但是我似乎想不出为结构体指针abc *foo
分配内存的方法......是否有其他方法可以为下面的SOA分配内存?
我不确定你遇到了什么困难。只要你没有一个由数组组成的结构体数组,为什么不直接使用:
abc *foo
cudaMalloc((void **)&foo, SOME_CONSTANT*sizeof(abc))
使用我上面提到的方法是一个好的做法吗?
AoS与SoA的选择取决于应用场景,关于这个主题在CUDA应用程序上有很多优秀的问题/答案(例如像
这个答案)。总之,当一个warp中的所有线程访问一段连续的内存块时,就会出现协同访存。因此,如果每个字段的访问可以协同,那么当使用SoA时可以期望看到更高的内存带宽。针对你给出的示例,让我们运行一个简单的测试来量化性能差异:
#include <stdio.h>
#include <stdlib.h>
#define CHECK_CUDA(call) \
{ \
const cudaError_t error = call; \
if (error != cudaSuccess) \
{ \
printf("ERROR:: File: %s, Line: %d, ", __FILE__, __LINE__); \
printf("code: %d, reason: %s\n", error, cudaGetErrorString(error)); \
exit(EXIT_FAILURE); \
} \
}
const int SOME_CONSTANT = 1024 * 1000;
struct soa_abc {
float *a;
float *b;
float *c;
};
struct aos_abc {
float a;
float b;
float c;
};
__global__ void kernel_soa(soa_abc foo) {
unsigned int tid = blockDim.x * blockIdx.x + threadIdx.x;
foo.a[tid] = 1.f;
foo.b[tid] = 2.f;
foo.c[tid] = 3.f;
}
__global__ void kernel_aos(aos_abc *bar) {
unsigned int tid = blockDim.x * blockIdx.x + threadIdx.x;
bar[tid].a = 1.f;
bar[tid].b = 2.f;
bar[tid].c = 3.f;
}
int main()
{
float milliseconds = 0;
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
soa_abc foo;
CHECK_CUDA(cudaMalloc((void **)&foo.a, SOME_CONSTANT * sizeof(float)));
CHECK_CUDA(cudaMalloc((void **)&foo.b, SOME_CONSTANT * sizeof(float)));
CHECK_CUDA(cudaMalloc((void **)&foo.c, SOME_CONSTANT * sizeof(float)));
cudaEventRecord(start);
kernel_soa <<<SOME_CONSTANT/1000, 1000 >>> (foo);
CHECK_CUDA(cudaDeviceSynchronize());
cudaEventRecord(stop);
cudaEventSynchronize(stop);
milliseconds = 0;
cudaEventElapsedTime(&milliseconds, start, stop);
printf("Time for SoA is %f ms.\n", milliseconds);
CHECK_CUDA(cudaFree(foo.a));
CHECK_CUDA(cudaFree(foo.b));
CHECK_CUDA(cudaFree(foo.c));
aos_abc *bar;
CHECK_CUDA(cudaMalloc((void **)&bar, SOME_CONSTANT*sizeof(aos_abc)));
cudaEventRecord(start);
kernel_aos <<<SOME_CONSTANT/1000, 1000 >>> (bar);
CHECK_CUDA(cudaDeviceSynchronize());
cudaEventRecord(stop);
cudaEventSynchronize(stop);
milliseconds = 0;
cudaEventElapsedTime(&milliseconds, start, stop);
printf("Time for AoS is %f ms.\n", milliseconds);
}
在Windows和CUDA 10上使用Quadro P400进行测试,结果如下:
Time for SoA is 0.492384 ms.
Time for AoS is 1.217568 ms.
这证实了SoA是更好的选择。
abc *foo = 0;
foo->a= (float *)malloc(SOME_CONSTANT * sizeof(float ));
foo->b= (float *)malloc(SOME_CONSTANT * sizeof(float ));
foo->c= (float *)malloc(SOME_CONSTANT * sizeof(float ));