Question

我想通过OpenACC优化此代码，但输出计算为零。我将非常感谢有机会以这种方式帮助我并使用您的指导来取得成功并解决我的麻烦。

金问候，穆罕默迪

  #include <stdio.h>
  #include <math.h>
  #include <stdlib.h>
  #include <assert.h>
  #include <openacc.h>
  #include<time.h>
  #include <string.h>
  #include <malloc.h>

  #define NX 4
  #define NY 4
  #define NZ 4
  int main(void)
  {
  int i, j,p, k,m;
  static double A[NX][NY][NZ]={0.} ,B[NX][NY][NZ]={0.},C[NX][NY][NZ]={0.},D[NX][NY][NZ]={0.};
  FILE *file;
  file = fopen("B-and-A.csv", "w");
#pragma acc data copyin(C,D),copy(A,B)
 {
for (p = 0; p <=5; p++) {
    #pragma acc kernels
    for ( i = 1; i < NX - 1; i++ ) {
            for ( j = 0; j < NY - 1; j++ ) {
                for ( k = 0; k < NZ - 1; k++ ) {
                 A[i][j][k] = A[i][j][k] + 1.*( B[i][j+1][k] + D[i][j][k] );
                        }
                    }
                }
#pragma acc kernels
 for ( i = 1; i < NX - 1; i++ ) {
        for ( j = 0; j < NY - 1; j++ ) {
            for ( k = 0; k < NZ - 1; k++ ) {
             B[i][j][k] = B[i][j][k]+ 1.*( A[i][j+1][k] + D[i][j][k] );
            }
        }
    }

    for (m = 0; m < NZ - 1; m++) {
        A[0][m][m] = -25. ;
        A[2][m][m] = 52. ;
        B[0][m][m] = 15. ;
        B[2][m][m] = -55. ;
                                }
#pragma acc update self(B)
fprintf(file,"%e\n",B[2][2][2]);
printf("%e\n",B[2][2][2]);
}
}
  fclose(file);
}

Answer 1

执行“更新自我（B）”时，您将B的设备值复制到B的主机副本。在此之前对主机副本所做的任何更改都将丢失。

对于此代码，您需要在主机上执行循环之前更新B，然后在主机循环后更新设备副本。或者，您可以将循环卸载到GPU，以便在设备上完成所有计算。

选项＃1：

// Update the host copies before executing on the host
#pragma acc update self(A,B)
                        // This loop is executed on the host
                        for (m = 0; m < NZ - 1; m++) {
                                A[0][m][m] = -25. ;
                                A[2][m][m] = 52. ;
                                B[0][m][m] = 15. ;
                                B[2][m][m] = -55. ;
                        }
// To keep the device and host copies coherent, update the device copies
#pragma acc update device(A,B)

选项＃2：

// Or offload the loop to the device
#pragma acc kernels
                        for (m = 0; m < NZ - 1; m++) {
                                A[0][m][m] = -25. ;
                                A[2][m][m] = 52. ;
                                B[0][m][m] = 15. ;
                                B[2][m][m] = -55. ;
                        }
#pragma acc update self(B)

使用OpenACC（数据子句）时输出错误

1 个答案: