[英]Why does perf show less floating point events than expected?
I'm writing a C program that does a 50x50 matrix multiplication. 我正在编写一个执行50x50矩阵乘法的C程序。
I compiled it to assembly and I see it has at least 2 floating point instructions (mulsd and addsd) in the most inner loop. 我将其编译为汇编,并且看到它在最内部的循环中至少有2个浮点指令(mulsd和addd)。 The loop is executed 125000 times, so I was expecting perf stat
to give me at least 250000 floating point operations counted. 该循环执行了125000次,因此我期望perf stat
至少可以计数250000个浮点运算。
However when I run the following command, 但是,当我运行以下命令时,
perf stat -e r530110 -e r531010 -e r532010 -e r534010 -e r538010 ./matmul
I only get the following counts 我只得到以下数
448 r530110
0 r531010
0 r532010
0 r534010
<not counted> r538010
0.001082287 seconds time elapsed
My compiler does not produce SSE instructions so the other zero counts are expected. 我的编译器不生成SSE指令,因此期望其他零计数。 However, I only got 448 floating point operations. 但是,我只有448个浮点运算。
What is happening? 怎么了?
I've figured it out. 我知道了。 I misunderstood: mulsd
and addsd
are actually SSE instructions. 我误会了: mulsd
和addsd
实际上是SSE指令。
The stat r530110
(FP_COMPS_OPS_EXE:X87) does not count mulsd
or addsd
. stat r530110
(FP_COMPS_OPS_EXE:X87)不计入mulsd
或addsd
。
I needed to look at r538010
(FP_COMPS_OPS:SSE_SCALAR_DOUBLE). 我需要查看r538010
(FP_COMPS_OPS:SSE_SCALAR_DOUBLE)。
My guess for the reason those were <not counted>
in my program was because the program finished too quickly. 我之所以认为这些在我的程序中<not counted>
是因为该程序完成得太快。 When I increased the size of the matrix (so the program runs longer), the floating point event count came out correct. 当我增加矩阵的大小(以便程序运行更长的时间)时,浮点事件计数正确。
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