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c++ - 二维数组CUDA问题

转载 作者:行者123 更新时间:2023-11-30 02:16:11 25 4
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我目前正在努力在我的 CUDA 内核中正确使用二维数组。 1D 很好,但到目前为止还没有转向 2D。这是我的宿主函数和内核:

__global__ void add_d2D(double *x, double *y,double *z, int n, int m){
for (int i = blockIdx.x * blockDim.x + threadIdx.x; i < n; i += blockDim.x * gridDim.x){
for(int j = blockIdx.y * blockDim.y + threadIdx.y; j < m; j += blockDim.y * gridDim.y){
z[i*m + j] = x[i*m + j] + y[i*m + j];
}
}
}

__host__ void add2D(double *a, double *b, double *result, int N, int M){
double *a_d, *b_d, *c_d;

size_t pitcha;
size_t pitchb;
size_t pitchc;

cudaErrchk(cudaMallocPitch(&a_d,&pitcha, M*sizeof(double),N));
cudaErrchk(cudaMallocPitch(&b_d,&pitchb, M*sizeof(double),N));
cudaErrchk(cudaMallocPitch(&c_d,&pitchc, M*sizeof(double),N));
cudaErrchk(cudaMemcpy2D(a_d,M*sizeof(double), a,pitcha, M*sizeof(double),N, cudaMemcpyHostToDevice));
cudaErrchk(cudaMemcpy2D(b_d,M*sizeof(double), b,pitchb, M*sizeof(double),N, cudaMemcpyHostToDevice));

dim3 threadsPerBlock(2, 2);
dim3 numBlocks(N/threadsPerBlock.x, M/threadsPerBlock.y);

add_d2D<<<numBlocks, threadsPerBlock>>>(a_d, b_d, c_d , N, M);

cudaDeviceSynchronize();

cudaErrchk(cudaMemcpy2D(result,M*sizeof(double), c_d,pitchc, M*sizeof(double),N, cudaMemcpyDeviceToHost));

cudaFree(a_d);
cudaFree(b_d);
cudaFree(c_d);
}

下面是我的例子来测试它。它正确地打印出 C 的前 10 个值,但所有其他值都保持为 0。我相信问题出在内核中。由于间距问题,它无法找到正确的值,但不确定如何正确解决它。

double a[4][10];
double b[4][10];
double c[4][10];
for (int i = 0; i < 4; i ++){
for (int j = 0; j < 10; j ++){
a[i][j] = 0 + rand() % 10;
b[i][j] = 0 + rand() % 10;
}
}
ertiscuda::add2D((double *)a, (double *)b, (double *)c, 4, 10);
for (int i = 0; i < 4; i ++){
for (int j = 0; j < 10; j ++){
std::cout << a[i][j] << " " << b[i][j] << " " << c[i][j] << std::endl;
}
}

最佳答案

你有两个错误

  1. 内核中的每个线程应该执行一个操作而不是所有操作。 (出于内存原因,您可能想要做更多的事情,我们会让这个示例保持简单)。

  2. 将数据加载到设备时,您切换了目标音高和源音高。

这是一个工作版本

#include <cuda_runtime.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>

#define CUDASAFECALL( err ) cuda_safe_call(err, __FILE__, __LINE__ )
void cuda_safe_call(const cudaError err, const char *file, const int line)
{
if (cudaSuccess != err)
{
std::stringstream error_msg;
error_msg << "cuda_safe_call() failed at " << file << ":" << line << ":" << cudaGetErrorString(err);
const auto error_msg_str = error_msg.str();
std::cout << error_msg_str << std::endl;
throw std::runtime_error(error_msg_str);
}
}

__global__ void add_d2D(const double *x, const double *y, double *z, int n, int m, int m_pitch_elements)
{
int row = blockIdx.x * blockDim.x + threadIdx.x;
int col = blockIdx.y * blockDim.y + threadIdx.y;
if (row< n && col <m )
{
auto idx = row*m_pitch_elements + col;
z[idx] = x[idx] + y[idx];
//z[idx] = idx;
}
}

__host__ void add2D(const double *a,const double *b, double *result, int N, int M) {
double *a_d, *b_d, *c_d;
size_t pitcha,pitchb,pitchc;

CUDASAFECALL(cudaMallocPitch(&a_d, &pitcha, M * sizeof(double), N));
CUDASAFECALL(cudaMallocPitch(&b_d, &pitchb, M * sizeof(double), N));
CUDASAFECALL(cudaMallocPitch(&c_d, &pitchc, M * sizeof(double), N));
CUDASAFECALL(cudaMemcpy2D(a_d, pitcha, a, M * sizeof(double), M * sizeof(double), N, cudaMemcpyHostToDevice));
CUDASAFECALL(cudaMemcpy2D(b_d, pitchb, b, M * sizeof(double), M * sizeof(double), N, cudaMemcpyHostToDevice));

dim3 threadsPerBlock(2, 2);
auto safediv = [](auto a, auto b) {return static_cast<unsigned int>(ceil(a / (b*1.0))); };
dim3 numBlocks(safediv(N, threadsPerBlock.x), safediv( M, threadsPerBlock.y));
//all the pitches should be the same
auto pitch_elements = pitcha / sizeof(double);
add_d2D << <numBlocks, threadsPerBlock >> >(a_d, b_d, c_d, N, M, pitch_elements);

CUDASAFECALL(cudaDeviceSynchronize());

CUDASAFECALL(cudaMemcpy2D(result, M * sizeof(double), c_d, pitchc, M * sizeof(double), N, cudaMemcpyDeviceToHost));

CUDASAFECALL(cudaFree(a_d));
CUDASAFECALL(cudaFree(b_d));
CUDASAFECALL(cudaFree(c_d));
}

int main()
{
double a[4][10];
double b[4][10];
double c[4][10];
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 10; j++) {
a[i][j] = 0 + rand() % 10;
b[i][j] = 0 + rand() % 10;
}
}
add2D((double *)a, (double *)b, (double *)c, 4, 10);
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 10; j++) {
std::cout << a[i][j] << " " << b[i][j] << " " << c[i][j]<< "|"<< a[i][j]+ b[i][j] << std::endl;
}
}
return 0;
}

关于c++ - 二维数组CUDA问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55212450/

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