我正在学习CUDA，试图解决一些标准问题。作为一个例子，我用以下代码在两个维度上求解扩散方程。但我的结果与标准结果不同，我无法弄明白。
//kernel definition
__global__ void diffusionSolver(double* A, double * old,int n_x,int n_y)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
int j = blockIdx.y * blockDim.y + threadIdx.y;
if(i*(n_x-i-1)*j*(n_y-j-1)!=0)
A[i+n_y*j] = A[i+n_y*j] + (old[i-1+n_y*j]+old[i+1+n_y*j]+
old[i+(j-1)*n_y]+old[i+(j+1)*n_y] -4*old[i+n_y*j])/40;
}
int main()
{
int i,j ,M;
M = n_y ;
phi = (double *) malloc( n_x*n_y* sizeof(double));
phi_old = (double *) malloc( n_x*n_y* sizeof(double));
dummy = (double *) malloc( n_x*n_y* sizeof(double));
int iterationMax =10;
//phase initialization
for(j=0;j<n_y ;j++)
{
for(i=0;i<n_x;i++)
{
if((.4*n_x-i)*(.6*n_x-i)<0)
phi[i+M*j] = -1;
else
phi[i+M*j] = 1;
phi_old[i+M*j] = phi[i+M*j];
}
}
double *dev_phi;
cudaMalloc((void **) &dev_phi, n_x*n_y*sizeof(double));
dim3 threadsPerBlock(100,10);
dim3 numBlocks(n_x*n_y / threadsPerBlock.x, n_x*n_y / threadsPerBlock.y);
//start iterating
for(int z=0; z<iterationMax; z++)
{
//copy array on host to device
cudaMemcpy(dev_phi, phi, n_x*n_y*sizeof(double),
cudaMemcpyHostToDevice);
//call kernel
diffusionSolver<<<numBlocks, threadsPerBlock>>>(dev_phi, phi_old,n_x,n_y);
//get updated array back on host
cudaMemcpy(phi, dev_phi,n_x*n_y*sizeof(double), cudaMemcpyDeviceToHost);
//old values will be assigned new values
for(j=0;j<n_y ;j++)
{
for(i=0;i<n_x;i++)
{
phi_old[i+n_y*j] = phi[i+n_y*j];
}
}
}
return 0;
}
有人能告诉我这个过程中是否有什么问题？任何帮助将不胜感激。
答案 0 :(得分：3)
你遇到的一个大错误是phi_old传递给内核并由内核使用，但这是一个主机指针。
使用cudaMalloc的malloc a dev_phi_old。将其设置为默认值并在进入z循环之前将其复制到GPU。
答案 1 :(得分：3)
talonmies，brano和huseyin已经指出了你的代码中的一些错误。
扩散（热）方程是可用CUDA求解的偏微分方程的经典例子之一。在CUDA by Example书的第7章中也有一个完整的例子。
作为对未来用户的参考，我在下面提供了一个完整的工作示例，包括CPU和GPU代码。而不是像talonmies所建议的那样交换指针，我只是在一个循环中凝聚两个Jacobi迭代。
#include <iostream>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include "Utilities.cuh"
#define BLOCK_SIZE_X 16
#define BLOCK_SIZE_Y 16
/***********************************/
/* JACOBI ITERATION FUNCTION - GPU */
/***********************************/
__global__ void Jacobi_Iterator_GPU(const float * __restrict__ T_old, float * __restrict__ T_new, const int NX, const int NY)
{
const int i = blockIdx.x * blockDim.x + threadIdx.x ;
const int j = blockIdx.y * blockDim.y + threadIdx.y ;
// N
int P = i + j*NX; // node (i,j) |
int N = i + (j+1)*NX; // node (i,j+1) |
int S = i + (j-1)*NX; // node (i,j-1) W ---- P ---- E
int E = (i+1) + j*NX; // node (i+1,j) |
int W = (i-1) + j*NX; // node (i-1,j) |
// S
// --- Only update "interior" (not boundary) node points
if (i>0 && i<NX-1 && j>0 && j<NY-1) T_new[P] = 0.25 * (T_old[E] + T_old[W] + T_old[N] + T_old[S]);
}
/***********************************/
/* JACOBI ITERATION FUNCTION - CPU */
/***********************************/
void Jacobi_Iterator_CPU(float * __restrict T, float * __restrict T_new, const int NX, const int NY, const int MAX_ITER)
{
for(int iter=0; iter<MAX_ITER; iter=iter+2)
{
// --- Only update "interior" (not boundary) node points
for(int j=1; j<NY-1; j++)
for(int i=1; i<NX-1; i++) {
float T_E = T[(i+1) + NX*j];
float T_W = T[(i-1) + NX*j];
float T_N = T[i + NX*(j+1)];
float T_S = T[i + NX*(j-1)];
T_new[i+NX*j] = 0.25*(T_E + T_W + T_N + T_S);
}
for(int j=1; j<NY-1; j++)
for(int i=1; i<NX-1; i++) {
float T_E = T_new[(i+1) + NX*j];
float T_W = T_new[(i-1) + NX*j];
float T_N = T_new[i + NX*(j+1)];
float T_S = T_new[i + NX*(j-1)];
T[i+NX*j] = 0.25*(T_E + T_W + T_N + T_S);
}
}
}
/******************************/
/* TEMPERATURE INITIALIZATION */
/******************************/
void Initialize(float * __restrict h_T, const int NX, const int NY)
{
// --- Set left wall to 1
for(int j=0; j<NY; j++) h_T[j * NX] = 1.0;
}
/********/
/* MAIN */
/********/
int main()
{
const int NX = 256; // --- Number of discretization points along the x axis
const int NY = 256; // --- Number of discretization points along the y axis
const int MAX_ITER = 1; // --- Number of Jacobi iterations
// --- CPU temperature distributions
float *h_T = (float *)calloc(NX * NY, sizeof(float));
float *h_T_old = (float *)calloc(NX * NY, sizeof(float));
Initialize(h_T, NX, NY);
Initialize(h_T_old, NX, NY);
float *h_T_GPU_result = (float *)malloc(NX * NY * sizeof(float));
// --- GPU temperature distribution
float *d_T; gpuErrchk(cudaMalloc((void**)&d_T, NX * NY * sizeof(float)));
float *d_T_old; gpuErrchk(cudaMalloc((void**)&d_T_old, NX * NY * sizeof(float)));
gpuErrchk(cudaMemcpy(d_T, h_T, NX * NY * sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_T_old, d_T, NX * NY * sizeof(float), cudaMemcpyDeviceToDevice));
// --- Grid size
dim3 dimBlock(BLOCK_SIZE_X, BLOCK_SIZE_Y);
dim3 dimGrid (iDivUp(NX, BLOCK_SIZE_X), iDivUp(NY, BLOCK_SIZE_Y));
// --- Jacobi iterations on the host
Jacobi_Iterator_CPU(h_T, h_T_old, NX, NY, MAX_ITER);
// --- Jacobi iterations on the device
for (int k=0; k<MAX_ITER; k=k+2) {
Jacobi_Iterator_GPU<<<dimGrid, dimBlock>>>(d_T, d_T_old, NX, NY); // --- Update d_T_old starting from data stored in d_T
Jacobi_Iterator_GPU<<<dimGrid, dimBlock>>>(d_T_old, d_T , NX, NY); // --- Update d_T starting from data stored in d_T_old
}
// --- Copy result from device to host
gpuErrchk(cudaMemcpy(h_T_GPU_result, d_T, NX * NY * sizeof(float), cudaMemcpyDeviceToHost));
// --- Calculate percentage root mean square error between host and device results
float sum = 0., sum_ref = 0.;
for (int j=0; j<NY; j++)
for (int i=0; i<NX; i++) {
sum = sum + (h_T_GPU_result[j * NX + i] - h_T[j * NX + i]) * (h_T_GPU_result[j * NX + i] - h_T[j * NX + i]);
sum_ref = sum_ref + h_T[j * NX + i] * h_T[j * NX + i];
}
printf("Percentage root mean square error = %f\n", 100.*sqrt(sum / sum_ref));
// --- Release host memory
free(h_T);
free(h_T_GPU_result);
// --- Release device memory
gpuErrchk(cudaFree(d_T));
gpuErrchk(cudaFree(d_T_old));
return 0;
}
运行此类示例所需的 Utilities.cu 和 Utilities.cuh 文件将保留在此github page。
答案 2 :(得分：2)
下面：
A[i+n_y*j] = A[i+n_y*j] + (old[i-1+n_y*j]+old[i+1+n_y*j]+old[i+(j-1)*n_y]+old[i+(j+1)*n_y] -4*old[i+n_y*j])/40;
您将除以40（整数），这可能导致错误的扩散率。实际上可以导致无扩散。
但是A是一系列双打。
将漫射率除以40.0并查看它是否有效。
如果这是来自Jos-Stam的求解器，则应该是4.0而不是40
还有另外一件事：
-4*old[i+n_y*j])/40;
这里乘以4（整数）。这也可能导致整体铸造！
此：
-4.0*old[i+n_y*j])/40.0;
减少了一些错误。
度过愉快的一天。