FMA指令显示为三个打包双操作? [英] FMA instruction showing up as three packed double operations?
问题描述
我正在分析一段线性代数代码，它是直接调用内在函数，例如
I'm analyzing a piece of linear algebra code which is calling intrinsics directly, e.g.
v_dot0 = _mm256_fmadd_pd( v_x0, v_y0, v_dot0 );
我的测试脚本计算了两个长度为4的双精度向量的点积(因此只需调用一次_mm256_fmadd_pd)，重复十亿次.当我用perf计算操作次数时，我得到如下信息:
My test script computes the dot product of two double precision vectors of length 4 (so only one call to _mm256_fmadd_pd needed), repeated 1 billion times. When I count the number of operations with perf I get something as follows:
Performance counter stats for './main':
0 r5380c7 (skl::FP_ARITH:512B_PACKED_SINGLE) (49.99%)
0 r5340c7 (skl::FP_ARITH:512B_PACKED_DOUBLE) (49.99%)
0 r5320c7 (skl::FP_ARITH:256B_PACKED_SINGLE) (49.99%)
2'998'943'659 r5310c7 (skl::FP_ARITH:256B_PACKED_DOUBLE) (50.01%)
0 r5308c7 (skl::FP_ARITH:128B_PACKED_SINGLE) (50.01%)
1'999'928'140 r5304c7 (skl::FP_ARITH:128B_PACKED_DOUBLE) (50.01%)
0 r5302c7 (skl::FP_ARITH:SCALAR_SINGLE) (50.01%)
1'000'352'249 r5301c7 (skl::FP_ARITH:SCALAR_DOUBLE) (49.99%)
256B_PACKED_DOUBLE操作的数量大约是令我感到惊讶. 30亿，而不是10亿，因为这是我的体系结构指令集中的一条指令. 为什么perf每次调用_mm256_fmadd_pd时都要计算3次压缩双精度运算?
I was surprised that the number of 256B_PACKED_DOUBLE operations is approx. 3 billion, instead of 1 billion, as this is an instruction from my architecture's instruction set. Why does perf count 3 packed double operations per call to _mm256_fmadd_pd?
注意:为了测试代码没有意外调用其他浮点运算，我注释了对上述内在函数的调用，并且perf完全计数了零个256B_PACKED_DOUBLE运算，正如预期的那样.
Note: to test that the code is not calling other floating point operations accidentally, I commented out the call to the above mentioned intrinsic, and perf counts exactly zero 256B_PACKED_DOUBLE operations, as expected.
根据要求MCVE:
ddot.c
#include <immintrin.h> // AVX
double ddot(int m, double *x, double *y) {
int ii;
double dot = 0.0;
__m128d u_dot0, u_x0, u_y0, u_tmp;
__m256d v_dot0, v_dot1, v_x0, v_x1, v_y0, v_y1, v_tmp;
v_dot0 = _mm256_setzero_pd();
v_dot1 = _mm256_setzero_pd();
u_dot0 = _mm_setzero_pd();
ii = 0;
for (; ii < m - 3; ii += 4) {
v_x0 = _mm256_loadu_pd(&x[ii + 0]);
v_y0 = _mm256_loadu_pd(&y[ii + 0]);
v_dot0 = _mm256_fmadd_pd(v_x0, v_y0, v_dot0);
}
// reduce
v_dot0 = _mm256_add_pd(v_dot0, v_dot1);
u_tmp = _mm_add_pd(_mm256_castpd256_pd128(v_dot0), _mm256_extractf128_pd(v_dot0, 0x1));
u_tmp = _mm_hadd_pd(u_tmp, u_tmp);
u_dot0 = _mm_add_sd(u_dot0, u_tmp);
_mm_store_sd(&dot, u_dot0);
return dot;
}
main.c:
#include <stdio.h>
double ddot(int, double *, double *);
int main(int argc, char const *argv[]) {
double x[4] = {1.0, 2.0, 3.0, 4.0}, y[4] = {5.0, 5.0, 5.0, 5.0};
double xTy;
for (int i = 0; i < 1000000000; ++i) {
ddot(4, x, y);
}
printf(" %f\n", xTy);
return 0;
}
我以
sudo perf stat -e r5380c7 -e r5340c7 -e r5320c7 -e r5310c7 -e r5308c7 -e r5304c7 -e r5302c7 -e r5301c7 ./a.out
ddot的反汇编如下:
0000000000000790 <ddot>:
790: 83 ff 03 cmp $0x3,%edi
793: 7e 6b jle 800 <ddot+0x70>
795: 8d 4f fc lea -0x4(%rdi),%ecx
798: c5 e9 57 d2 vxorpd %xmm2,%xmm2,%xmm2
79c: 31 c0 xor %eax,%eax
79e: c1 e9 02 shr $0x2,%ecx
7a1: 48 83 c1 01 add $0x1,%rcx
7a5: 48 c1 e1 05 shl $0x5,%rcx
7a9: 0f 1f 80 00 00 00 00 nopl 0x0(%rax)
7b0: c5 f9 10 0c 06 vmovupd (%rsi,%rax,1),%xmm1
7b5: c5 f9 10 04 02 vmovupd (%rdx,%rax,1),%xmm0
7ba: c4 e3 75 18 4c 06 10 vinsertf128 $0x1,0x10(%rsi,%rax,1),%ymm1,%ymm1
7c1: 01
7c2: c4 e3 7d 18 44 02 10 vinsertf128 $0x1,0x10(%rdx,%rax,1),%ymm0,%ymm0
7c9: 01
7ca: 48 83 c0 20 add $0x20,%rax
7ce: 48 39 c1 cmp %rax,%rcx
7d1: c4 e2 f5 b8 d0 vfmadd231pd %ymm0,%ymm1,%ymm2
7d6: 75 d8 jne 7b0 <ddot+0x20>
7d8: c5 f9 57 c0 vxorpd %xmm0,%xmm0,%xmm0
7dc: c5 ed 58 d0 vaddpd %ymm0,%ymm2,%ymm2
7e0: c4 e3 7d 19 d0 01 vextractf128 $0x1,%ymm2,%xmm0
7e6: c5 f9 58 d2 vaddpd %xmm2,%xmm0,%xmm2
7ea: c5 f9 57 c0 vxorpd %xmm0,%xmm0,%xmm0
7ee: c5 e9 7c d2 vhaddpd %xmm2,%xmm2,%xmm2
7f2: c5 fb 58 d2 vaddsd %xmm2,%xmm0,%xmm2
7f6: c5 f9 28 c2 vmovapd %xmm2,%xmm0
7fa: c5 f8 77 vzeroupper
7fd: c3 retq
7fe: 66 90 xchg %ax,%ax
800: c5 e9 57 d2 vxorpd %xmm2,%xmm2,%xmm2
804: eb da jmp 7e0 <ddot+0x50>
806: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1)
80d: 00 00 00
推荐答案
我刚刚在SKL上使用asm循环进行了测试.像vfmadd231pd ymm0, ymm1, ymm3这样的FMA指令将2个fp_arith_inst_retired.256b_packed_double计数，即使它是单个uop！
I just tested with an asm loop on SKL. An FMA instructions like vfmadd231pd ymm0, ymm1, ymm3 counts for 2 counts of fp_arith_inst_retired.256b_packed_double, even though it's a single uop!
我猜英特尔真的想要FLOP计数器，而不是指令或uop计数器.
您的第3个256位FP uop可能来自您正在做的其他事情，例如，水平和开始时先进行256位改组和另一个256位加法，而不是先减少到128位.希望您不要使用_mm256_hadd_pd！
Your 3rd 256-bit FP uop is probably coming from something else you're doing, like a horizontal sum that starts out doing a 256-bit shuffle and another 256-bit add, instead of reducing to 128-bit first. I hope you're not using _mm256_hadd_pd!
测试代码内循环:
$ asm-link -d -n "testloop.asm" # assemble with NASM -felf64 and link with ld into a static binary
mov ebp, 100000000 # setup stuff outside the loop
vzeroupper
0000000000401040 <_start.loop>:
401040: c4 e2 f5 b8 c3 vfmadd231pd ymm0,ymm1,ymm3
401045: c4 e2 f5 b8 e3 vfmadd231pd ymm4,ymm1,ymm3
40104a: ff cd dec ebp
40104c: 75 f2 jne 401040 <_start.loop>
$ taskset -c 3 perf stat -etask-clock,context-switches,cpu-migrations,page-faults,cycles,branches,instructions,uops_issued.any,uops_executed.thread,fp_arith_inst_retired.256b_packed_double -r4 ./"$t"
Performance counter stats for './testloop-cvtss2sd' (4 runs):
102.67 msec task-clock # 0.999 CPUs utilized ( +- 0.00% )
2 context-switches # 24.510 M/sec ( +- 20.00% )
0 cpu-migrations # 0.000 K/sec
2 page-faults # 22.059 M/sec ( +- 11.11% )
400,388,898 cycles # 3925381.355 GHz ( +- 0.00% )
100,050,708 branches # 980889291.667 M/sec ( +- 0.00% )
400,256,258 instructions # 1.00 insn per cycle ( +- 0.00% )
300,377,737 uops_issued.any # 2944879772.059 M/sec ( +- 0.00% )
300,389,230 uops_executed.thread # 2944992450.980 M/sec ( +- 0.00% )
400,000,000 fp_arith_inst_retired.256b_packed_double # 3921568627.451 M/sec
0.1028042 +- 0.0000170 seconds time elapsed ( +- 0.02% )
对于200M FMA指令/100M循环迭代，fp_arith_inst_retired.256b_packed_double的计数为400M.
400M counts of fp_arith_inst_retired.256b_packed_double for 200M FMA instructions / 100M loop iterations.
(IDK用perf 4.20.g8fe28c +内核4.20.3-arch1-1-ARCH处理.它们以秒为单位计算每秒的内容，用小数点表示的单位错误.例如3925381.355 kHz是正确的，而不是GHz.不确定是否是性能或内核中的错误.
(IDK what up with perf 4.20.g8fe28c + kernel 4.20.3-arch1-1-ARCH. They calculate per-second stuff with the decimal in the wrong place for the unit. e.g. 3925381.355 kHz is correct, not GHz. Not sure if it's a bug in perf or the kernel.
在没有vzeroupper的情况下，对于FMA，我有时会看到5个周期的延迟，而不是4个周期.如果内核将寄存器保持在污染状态或其他状态，则为IDK.
Without vzeroupper, I'd sometimes see a latency of 5 cycles, not 4, for FMA. IDK if the kernel left a register in a polluted state or something.
为什么我得到三个，而不是两个? (请参阅添加到原始帖子中的MCVE)
Why do I get three though, and not two? (see MCVE added to original post)
您的ddot4在清理开始时运行_mm256_add_pd(v_dot0, v_dot1); ，由于您使用size = 4调用它，因此每个FMA都会进行一次清理.
Your ddot4 runs _mm256_add_pd(v_dot0, v_dot1); at the start of the cleanup, and since you call it with size=4, you get the cleanup once per FMA.
请注意，您的v_dot1始终为零(因为您实际上并未像您计划的那样使用2个累加器展开?)这毫无意义，但CPU并不知道.我的猜测是错误的，它不是256位的hadd，而是无用的256位垂直添加.
Note that your v_dot1 is always zero (because you didn't actually unroll with 2 accumulators like you're planning to?) So this is pointless, but the CPU doesn't know that. My guess was wrong, it's not a 256-bit hadd, it's just a useless 256-bit vertical add.
(对于更大的向量，是许多累加器对于隐藏FMA延迟非常有价值.您至少需要8个向量..请参见
(For larger vectors, yes multiple accumulators are very valuable to hide FMA latency. You'll want at least 8 vectors. See Why does mulss take only 3 cycles on Haswell, different from Agner's instruction tables? for more about unrolling with multiple accumulators. But then you'll want a cleanup loop that does 1 vector at a time until you're down to the last up-to-3 elements.)
此外，我认为您最终的_mm_add_sd(u_dot0, u_tmp);实际上是一个错误:您已经添加了最后一对具有低效128位hadd的元素，因此这将最低元素加倍了.
Also, I think your final _mm_add_sd(u_dot0, u_tmp); is actually a bug: you've already added the last pair of elements with an inefficient 128-bit hadd, so this double-counts the lowest element.
请参见使用SSE/AVX获取存储在__m256d中的值的总和，以确保不会丢失.
See Get sum of values stored in __m256d with SSE/AVX for a way that doesn't suck.
还请注意，GCC使用vinsertf128将未对齐的负载分成128位半部分，因为您是使用默认的-mtune=generic(支持Sandybridge)编译的，而不是使用-march=haswell启用AVX + FMA并设置了-march=native)
Also note that GCC is splitting your unaligned loads into 128-bit halves with vinsertf128 because you compiled with the default -mtune=generic (which favours Sandybridge) instead of using -march=haswell to enable AVX+FMA and set -mtune=haswell. (Or use -march=native)
这篇关于FMA指令显示为三个打包双操作?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
