假设我们有一个函数将两个数组相乘，每个数组有 1000000 个 double 值。在 C/C++ 中，该函数如下所示:
void mul_c(double* a, double* b)
{
for (int i = 0; i != 1000000; ++i)
{
a[i] = a[i] * b[i];
}
}
编译器使用
-O2 生成以下程序集:mul_c(double*, double*):
xor eax, eax
.L2:
movsd xmm0, QWORD PTR [rdi+rax]
mulsd xmm0, QWORD PTR [rsi+rax]
movsd QWORD PTR [rdi+rax], xmm0
add rax, 8
cmp rax, 8000000
jne .L2
rep ret
从上面的汇编看来，编译器使用了 SIMD 指令，但每次迭代只乘以一倍。所以我决定在内联汇编中编写相同的函数，我充分利用了
xmm0一次注册并乘以两个 double :void mul_asm(double* a, double* b)
{
asm volatile
(
".intel_syntax noprefix \n\t"
"xor rax, rax \n\t"
"0: \n\t"
"movupd xmm0, xmmword ptr [rdi+rax] \n\t"
"mulpd xmm0, xmmword ptr [rsi+rax] \n\t"
"movupd xmmword ptr [rdi+rax], xmm0 \n\t"
"add rax, 16 \n\t"
"cmp rax, 8000000 \n\t"
"jne 0b \n\t"
".att_syntax noprefix \n\t"
:
: "D" (a), "S" (b)
: "memory", "cc"
);
}
在分别测量这两个函数的执行时间后，它们似乎都需要 1 ms 才能完成:
> gcc -O2 main.cpp
> ./a.out < input
mul_c: 1 ms
mul_asm: 1 ms
[a lot of doubles...]
我预计 SIMD 实现的速度至少是两倍(0 毫秒)，因为只有一半的乘法/内存指令。
所以我的问题是:当 SIMD 实现只执行一半的乘法/内存指令时，为什么 SIMD 实现不比普通的 C/C++ 实现快？
这是完整的程序:
#include <stdio.h>
#include <stdlib.h>
#include <sys/time.h>
void mul_c(double* a, double* b)
{
for (int i = 0; i != 1000000; ++i)
{
a[i] = a[i] * b[i];
}
}
void mul_asm(double* a, double* b)
{
asm volatile
(
".intel_syntax noprefix \n\t"
"xor rax, rax \n\t"
"0: \n\t"
"movupd xmm0, xmmword ptr [rdi+rax] \n\t"
"mulpd xmm0, xmmword ptr [rsi+rax] \n\t"
"movupd xmmword ptr [rdi+rax], xmm0 \n\t"
"add rax, 16 \n\t"
"cmp rax, 8000000 \n\t"
"jne 0b \n\t"
".att_syntax noprefix \n\t"
:
: "D" (a), "S" (b)
: "memory", "cc"
);
}
int main()
{
struct timeval t1;
struct timeval t2;
unsigned long long time;
double* a = (double*)malloc(sizeof(double) * 1000000);
double* b = (double*)malloc(sizeof(double) * 1000000);
double* c = (double*)malloc(sizeof(double) * 1000000);
for (int i = 0; i != 1000000; ++i)
{
double v;
scanf("%lf", &v);
a[i] = v;
b[i] = v;
c[i] = v;
}
gettimeofday(&t1, NULL);
mul_c(a, b);
gettimeofday(&t2, NULL);
time = 1000 * (t2.tv_sec - t1.tv_sec) + (t2.tv_usec - t1.tv_usec) / 1000;
printf("mul_c: %llu ms\n", time);
gettimeofday(&t1, NULL);
mul_asm(b, c);
gettimeofday(&t2, NULL);
time = 1000 * (t2.tv_sec - t1.tv_sec) + (t2.tv_usec - t1.tv_usec) / 1000;
printf("mul_asm: %llu ms\n\n", time);
for (int i = 0; i != 1000000; ++i)
{
printf("%lf\t\t\t%lf\n", a[i], b[i]);
}
return 0;
}
我也试图利用所有
xmm寄存器 (0-7) 并删除指令依赖性以获得更好的并行计算:void mul_asm(double* a, double* b)
{
asm volatile
(
".intel_syntax noprefix \n\t"
"xor rax, rax \n\t"
"0: \n\t"
"movupd xmm0, xmmword ptr [rdi+rax] \n\t"
"movupd xmm1, xmmword ptr [rdi+rax+16] \n\t"
"movupd xmm2, xmmword ptr [rdi+rax+32] \n\t"
"movupd xmm3, xmmword ptr [rdi+rax+48] \n\t"
"movupd xmm4, xmmword ptr [rdi+rax+64] \n\t"
"movupd xmm5, xmmword ptr [rdi+rax+80] \n\t"
"movupd xmm6, xmmword ptr [rdi+rax+96] \n\t"
"movupd xmm7, xmmword ptr [rdi+rax+112] \n\t"
"mulpd xmm0, xmmword ptr [rsi+rax] \n\t"
"mulpd xmm1, xmmword ptr [rsi+rax+16] \n\t"
"mulpd xmm2, xmmword ptr [rsi+rax+32] \n\t"
"mulpd xmm3, xmmword ptr [rsi+rax+48] \n\t"
"mulpd xmm4, xmmword ptr [rsi+rax+64] \n\t"
"mulpd xmm5, xmmword ptr [rsi+rax+80] \n\t"
"mulpd xmm6, xmmword ptr [rsi+rax+96] \n\t"
"mulpd xmm7, xmmword ptr [rsi+rax+112] \n\t"
"movupd xmmword ptr [rdi+rax], xmm0 \n\t"
"movupd xmmword ptr [rdi+rax+16], xmm1 \n\t"
"movupd xmmword ptr [rdi+rax+32], xmm2 \n\t"
"movupd xmmword ptr [rdi+rax+48], xmm3 \n\t"
"movupd xmmword ptr [rdi+rax+64], xmm4 \n\t"
"movupd xmmword ptr [rdi+rax+80], xmm5 \n\t"
"movupd xmmword ptr [rdi+rax+96], xmm6 \n\t"
"movupd xmmword ptr [rdi+rax+112], xmm7 \n\t"
"add rax, 128 \n\t"
"cmp rax, 8000000 \n\t"
"jne 0b \n\t"
".att_syntax noprefix \n\t"
:
: "D" (a), "S" (b)
: "memory", "cc"
);
}
但它仍然以 1 ms 运行，与普通 C/C++ 实现的速度相同。
更新
正如答案/评论所建议的那样，我已经实现了另一种测量执行时间的方法:
#include <stdio.h>
#include <stdlib.h>
void mul_c(double* a, double* b)
{
for (int i = 0; i != 1000000; ++i)
{
a[i] = a[i] * b[i];
}
}
void mul_asm(double* a, double* b)
{
asm volatile
(
".intel_syntax noprefix \n\t"
"xor rax, rax \n\t"
"0: \n\t"
"movupd xmm0, xmmword ptr [rdi+rax] \n\t"
"mulpd xmm0, xmmword ptr [rsi+rax] \n\t"
"movupd xmmword ptr [rdi+rax], xmm0 \n\t"
"add rax, 16 \n\t"
"cmp rax, 8000000 \n\t"
"jne 0b \n\t"
".att_syntax noprefix \n\t"
:
: "D" (a), "S" (b)
: "memory", "cc"
);
}
void mul_asm2(double* a, double* b)
{
asm volatile
(
".intel_syntax noprefix \n\t"
"xor rax, rax \n\t"
"0: \n\t"
"movupd xmm0, xmmword ptr [rdi+rax] \n\t"
"movupd xmm1, xmmword ptr [rdi+rax+16] \n\t"
"movupd xmm2, xmmword ptr [rdi+rax+32] \n\t"
"movupd xmm3, xmmword ptr [rdi+rax+48] \n\t"
"movupd xmm4, xmmword ptr [rdi+rax+64] \n\t"
"movupd xmm5, xmmword ptr [rdi+rax+80] \n\t"
"movupd xmm6, xmmword ptr [rdi+rax+96] \n\t"
"movupd xmm7, xmmword ptr [rdi+rax+112] \n\t"
"mulpd xmm0, xmmword ptr [rsi+rax] \n\t"
"mulpd xmm1, xmmword ptr [rsi+rax+16] \n\t"
"mulpd xmm2, xmmword ptr [rsi+rax+32] \n\t"
"mulpd xmm3, xmmword ptr [rsi+rax+48] \n\t"
"mulpd xmm4, xmmword ptr [rsi+rax+64] \n\t"
"mulpd xmm5, xmmword ptr [rsi+rax+80] \n\t"
"mulpd xmm6, xmmword ptr [rsi+rax+96] \n\t"
"mulpd xmm7, xmmword ptr [rsi+rax+112] \n\t"
"movupd xmmword ptr [rdi+rax], xmm0 \n\t"
"movupd xmmword ptr [rdi+rax+16], xmm1 \n\t"
"movupd xmmword ptr [rdi+rax+32], xmm2 \n\t"
"movupd xmmword ptr [rdi+rax+48], xmm3 \n\t"
"movupd xmmword ptr [rdi+rax+64], xmm4 \n\t"
"movupd xmmword ptr [rdi+rax+80], xmm5 \n\t"
"movupd xmmword ptr [rdi+rax+96], xmm6 \n\t"
"movupd xmmword ptr [rdi+rax+112], xmm7 \n\t"
"add rax, 128 \n\t"
"cmp rax, 8000000 \n\t"
"jne 0b \n\t"
".att_syntax noprefix \n\t"
:
: "D" (a), "S" (b)
: "memory", "cc"
);
}
unsigned long timestamp()
{
unsigned long a;
asm volatile
(
".intel_syntax noprefix \n\t"
"xor rax, rax \n\t"
"xor rdx, rdx \n\t"
"RDTSCP \n\t"
"shl rdx, 32 \n\t"
"or rax, rdx \n\t"
".att_syntax noprefix \n\t"
: "=a" (a)
:
: "memory", "cc"
);
return a;
}
int main()
{
unsigned long t1;
unsigned long t2;
double* a;
double* b;
a = (double*)malloc(sizeof(double) * 1000000);
b = (double*)malloc(sizeof(double) * 1000000);
for (int i = 0; i != 1000000; ++i)
{
double v;
scanf("%lf", &v);
a[i] = v;
b[i] = v;
}
t1 = timestamp();
mul_c(a, b);
//mul_asm(a, b);
//mul_asm2(a, b);
t2 = timestamp();
printf("mul_c: %lu cycles\n\n", t2 - t1);
for (int i = 0; i != 1000000; ++i)
{
printf("%lf\t\t\t%lf\n", a[i], b[i]);
}
return 0;
}
当我用这个测量运行程序时，我得到了这个结果:
mul_c: ~2163971628 cycles
mul_asm: ~2532045184 cycles
mul_asm2: ~5230488 cycles <-- what???
这里有两件事值得注意，首先，周期数变化很大，我认为这是因为操作系统允许其他进程在其间运行。有什么办法可以防止这种情况发生，或者只在我的程序执行时计算周期数？另外，
mul_asm2与其他两个相比产生相同的输出，但速度要快得多，如何？我在我的系统上尝试了 Z boson 的程序以及我的 2 个实现，并得到了以下结果:
> g++ -O2 -fopenmp main.cpp
> ./a.out
mul time 1.33, 18.08 GB/s
mul_SSE time 1.13, 21.24 GB/s
mul_SSE_NT time 1.51, 15.88 GB/s
mul_SSE_OMP time 0.79, 30.28 GB/s
mul_SSE_v2 time 1.12, 21.49 GB/s
mul_v2 time 1.26, 18.99 GB/s
mul_asm time 1.12, 21.50 GB/s
mul_asm2 time 1.09, 22.08 GB/s
最佳答案
有 a major bug in the timing function I used对于以前的基准。这严重低估了没有矢量化和其他测量的带宽。此外，还有一个问题是高估了带宽due to COW在已读取但未写入的数组上。最后，我使用的最大带宽不正确。我已经用更正更新了我的答案，并在此答案的末尾留下了旧答案。
您的操作受内存带宽限制。这意味着 CPU 大部分时间都在等待缓慢的内存读取和写入。对此的一个很好的解释可以在这里找到:Why vectorizing the loop does not have performance improvement .
但是，我不得不稍微不同意该答案中的一个陈述。
So regardless of how it's optimized, (vectorized, unrolled, etc...) it isn't gonna get much faster.
事实上，即使在内存带宽受限的操作中，矢量化、展开和多线程也可以显着增加带宽。原因是很难获得最大的内存带宽。对此的一个很好的解释可以在这里找到:https://stackoverflow.com/a/25187492/2542702 .
我的其余答案将展示矢量化和多线程如何接近最大内存带宽。
我的测试系统:Ubuntu 16.10、Skylake (i7-6700HQ@2.60GHz)、32GB RAM、双 channel DDR4@2400 GHz。我的系统的最大带宽是 38.4 GB/s。
从下面的代码中，我生成了下表。我使用 OMP_NUM_THREADS 设置线程数，例如
export OMP_NUM_THREADS=4 .效率是bandwidth/max_bandwidth .-O2 -march=native -fopenmp
Threads Efficiency
1 59.2%
2 76.6%
4 74.3%
8 70.7%
-O2 -march=native -fopenmp -funroll-loops
1 55.8%
2 76.5%
4 72.1%
8 72.2%
-O3 -march=native -fopenmp
1 63.9%
2 74.6%
4 63.9%
8 63.2%
-O3 -march=native -fopenmp -mprefer-avx128
1 67.8%
2 76.0%
4 63.9%
8 63.2%
-O3 -march=native -fopenmp -mprefer-avx128 -funroll-loops
1 68.8%
2 73.9%
4 69.0%
8 66.8%
由于测量的不确定性，经过多次迭代运行，我得出了以下结论:
提供最佳带宽的解决方案是使用两个线程进行标量运算。
我用来基准测试的代码:
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <omp.h>
#define N 10000000
#define R 100
void mul(double *a, double *b) {
#pragma omp parallel for
for (int i = 0; i<N; i++) a[i] *= b[i];
}
int main() {
double maxbw = 2.4*2*8; // 2.4GHz * 2-channels * 64-bits * 1-byte/8-bits
double mem = 3*sizeof(double)*N*R*1E-9; // GB
double *a = (double*)malloc(sizeof *a * N);
double *b = (double*)malloc(sizeof *b * N);
//due to copy-on-write b must be initialized to get the correct bandwidth
//also, GCC will convert malloc + memset(0) to calloc so use memset(1)
memset(b, 1, sizeof *b * N);
double dtime = -omp_get_wtime();
for(int i=0; i<R; i++) mul(a,b);
dtime += omp_get_wtime();
printf("%.2f s, %.1f GB/s, %.1f%%\n", dtime, mem/dtime, 100*mem/dtime/maxbw);
free(a), free(b);
}
带有计时错误的旧解决方案
内联汇编的现代解决方案是使用内在函数。仍然存在需要内联汇编的情况，但这不是其中之一。
内联汇编方法的一种内在解决方案很简单:
void mul_SSE(double* a, double* b) {
for (int i = 0; i<N/2; i++)
_mm_store_pd(&a[2*i], _mm_mul_pd(_mm_load_pd(&a[2*i]),_mm_load_pd(&b[2*i])));
}
让我定义一些测试代码
#include <x86intrin.h>
#include <string.h>
#include <stdio.h>
#include <x86intrin.h>
#include <omp.h>
#define N 1000000
#define R 1000
typedef __attribute__(( aligned(32))) double aligned_double;
void (*fp)(aligned_double *a, aligned_double *b);
void mul(aligned_double* __restrict a, aligned_double* __restrict b) {
for (int i = 0; i<N; i++) a[i] *= b[i];
}
void mul_SSE(double* a, double* b) {
for (int i = 0; i<N/2; i++) _mm_store_pd(&a[2*i], _mm_mul_pd(_mm_load_pd(&a[2*i]),_mm_load_pd(&b[2*i])));
}
void mul_SSE_NT(double* a, double* b) {
for (int i = 0; i<N/2; i++) _mm_stream_pd(&a[2*i], _mm_mul_pd(_mm_load_pd(&a[2*i]),_mm_load_pd(&b[2*i])));
}
void mul_SSE_OMP(double* a, double* b) {
#pragma omp parallel for
for (int i = 0; i<N; i++) a[i] *= b[i];
}
void test(aligned_double *a, aligned_double *b, const char *name) {
double dtime;
const double mem = 3*sizeof(double)*N*R/1024/1024/1024;
const double maxbw = 34.1;
dtime = -omp_get_wtime();
for(int i=0; i<R; i++) fp(a,b);
dtime += omp_get_wtime();
printf("%s \t time %.2f s, %.1f GB/s, efficency %.1f%%\n", name, dtime, mem/dtime, 100*mem/dtime/maxbw);
}
int main() {
double *a = (double*)_mm_malloc(sizeof *a * N, 32);
double *b = (double*)_mm_malloc(sizeof *b * N, 32);
//b must be initialized to get the correct bandwidth!!!
memset(a, 1, sizeof *a * N);
memset(b, 1, sizeof *a * N);
fp = mul, test(a,b, "mul ");
fp = mul_SSE, test(a,b, "mul_SSE ");
fp = mul_SSE_NT, test(a,b, "mul_SSE_NT ");
fp = mul_SSE_OMP, test(a,b, "mul_SSE_OMP");
_mm_free(a), _mm_free(b);
}
现在第一次测试
g++ -O2 -fopenmp test.cpp
./a.out
mul time 1.67 s, 13.1 GB/s, efficiency 38.5%
mul_SSE time 1.00 s, 21.9 GB/s, efficiency 64.3%
mul_SSE_NT time 1.05 s, 20.9 GB/s, efficiency 61.4%
mul_SSE_OMP time 0.74 s, 29.7 GB/s, efficiency 87.0%
所以与
-O2它不向量化循环，我们看到内在的 SSE 版本比普通的 C 解决方案快得多 mul . efficiency = bandwith_measured/max_bandwidth我的系统的最大值为 34.1 GB/s。第二次测试
g++ -O3 -fopenmp test.cpp
./a.out
mul time 1.05 s, 20.9 GB/s, efficiency 61.2%
mul_SSE time 0.99 s, 22.3 GB/s, efficiency 65.3%
mul_SSE_NT time 1.01 s, 21.7 GB/s, efficiency 63.7%
mul_SSE_OMP time 0.68 s, 32.5 GB/s, efficiency 95.2%
与
-O3将循环向量化，而内在函数基本上没有任何优势。第三次测试
g++ -O3 -fopenmp -funroll-loops test.cpp
./a.out
mul time 0.85 s, 25.9 GB/s, efficency 76.1%
mul_SSE time 0.84 s, 26.2 GB/s, efficency 76.7%
mul_SSE_NT time 1.06 s, 20.8 GB/s, efficency 61.0%
mul_SSE_OMP time 0.76 s, 29.0 GB/s, efficency 85.0%
与
-funroll-loops GCC 将循环展开八次，我们看到除了非临时存储解决方案和OpenMP 解决方案的真正优势之外的显着改进。在展开循环之前
mul 的组件与 -O3是 xor eax, eax
.L2:
movupd xmm0, XMMWORD PTR [rsi+rax]
mulpd xmm0, XMMWORD PTR [rdi+rax]
movaps XMMWORD PTR [rdi+rax], xmm0
add rax, 16
cmp rax, 8000000
jne .L2
rep ret
与
-O3 -funroll-loops mul 的组件是: xor eax, eax
.L2:
movupd xmm0, XMMWORD PTR [rsi+rax]
movupd xmm1, XMMWORD PTR [rsi+16+rax]
mulpd xmm0, XMMWORD PTR [rdi+rax]
movupd xmm2, XMMWORD PTR [rsi+32+rax]
mulpd xmm1, XMMWORD PTR [rdi+16+rax]
movupd xmm3, XMMWORD PTR [rsi+48+rax]
mulpd xmm2, XMMWORD PTR [rdi+32+rax]
movupd xmm4, XMMWORD PTR [rsi+64+rax]
mulpd xmm3, XMMWORD PTR [rdi+48+rax]
movupd xmm5, XMMWORD PTR [rsi+80+rax]
mulpd xmm4, XMMWORD PTR [rdi+64+rax]
movupd xmm6, XMMWORD PTR [rsi+96+rax]
mulpd xmm5, XMMWORD PTR [rdi+80+rax]
movupd xmm7, XMMWORD PTR [rsi+112+rax]
mulpd xmm6, XMMWORD PTR [rdi+96+rax]
movaps XMMWORD PTR [rdi+rax], xmm0
mulpd xmm7, XMMWORD PTR [rdi+112+rax]
movaps XMMWORD PTR [rdi+16+rax], xmm1
movaps XMMWORD PTR [rdi+32+rax], xmm2
movaps XMMWORD PTR [rdi+48+rax], xmm3
movaps XMMWORD PTR [rdi+64+rax], xmm4
movaps XMMWORD PTR [rdi+80+rax], xmm5
movaps XMMWORD PTR [rdi+96+rax], xmm6
movaps XMMWORD PTR [rdi+112+rax], xmm7
sub rax, -128
cmp rax, 8000000
jne .L2
rep ret
第四次测试
g++ -O3 -fopenmp -mavx test.cpp
./a.out
mul time 0.87 s, 25.3 GB/s, efficiency 74.3%
mul_SSE time 0.88 s, 24.9 GB/s, efficiency 73.0%
mul_SSE_NT time 1.07 s, 20.6 GB/s, efficiency 60.5%
mul_SSE_OMP time 0.76 s, 29.0 GB/s, efficiency 85.2%
现在非内在函数是最快的(不包括 OpenMP 版本)。
所以在这种情况下没有理由使用内部函数或内联汇编，因为我们可以通过适当的编译器选项(例如
-O3 、 -funroll-loops 、 -mavx )获得最佳性能。测试系统:Ubuntu 16.10、Skylake (i7-6700HQ@2.60GHz)、32GB RAM。最大内存带宽 (34.1 GB/s) https://ark.intel.com/products/88967/Intel-Core-i7-6700HQ-Processor-6M-Cache-up-to-3_50-GHz
这是另一个值得考虑的解决方案。 The
cmp instruction is not necessary如果我们从 -N 数到零并以 N+i 访问数组. GCC 早就应该解决这个问题了。它消除了一条指令(尽管由于宏操作融合，cmp 和 jmp 通常算作一个微操作)。void mul_SSE_v2(double* a, double* b) {
for (ptrdiff_t i = -N; i<0; i+=2)
_mm_store_pd(&a[N + i], _mm_mul_pd(_mm_load_pd(&a[N + i]),_mm_load_pd(&b[N + i])));
用
-O3 组装mul_SSE_v2(double*, double*):
mov rax, -1000000
.L9:
movapd xmm0, XMMWORD PTR [rdi+8000000+rax*8]
mulpd xmm0, XMMWORD PTR [rsi+8000000+rax*8]
movaps XMMWORD PTR [rdi+8000000+rax*8], xmm0
add rax, 2
jne .L9
rep ret
}
这种优化可能只会对适合的数组有帮助，例如L1 缓存，即不从主内存读取。
我终于找到了一种方法来让普通的 C 解决方案不生成
cmp指令。void mul_v2(aligned_double* __restrict a, aligned_double* __restrict b) {
for (int i = -N; i<0; i++) a[i] *= b[i];
}
然后从一个单独的目标文件中调用该函数，如
mul_v2(&a[N],&b[N])所以这也许是最好的解决方案。但是，如果您从与 GCC 中定义的目标文件(翻译单元)相同的目标文件(翻译单元)调用该函数，则会生成 cmp再次指导。还有，
void mul_v3(aligned_double* __restrict a, aligned_double* __restrict b) {
for (int i = -N; i<0; i++) a[N+i] *= b[N+i];
}
仍然生成
cmp指令并生成与 mul 相同的程序集功能。函数
mul_SSE_NT很傻。它使用非临时存储，仅在写入内存时才有用，但由于函数读取和写入同一地址，非临时存储不仅无用，而且会产生较差的结果。此答案的先前版本获得了错误的带宽。原因是数组未初始化时。
关于c++ - 为什么这种 SIMD 乘法不如非 SIMD 乘法快？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42964820/