有可能加速这个简单的内核功能吗?我曾经考虑过使用共享内存,但是N等于507904,所以它远远超过了共享内存数组。
我的程序创建每个256个线程的块。
__global__ void compute(COMPLEX_TYPE *a, COMPLEX_TYPE *b,
FLOAT_TYPE *F, FLOAT_TYPE f, int N)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < N)
{
F[i] = ( a[i].x*a[i].x + a[i].y*a[i].y + b[i].x*b[i].x + b[i].y*b[i].y) / (f);
}
}
最简单的常规优化如下所示:
__global__ void compute(const COMPLEX_TYPE * __restrict__ a,
const COMPLEX_TYPE * __restrict__ b,
FLOAT_TYPE *F, FLOAT_TYPE f, int N)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
#pragma unroll 8
for(; i < N; i += blockDim.x * gridDim.x;)
{
COMPLEX_TYPE aval = a[i], bval = b[i]
FLOAT_TYPE Fval;
Fval = ( aval.x*aval.x + aval.y*aval.y + bval.x*bval.x + bval.y*bval.y) / (f);
F[i] = Fval;
}
}
[免责声明:在浏览器中编写,未经测试,使用风险自负]
这里的想法是只启动与在目标GPU上并发执行的线程一样多的线程,然后让每个线程执行多个操作,而不是一个。这有助于在块调度程序和设置代码级别上分摊大量固定开销,并提高整体效率。在大多数体系结构上,无论如何这都可能会限制内存带宽,因此内存合并和事务优化是您将能够进行的最重要的性能优化。
编辑:由于此答案被标记为CW,所以我选择在此处添加测试,而不是创建自己的答案。如果有人对此表示反对,请仅将编辑回滚到以前可接受的版本。我没有添加任何新想法,只是测试@talonmies和@JanLucas提供的想法
在我的测试案例中,@ talonmies提出的建议(除展开实用性外)似乎使性能提高了约10%。@JanLucas的建议是,如果可以接受的话,用浮点乘法代替浮点除法,似乎可以使性能提高一倍。显然,这将取决于GPU和其他细节。这是我的测试:
$ cat t891.cu
#include <cuComplex.h>
#include <stdio.h>
#include <stdlib.h>
#define DSIZE 507904
#define nTPB 256
#define nBLK 256
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
typedef cuFloatComplex COMPLEX_TYPE;
typedef float FLOAT_TYPE;
__global__ void compute(COMPLEX_TYPE *a, COMPLEX_TYPE *b,
FLOAT_TYPE *F, FLOAT_TYPE f, int N)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < N)
{
F[i] = ( a[i].x*a[i].x + a[i].y*a[i].y + b[i].x*b[i].x + b[i].y*b[i].y) / (f);
}
}
__global__ void compute_imp(const COMPLEX_TYPE * __restrict__ a,
const COMPLEX_TYPE * __restrict__ b,
FLOAT_TYPE *F, FLOAT_TYPE f, int N)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
// #pragma unroll 8
for(; i < N; i += blockDim.x * gridDim.x)
{
COMPLEX_TYPE aval = a[i];
COMPLEX_TYPE bval = b[i];
FLOAT_TYPE Fval = ( aval.x*aval.x + aval.y*aval.y + bval.x*bval.x + bval.y*bval.y) / (f);
F[i] = Fval;
}
}
__global__ void compute_imp2(const COMPLEX_TYPE * __restrict__ a,
const COMPLEX_TYPE * __restrict__ b,
FLOAT_TYPE *F, FLOAT_TYPE f, int N)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
// #pragma unroll 8
for(; i < N; i += blockDim.x * gridDim.x)
{
COMPLEX_TYPE aval = a[i];
COMPLEX_TYPE bval = b[i];
FLOAT_TYPE Fval = ( aval.x*aval.x + aval.y*aval.y + bval.x*bval.x + bval.y*bval.y) * (f);
F[i] = Fval;
}
}
int main(){
COMPLEX_TYPE *d_A, *d_B;
FLOAT_TYPE *d_F, f = 4.0f;
cudaMalloc(&d_A, DSIZE*sizeof(COMPLEX_TYPE));
cudaMalloc(&d_B, DSIZE*sizeof(COMPLEX_TYPE));
cudaMalloc(&d_F, DSIZE*sizeof(FLOAT_TYPE));
//warm-up
compute<<<(DSIZE+nTPB-1)/nTPB,nTPB>>>(d_A, d_B, d_F, f, DSIZE);
cudaDeviceSynchronize();
unsigned long long t1 = dtime_usec(0);
compute<<<(DSIZE+nTPB-1)/nTPB,nTPB>>>(d_A, d_B, d_F, f, DSIZE);
cudaDeviceSynchronize();
t1 = dtime_usec(t1);
//warm-up
compute_imp<<<DSIZE/(8*nTPB),nTPB>>>(d_A, d_B, d_F, f, DSIZE);
cudaDeviceSynchronize();
unsigned long long t2 = dtime_usec(0);
compute_imp<<<nBLK,nTPB>>>(d_A, d_B, d_F, f, DSIZE);
cudaDeviceSynchronize();
t2 = dtime_usec(t2);
//warm-up
compute_imp2<<<(DSIZE+nTPB-1)/nTPB,nTPB>>>(d_A, d_B, d_F, 1/f, DSIZE);
cudaDeviceSynchronize();
unsigned long long t3 = dtime_usec(0);
compute_imp2<<<nBLK,nTPB>>>(d_A, d_B, d_F, 1/f, DSIZE);
cudaDeviceSynchronize();
t3 = dtime_usec(t3);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs, t3: %fs\n", t1/(float)USECPSEC, t2/(float)(USECPSEC), t3/(float)USECPSEC);
}
$ nvcc -O3 -o t891 t891.cu
$ ./t891
t1: 0.000226s, t2: 0.000209s, t3: 0.000110s
$
笔记:
nBLK
这段代码中的参数是另一个“可调”参数,但是在大约64左右时,此参数几乎没有变化。最好的情况是使网格的大小等于数据的大小。本文收集自互联网,转载请注明来源。
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我来说两句