【发布时间】:2014-12-01 14:33:34
【问题描述】:
我正在尝试进行并行归约以对 CUDA 中的数组求和。目前我传递一个数组来存储每个块中元素的总和。这是我的代码:
#include <cstdlib>
#include <iostream>
#include <cuda.h>
#include <cuda_runtime_api.h>
#include <helper_cuda.h>
#include <host_config.h>
#define THREADS_PER_BLOCK 256
#define CUDA_ERROR_CHECK(ans) { gpuAssert((ans), __FILE__, __LINE__); }
using namespace std;
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
struct double3c {
double x;
double y;
double z;
__host__ __device__ double3c() : x(0), y(0), z(0) {}
__host__ __device__ double3c(int x_, int y_, int z_) : x(x_), y(y_), z(z_) {}
__host__ __device__ double3c& operator+=(const double3c& rhs) { x += rhs.x; y += rhs.y; z += rhs.z;}
__host__ __device__ double3c& operator/=(const double& rhs) { x /= rhs; y /= rhs; z /= rhs;}
};
class VectorField {
public:
double3c *data;
int size_x, size_y, size_z;
bool is_copy;
__host__ VectorField () {}
__host__ VectorField (int x, int y, int z) {
size_x = x; size_y = y; size_z = z;
is_copy = false;
CUDA_ERROR_CHECK (cudaMalloc(&data, x * y * z * sizeof(double3c)));
}
__host__ VectorField (const VectorField& other) {
size_x = other.size_x; size_y = other.size_y; size_z = other.size_z;
this->data = other.data;
is_copy = true;
}
__host__ ~VectorField() {
if (!is_copy) CUDA_ERROR_CHECK (cudaFree(data));
}
};
__global__ void KernelCalculateMeanFieldBlock (VectorField m, double3c* result) {
__shared__ double3c blockmean[THREADS_PER_BLOCK];
int index = threadIdx.x + blockIdx.x * blockDim.x;
if (index < m.size_x * m.size_y * m.size_z) blockmean[threadIdx.x] = m.data[index] = double3c(0, 1, 0);
else blockmean[threadIdx.x] = double3c(0,0,0);
__syncthreads();
for(int s = THREADS_PER_BLOCK / 2; s > 0; s /= 2) {
if (threadIdx.x < s) blockmean[threadIdx.x] += blockmean[threadIdx.x + s];
__syncthreads();
}
if(threadIdx.x == 0) result[blockIdx.x] = blockmean[0];
}
double3c CalculateMeanField (VectorField& m) {
int blocknum = (m.size_x * m.size_y * m.size_z - 1) / THREADS_PER_BLOCK + 1;
double3c *mean = new double3c[blocknum]();
double3c *cu_mean;
CUDA_ERROR_CHECK (cudaMalloc(&cu_mean, sizeof(double3c) * blocknum));
CUDA_ERROR_CHECK (cudaMemset (cu_mean, 0, sizeof(double3c) * blocknum));
KernelCalculateMeanFieldBlock <<<blocknum, THREADS_PER_BLOCK>>> (m, cu_mean);
CUDA_ERROR_CHECK (cudaPeekAtLastError());
CUDA_ERROR_CHECK (cudaDeviceSynchronize());
CUDA_ERROR_CHECK (cudaMemcpy(mean, cu_mean, sizeof(double3c) * blocknum, cudaMemcpyDeviceToHost));
CUDA_ERROR_CHECK (cudaFree(cu_mean));
for (int i = 1; i < blocknum; i++) {mean[0] += mean[i];}
mean[0] /= m.size_x * m.size_y * m.size_z;
double3c aux = mean[0];
delete[] mean;
return aux;
}
int main() {
VectorField m(100,100,100);
double3c sum = CalculateMeanField (m);
cout << sum.x << '\t' << sum.y << '\t' <<sum.z;
return 0;
}
编辑
贴出功能代码。用 10x10x10 元素构造 VectorField 可以正常工作并给出平均值 1,但用 100x100x100 元素构造它的平均值约为 0.97(它因运行而异)。这是进行并行缩减的正确方法,还是我应该坚持每个块启动一个内核?
【问题讨论】:
-
您应该提供完整的代码。
m是否已在设备上分配?在计算平均值之前,您需要在主机代码中使用一个循环来将每个块的结果相加(即将mean的blocknum元素相加到mean[0])。 -
我有循环,但未能在问题中复制它。是的,m 是在设备上分配的。编辑问题
-
请提供完整的MCVE。其他人可以复制、粘贴、编译和运行并观察问题的代码,无需添加任何内容或更改任何内容。
VectorField的构造函数正在对类成员(属性)m执行cudaMalloc,但您在传递给设备的类中引用了另一个类成员data。我宁愿不玩 20 个问题来尝试梳理这一切。请提供 MCVE。所以expects that.