On Thu, 25 Mar 2010, stephane eranian wrote:
> vince,
>
> Could you make your program such it has only 1 billion instructions?
> I want to see if there is some correlation with counter overflows.
> They are actually only 31 bit on Intel.
sure, I've attached the 1 billion instruction version.
I'm also working on a few memory microbenchmarks too to see if I can maybe
find any patterns in the retired_stores cases.
Vince
# Test case with 1 billion instructions
# by Vince Weaver, vweaver1 _at_ eecs.utk.edu
# To build on x86_64
# as -o one_billion.o one_billion.s ; ld -o one_billion one_billion.o
# To build on i686
# as --defsym i386=1 one_billion.o one_billion.s ; ld -o one_billion
one_billion.o
# With deterministic performance counters, this should get you
# 1,000,000,000 retired instructions
# 499,998,996 retired branches
# 0 retired stores
# 0 retired loads
# retired_instructions is
# 2 + (( (inside) + 2) * 1000) + 2 + (996 * 2) + 4
# inside = ( 2 + (499997 * 2)) = 999996
# total = 2 + ((999996+2) * 1000) + 2 + 1992 + 4
# total = 10 billion
# retired branches is
# ((499,997 + 1) * 1,000) + 996 = 499,998,996
.globl _start
_start:
xor %edx,%edx # outer counter
mov $1000,%edx #
outside_loop:
xor %ecx,%ecx # not needed, pads total to 1M
mov $499997,%ecx # load counter
inside_loop:
dec %ecx # repeat count times
jnz inside_loop
dec %edx
jnz outside_loop
# above gets us to within 1,994
xor %ecx,%ecx
mov $996,%ecx
final_loop:
dec %ecx
jnz final_loop
#================================
# Exit
#================================
exit:
nop # make it an even number of insn
.ifdef i386
xor %ebx,%ebx # we return 0
mov $1,%eax
int $0x80
.else
xor %rdi,%rdi # we return 0
mov $60,%rax
syscall
.endif
------------------------------------------------------------------------------
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