operands are ready, instruction can
be issued for execution. So that delay from dispatch to issue should
not be constant, which depends on the number of ready instructions and
the available execution units. Can anybody give me some hints
regarding gem5 implementation of the o3 scheduler part? Really
appreciate for any help.
[system.cpu.fuPool.FUList1]
type=FUDesc
children=opList0 opList1
count=2
opList=system.cpu.fuPool.FUList1.opList0
system.cpu.fuPool.FUList1.opList1
[system.cpu.fuPool.FUList1.opList0]
type=OpDesc
issueLat=1
opClass=IntMult
opLat=3
[system.cpu.fuPool.FUList1.opList1]
type=OpDesc
issueLat=12
opClass=IntDiv
opLat=12
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