在 CUDA 内核中实现 memcpy 的正确方法是什么？

Question

我正在 CUDA 中实现 PDE 求解器 (Lax-Friedrichs)，这是我之前用 C 编写的。请在下面找到 C 代码：

void solve(int M, double u[M+3][M+3], double unp1[M+3][M+3], double params[3]){
 int i;
 int j;
 int n;


 for (n=0; n<params[0]; n++){
        for (i=0; i<M+2; i++)
           for(j=0; j<M+2; j++){
              unp1[i][j] = 0.25*(u[i+1][j] + u[i-1][j] + u[i][j+1] + u[i][j-1])
                 - params[1]*(u[i+1][j] - u[i-1][j])
                 - params[2]*(u[i][j+1] - u[i][j-1]);
           }
 
        memcpy(u, unp1, pow(M+3,2)*sizeof(double));
 
 
        /*Periodic Boundary Conditions*/
        for (i=0; i<M+2; i++){
           u[0][i] = u[N+1][i];
           u[i][0] = u[i][N+1];
           u[N+2][i] = u[1][i];
           u[i][N+2] = u[i][1];
        }
     }
  }

而且效果很好。但是当我试图在 CUDA 中实现它时，我没有得到相同的数据。不幸的是，我无法准确指出确切的问题，因为我是整个并行编程的初学者，但我认为这可能与求解器上的 u[i*(N+3) + j] = unp1[i*(N+3) + j] 有关，因为我无法真正在内核中执行 memcpy ，因为它没有改变任何东西，我不知道如何进行。我看了一下This previous answer，可惜没能解决我的问题。这是我正在尝试编码的 CUDA 求解器：

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

/*Configuration of the grid*/
const int N = 100; //Number of nodes  
const double xmin = -1.0;
const double ymin = -1.0;
const double xmax = 1.0;
const double ymax = 1.0;
const double tmax = 0.5;

/*Configuration of the simulation physics*/
const double dx = (xmax - xmin)/N;
const double dy = (ymax - ymin)/N;
const double dt = 0.009;
const double vx = 1.0;
const double vy = 1.0;


__global__ void initializeDomain(double *x, double *y){
    /*Initializes the grid of size (N+3)x(N+3) to better accomodate Boundary Conditions*/
    int index = blockIdx.x * blockDim.x + threadIdx.x;
    int stride = blockDim.x * gridDim.x;

    for (int j=index ; j<N+3; j+=stride){
        x[j] = xmin + (j-1)*dx;
        y[j] = ymin + (j-1)*dy;
    }
}


__global__ void initializeU(double *x, double *y, double *u0){
    
    double sigma_x = 2.0;
    double sigma_y = 6.0;
    
    int index_x = blockIdx.x * blockDim.x + threadIdx.x;
    int stride_x = blockDim.x * gridDim.x;
    int index_y = blockIdx.y * blockDim.y + threadIdx.y;
    int stride_y = blockDim.y * gridDim.y;
    
    for (int i = index_x; i < N+3; i += stride_x)
        for (int j = index_y; j < N+3; j+= stride_y){
            u0[i*(N+3) + j] = exp(-200*(pow(x[i],2)/(2*pow(sigma_x,2)) + pow(y[j],2)/(2*pow(sigma_y,2))));
            u0[i*(N+3) + j] *= 1/(2*M_PI*sigma_x*sigma_y);
               //u[i*(N+3) + j] = u0[i*(N+3) + j];
               //unp1[i*(N+3) + j] = u0[i*(N+3) + j];
    }
}


void initializeParams(double params[3]){
    params[0] = round(tmax/dt);
    params[1] = vx*dt/(2*dx);
    params[2] = vy*dt/(2*dy);
}


__global__ void solve(double *u, double *unp1, double params[3]){

     int index_x = blockIdx.x * blockDim.x + threadIdx.x;
     int stride_x = blockDim.x * gridDim.x;
     int index_y = blockIdx.y * blockDim.y + threadIdx.y;
     int stride_y = blockDim.y * gridDim.y; 

     for (int i = index_x; i < N+2; i += stride_x)
          for (int j = index_y; j < N+2; j += stride_y){
               unp1[i*(N+3) + j] = 0.25*(u[(i+1)*(N+3) + j] + u[(i-1)*(N+3) + j] + u[i*(N+3) + (j+1)] + u[i*(N+3) + (j-1)]) \
               - params[1]*(u[(i+1)*(N+3) + j] - u[(i-1)*(N+3) + j]) \
               - params[2]*(u[i*(N+3) + (j+1)] - u[i*(N+3) + (j-1)]);
          }

}


__global__ void bc(double *u){
     int index_x = blockIdx.x * blockDim.x + threadIdx.x;
     int stride_x = blockDim.x * gridDim.x;


     /*Also BC are set on parallel */
     for (int i = index_x; i < N+2; i += stride_x){
          u[0*(N+3) + i] = u[(N+1)*(N+3) + i];
          u[i*(N+3) + 0] = u[i*(N+3) + (N+1)];
          u[(N+2)*(N+3) + i] = u[1*(N+3) + i];
          u[i*(N+3) + (N+2)] = u[i*(N+3) + 1];
     }
}


int main(){
     int i;
     int j;

     double *x = (double *)malloc((N+3)*sizeof(double));
     double *y = (double *)malloc((N+3)*sizeof(double));

     double *d_x, *d_y;
     cudaMalloc(&d_x, (N+3)*sizeof(double));
     cudaMalloc(&d_y, (N+3)*sizeof(double));

     initializeDomain<<<1, 1>>>(d_x, d_y);
     cudaDeviceSynchronize();

     cudaMemcpy(x, d_x, (N+3)*sizeof(double), cudaMemcpyDeviceToHost);
     cudaMemcpy(y, d_y, (N+3)*sizeof(double), cudaMemcpyDeviceToHost);

     FILE *fout1 = fopen("data_x.csv", "w");
    FILE *fout2 = fopen("data_y.csv", "w");

    for (i=0; i<N+3; i++){
        if (i==N+2){
            fprintf(fout1, "%.5f", x[i]);
            fprintf(fout2, "%.5f", y[i]);
        }
        else{
            fprintf(fout1, "%.5f, ", x[i]);
            fprintf(fout2, "%.5f, ", y[i]);
        }
    }


     dim3 Block2D(1,1);
     dim3 ThreadsPerBlock(1,1);

     double *d_u0;
     double *u0 = (double *)malloc((N+3)*(N+3)*sizeof(double));
     cudaMalloc(&d_u0, (N+3)*(N+3)*sizeof(double));

     initializeU<<<Block2D, ThreadsPerBlock>>>(d_x, d_y, d_u0);
     cudaDeviceSynchronize();
     cudaMemcpy(u0, d_u0, (N+3)*(N+3)*sizeof(double), cudaMemcpyDeviceToHost);

     /*Initialize parameters*/
     double params[3];
     initializeParams(params);

     /*Allocate memory for u and unp1 on device for the solver*/
     double *d_u, *d_unp1;
     cudaMalloc(&d_u, (N+3)*(N+3)*sizeof(double));
     cudaMalloc(&d_unp1, (N+3)*(N+3)*sizeof(double));
     cudaMemcpy(d_u, d_u0, (N+3)*(N+3)*sizeof(double), cudaMemcpyDeviceToDevice);
     cudaMemcpy(d_unp1, d_u0, (N+3)*(N+3)*sizeof(double), cudaMemcpyDeviceToDevice);
     
     /*Solve*/
     for (int n=0; n<params[0]; n++){
          solve<<<Block2D, ThreadsPerBlock>>>(d_u, d_unp1, params);
          double *temp = d_u;
          d_u = d_unp1;
          d_unp1 = temp;
          bc<<<1,1>>>(d_u);
          cudaDeviceSynchronize();
     }

     /*Copy results on host*/
     double *u = (double *)malloc((N+3)*(N+3)*sizeof(double));
     cudaMemcpy(u, d_u, (N+3)*(N+3)*sizeof(double), cudaMemcpyDeviceToHost);

     FILE *fu = fopen("data_u.csv", "w");
    for (i=0; i<N+3; i++){
          for(j=0; j<N+3; j++)
            if (j==N+2)
                fprintf(fu, "%.5f", u[i*(N+3) + j]);
            else
                fprintf(fu, "%.5f, ", u[i*(N+3) + j]);
        fprintf(fu, "\n");
    }
     fclose(fu);

     free(x);
     free(y);
     free(u0);
     free(u);
     cudaFree(d_x);
     cudaFree(d_y);
     cudaFree(d_u0);
     cudaFree(d_u);
     cudaFree(d_unp1);

     return 0;
}

不幸的是，我一直遇到同样的问题：我得到的数据是 0.0000。

Answer 1

让您感到困惑的一件事是您的原始算法具有正确性所需的顺序：

从 u

unp

复制unp到u
执行边界条件
重复

您的算法要求第 1 步完全在第 2 步开始之前完成，同样第 2 步在第 3 步之前完成。您的 CUDA 实现（将第 1 步和第 3 步，或 1 ,2,3) 在单个内核中不保留或保证该顺序。 CUDA 线程可以以任何顺序执行。如果您严格地将其应用于您的代码（例如，假设索引为 0 的线程在任何其他线程开始之前 完全执行 。这将是有效的 CUDA 执行）那么您将看到您的内核设计不保留所需的顺序。

所以做这样的事情：

创建求解核只是第一步：

 __global__ void solve(double *u, double *unp1, double params[3]){

      int index_x = blockIdx.x * blockDim.x + threadIdx.x;
      int stride_x = blockDim.x * gridDim.x;
      int index_y = blockIdx.y * blockDim.y + threadIdx.y;
      int stride_y = blockDim.y * gridDim.y;

      for (int i = index_x; i < N+2; i += stride_x)
          for (int j = index_y; j < N+2; j += stride_y){
               unp1[i*(N+3) + j] = 0.25*(u[(i+1)*(N+3) + j] + u[(i-1)*(N+3) + j] + u[i*(N+3) + (j+1)] + u[i*(N+3) + (j-1)]) \
               - params[1]*(u[(i+1)*(N+3) + j] - u[(i-1)*(N+3) + j]) \
               - params[2]*(u[i*(N+3) + (j+1)] - u[i*(N+3) + (j-1)]);
               u[i*(N+3) + j] = unp1[i*(N+3) + j];
          }
 }

不用理会memcpy操作。更好的方法是交换指针（在 主机代码 中）。

创建一个单独的内核来强制执行边界：

 __global__ void bc(double *u, double *unp1, double params[3]){

      int index_x = blockIdx.x * blockDim.x + threadIdx.x;
      int stride_x = blockDim.x * gridDim.x;
      int index_y = blockIdx.y * blockDim.y + threadIdx.y;
      int stride_y = blockDim.y * gridDim.y;
      /*Also BC are set on parallel */
      for (int i = index_x; i < N+2; i += stride_x){
           u[0*(N+3) + i] = u[(N+1)*(N+3) + i];
           u[i*(N+3) + 0] = u[i*(N+3) + (N+1)];
           u[(N+2)*(N+3) + i] = u[1*(N+3) + i];
           u[i*(N+3) + (N+2)] = u[i*(N+3) + 1];
      }
 }

修改您的主机代码以按顺序调用这些内核，中间交换指针：

   /*Solve*/
   for(int n = 0; n<params[0]; n++){
        solve<<<Block2D, ThreadsPerBlock>>>(d_u, d_unp1, params);
        double *temp = d_u;
        d_u = d_unp1;
        d_unp1 = temp;
        bc<<<Block2D, ThreadsPerBlock>>>(d_u, d_unp1, params);
        cudaDeviceSynchronize();
   }

（在浏览器中编码，未经测试）

这将强制执行您的算法所需的顺序。

注意：正如下面评论中所指出的，solve 内核如上所述（以及在 OP 的原始 post 和他们的 posted CPU 代码中version) 有至少与 i-1 和 j-1 索引模式相关的索引错误。这些应该被修复，否则代码损坏。修复它们需要决定如何处理边缘情况，OP 没有提供指导，因此我保留了该代码。

在 CUDA 内核中实现 memcpy 的正确方法是什么？

What is a correct way to implement memcpy inside a CUDA kernel?

cuda

numerical-methods