使用 OpenACC 并行化嵌套循环
Using OpenACC to parallelize nested loops
我是 openacc 的新手，只有高级知识，所以任何帮助和解释我做错的事情都将不胜感激。
我正在尝试加速（并行化）一个不太直接的嵌套循环，该循环使用 openacc 指令更新展平（3D 到 1D）数组。当使用
编译时，我在下面发布了一个简化的示例代码
pgcc -acc -Minfo=accel test.c
给出以下错误：
call to cuStreamSynchronize returned error 700: Illegal address during kernel execution
代码：
#include <stdio.h>
#include <stdlib.h>
#define min(a,b) (a > b) ? b : a
#define max(a,b) (a < b) ? b : a
#define NX 10
#define NY 10
#define NZ 10
struct phiType {
double dx, dy, dz;
double * distance;
};
typedef struct phiType Phi;
#pragma acc routine seq
double solve(Phi *p, int index) {
// for simplicity just returning a value
return 2;
}
void fast_sweep(Phi *p) {
// removing boundaries
int x = NX - 2;
int y = NY - 2;
int z = NZ - 2;
int startLevel = 3;
int endLevel = x + y + z;
#pragma acc data copy(p->distance[0:NX*NY*NZ])
for(int level = startLevel; level <= endLevel; level++){
int ks = max(1, level-(y + z));
int ke = min(x, level-2);
int js = max(1, level-(x + z));
int je = min(y, level-2);
#pragma acc region
{
#pragma acc loop independent
for(int k = ks; k <= ke; k++){
#pragma acc loop independent
for(int j = js; j <= je; j++){
int i = level - (k + j);
if(i > 0 && i <= z){
int index = i * NX * NY + j * NX + k;
p->distance[index] = solve(p, index);
}
}
}
}
}
}
void create_phi(Phi *p){
p->dx = 1;
p->dy = 1;
p->dz = 1;
p->distance = (double *) malloc(sizeof(double) * NX * NY * NZ);
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
p->distance[index] = (i*j*k == 0) ? 0 : 1;
}
}
}
}
int main()
{
printf("start \n");
Phi *p = (Phi *) malloc(sizeof(Phi));
create_phi(p);
printf("calling fast sweep \n");
fast_sweep(p);
printf(" print the results \n");
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
printf("%f ", p->distance[index]);
}
printf("\n");
}
printf("\n");
}
return 0;
}
不使用 region 和 loop 指令，而是使用
#pragma acc kernels
产生以下错误：
solve:
19, Generating acc routine seq
fast_sweep:
34, Generating copy(p->distance[:1000])
42, Generating copy(p[:1])
45, Loop carried dependence due to exposed use of p[:1] prevents parallelization
Accelerator scalar kernel generated
47, Loop carried dependence due to exposed use of p[:i1+1] prevents parallelization
我运行这个代码在
GNU/Linux
CentOS release 6.7 (Final)
GeForce GTX Titan
pgcc 15.7-0 64-bit target on x86-64 Linux -tp sandybridge
错误来自 GPU 上的计算内核取消引用 CPU 指针。这是一个非常普遍的问题，OpenACC 委员会正在努力解决这个问题。像这样的动态数据结构确实会导致很多问题，所以我们想修复它。这里有两种可能的解决方法。
1) 在编译器安装期间通过 PGI "unified memory evaluation package" 选项使用 "managed memory"。这是一个 beta 功能，但它会将您的所有数据放入一种特殊类型的内存中，这种内存对 CPU 和 GPU 都可见。您应该在文档中阅读很多注意事项，最重要的是，您受限于 GPU 上可用的内存量，并且无法从 CPU 访问正在使用的内存GPU，但这是一种可能的解决方法。假设您在安装期间启用了该选项，只需将 -ta=tesla:managed 添加到您的编译器标志中即可将其打开。我用你的代码试过了，它成功了。
2) 添加指向代码的指针，这样您就不会通过 p 访问 distance，而是直接访问它，如下所示：
double *distance = p->distance;
#pragma acc data copy(p[0:1],distance[0:NX*NY*NZ])
for(int level = startLevel; level <= endLevel; level++){
int ks = max(1, level-(y + z));
int ke = min(x, level-2);
int js = max(1, level-(x + z));
int je = min(y, level-2);
#pragma acc parallel
{
#pragma acc loop independent
for(int k = ks; k <= ke; k++){
#pragma acc loop independent
for(int j = js; j <= je; j++){
int i = level - (k + j);
if(i > 0 && i <= z){
int index = i * NX * NY + j * NX + k;
distance[index] = solve(p, index);
}
}
}
}
我知道当有很多数据数组要执行此操作时，这可能会很痛苦，但这是我在很多代码中成功使用的解决方法。不幸的是，这是必要的，这就是为什么我们希望在未来版本的 OpenACC 中提供更好的解决方案。
希望对您有所帮助！如果我能想出一个不需要额外指针的解决方案，我会更新这个答案。
Jeff 是正确的，OpenACC 委员会仍在研究如何标准化对具有动态数据成员的聚合数据类型的支持。但是，对于 PGI 14.9 或更高版本，我们添加了对结构和 C++ 类 的更好支持，因此在这种情况下，您只需添加 create(p[0:1]) 即可简化代码。将会发生的是，编译器将创建一个 p 的设备副本，其中只为数据成员分配内存。那么当你做p->distance的copy时，内存会分配给"distance"然后attach到p。 （即 运行 时间将填充结构中的设备指针）。
有一些注意事项。首先是此行为尚未标准化，因此其他编译器（如 Cray、Pathscale、GCC 等）可能具有不同的行为。第二，顺序很重要。 p 需要在附加 distance 之前创建。第三，更复杂的数据结构变得很难管理。正如 Jeff 所建议的，使用 CUDA 统一内存是管理复杂数据结构的一个很好的选择。
如果您有兴趣，我的 GTC2015 演示文稿的大部分内容都在讨论这个主题 (link)。演讲的重点是 C++ Class 数据管理，但也适用于 C 结构。
希望这对您有所帮助，
垫子
% cat test1.c
#include <stdio.h>
#include <stdlib.h>
#define min(a,b) (a > b) ? b : a
#define max(a,b) (a < b) ? b : a
#define NX 10
#define NY 10
#define NZ 10
struct phiType {
double dx, dy, dz;
double * distance;
};
typedef struct phiType Phi;
#pragma acc routine seq
double solve(Phi *p, int index) {
// for simplicity just returning a value
return 2;
}
void fast_sweep(Phi *p) {
// removing boundaries
int x = NX - 2;
int y = NY - 2;
int z = NZ - 2;
int startLevel = 3;
int endLevel = x + y + z;
#pragma acc data create(p[0:1]) copy(p->distance[0:NX*NY*NZ])
for(int level = startLevel; level <= endLevel; level++){
int ks = max(1, level-(y + z));
int ke = min(x, level-2);
int js = max(1, level-(x + z));
int je = min(y, level-2);
#pragma acc region
{
#pragma acc loop independent
for(int k = ks; k <= ke; k++){
#pragma acc loop independent
for(int j = js; j <= je; j++){
int i = level - (k + j);
if(i > 0 && i <= z){
int index = i * NX * NY + j * NX + k;
p->distance[index] = solve(p, index);
}
}
}
}
}
}
void create_phi(Phi *p){
p->dx = 1;
p->dy = 1;
p->dz = 1;
p->distance = (double *) malloc(sizeof(double) * NX * NY * NZ);
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
p->distance[index] = (i*j*k == 0) ? 0 : 1;
}
}
}
}
int main()
{
printf("start \n");
Phi *p = (Phi *) malloc(sizeof(Phi));
create_phi(p);
printf("calling fast sweep \n");
fast_sweep(p);
printf(" print the results \n");
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
printf("%f ", p->distance[index]);
}
printf("\n");
}
printf("\n");
}
return 0;
}
% pgcc -acc -ta=tesla:cc35 -Minfo=accel test1.c -V15.7 ; a.out
solve:
19, Generating acc routine seq
fast_sweep:
34, Generating create(p[:1])
Generating copy(p->distance[:1000])
45, Loop is parallelizable
47, Loop is parallelizable
Accelerator kernel generated
Generating Tesla code
45, #pragma acc loop gang /* blockIdx.y */
47, #pragma acc loop gang, vector(128) /* blockIdx.x threadIdx.x */
start
calling fast sweep
print the results
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我是 openacc 的新手，只有高级知识，所以任何帮助和解释我做错的事情都将不胜感激。
我正在尝试加速（并行化）一个不太直接的嵌套循环，该循环使用 openacc 指令更新展平（3D 到 1D）数组。当使用
编译时，我在下面发布了一个简化的示例代码pgcc -acc -Minfo=accel test.c
给出以下错误：
call to cuStreamSynchronize returned error 700: Illegal address during kernel execution
代码：
#include <stdio.h>
#include <stdlib.h>
#define min(a,b) (a > b) ? b : a
#define max(a,b) (a < b) ? b : a
#define NX 10
#define NY 10
#define NZ 10
struct phiType {
double dx, dy, dz;
double * distance;
};
typedef struct phiType Phi;
#pragma acc routine seq
double solve(Phi *p, int index) {
// for simplicity just returning a value
return 2;
}
void fast_sweep(Phi *p) {
// removing boundaries
int x = NX - 2;
int y = NY - 2;
int z = NZ - 2;
int startLevel = 3;
int endLevel = x + y + z;
#pragma acc data copy(p->distance[0:NX*NY*NZ])
for(int level = startLevel; level <= endLevel; level++){
int ks = max(1, level-(y + z));
int ke = min(x, level-2);
int js = max(1, level-(x + z));
int je = min(y, level-2);
#pragma acc region
{
#pragma acc loop independent
for(int k = ks; k <= ke; k++){
#pragma acc loop independent
for(int j = js; j <= je; j++){
int i = level - (k + j);
if(i > 0 && i <= z){
int index = i * NX * NY + j * NX + k;
p->distance[index] = solve(p, index);
}
}
}
}
}
}
void create_phi(Phi *p){
p->dx = 1;
p->dy = 1;
p->dz = 1;
p->distance = (double *) malloc(sizeof(double) * NX * NY * NZ);
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
p->distance[index] = (i*j*k == 0) ? 0 : 1;
}
}
}
}
int main()
{
printf("start \n");
Phi *p = (Phi *) malloc(sizeof(Phi));
create_phi(p);
printf("calling fast sweep \n");
fast_sweep(p);
printf(" print the results \n");
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
printf("%f ", p->distance[index]);
}
printf("\n");
}
printf("\n");
}
return 0;
}
不使用 region 和 loop 指令，而是使用
#pragma acc kernels
产生以下错误：
solve:
19, Generating acc routine seq
fast_sweep:
34, Generating copy(p->distance[:1000])
42, Generating copy(p[:1])
45, Loop carried dependence due to exposed use of p[:1] prevents parallelization
Accelerator scalar kernel generated
47, Loop carried dependence due to exposed use of p[:i1+1] prevents parallelization
我运行这个代码在
GNU/Linux
CentOS release 6.7 (Final)
GeForce GTX Titan
pgcc 15.7-0 64-bit target on x86-64 Linux -tp sandybridge
错误来自 GPU 上的计算内核取消引用 CPU 指针。这是一个非常普遍的问题，OpenACC 委员会正在努力解决这个问题。像这样的动态数据结构确实会导致很多问题，所以我们想修复它。这里有两种可能的解决方法。
1) 在编译器安装期间通过 PGI "unified memory evaluation package" 选项使用 "managed memory"。这是一个 beta 功能，但它会将您的所有数据放入一种特殊类型的内存中，这种内存对 CPU 和 GPU 都可见。您应该在文档中阅读很多注意事项，最重要的是，您受限于 GPU 上可用的内存量，并且无法从 CPU 访问正在使用的内存GPU，但这是一种可能的解决方法。假设您在安装期间启用了该选项，只需将 -ta=tesla:managed 添加到您的编译器标志中即可将其打开。我用你的代码试过了，它成功了。
2) 添加指向代码的指针，这样您就不会通过 p 访问 distance，而是直接访问它，如下所示：
double *distance = p->distance;
#pragma acc data copy(p[0:1],distance[0:NX*NY*NZ])
for(int level = startLevel; level <= endLevel; level++){
int ks = max(1, level-(y + z));
int ke = min(x, level-2);
int js = max(1, level-(x + z));
int je = min(y, level-2);
#pragma acc parallel
{
#pragma acc loop independent
for(int k = ks; k <= ke; k++){
#pragma acc loop independent
for(int j = js; j <= je; j++){
int i = level - (k + j);
if(i > 0 && i <= z){
int index = i * NX * NY + j * NX + k;
distance[index] = solve(p, index);
}
}
}
}
我知道当有很多数据数组要执行此操作时，这可能会很痛苦，但这是我在很多代码中成功使用的解决方法。不幸的是，这是必要的，这就是为什么我们希望在未来版本的 OpenACC 中提供更好的解决方案。
希望对您有所帮助！如果我能想出一个不需要额外指针的解决方案，我会更新这个答案。
Jeff 是正确的，OpenACC 委员会仍在研究如何标准化对具有动态数据成员的聚合数据类型的支持。但是，对于 PGI 14.9 或更高版本，我们添加了对结构和 C++ 类 的更好支持，因此在这种情况下，您只需添加 create(p[0:1]) 即可简化代码。将会发生的是，编译器将创建一个 p 的设备副本，其中只为数据成员分配内存。那么当你做p->distance的copy时，内存会分配给"distance"然后attach到p。 （即 运行 时间将填充结构中的设备指针）。
有一些注意事项。首先是此行为尚未标准化，因此其他编译器（如 Cray、Pathscale、GCC 等）可能具有不同的行为。第二，顺序很重要。 p 需要在附加 distance 之前创建。第三，更复杂的数据结构变得很难管理。正如 Jeff 所建议的，使用 CUDA 统一内存是管理复杂数据结构的一个很好的选择。
如果您有兴趣，我的 GTC2015 演示文稿的大部分内容都在讨论这个主题 (link)。演讲的重点是 C++ Class 数据管理，但也适用于 C 结构。
希望这对您有所帮助， 垫子
% cat test1.c
#include <stdio.h>
#include <stdlib.h>
#define min(a,b) (a > b) ? b : a
#define max(a,b) (a < b) ? b : a
#define NX 10
#define NY 10
#define NZ 10
struct phiType {
double dx, dy, dz;
double * distance;
};
typedef struct phiType Phi;
#pragma acc routine seq
double solve(Phi *p, int index) {
// for simplicity just returning a value
return 2;
}
void fast_sweep(Phi *p) {
// removing boundaries
int x = NX - 2;
int y = NY - 2;
int z = NZ - 2;
int startLevel = 3;
int endLevel = x + y + z;
#pragma acc data create(p[0:1]) copy(p->distance[0:NX*NY*NZ])
for(int level = startLevel; level <= endLevel; level++){
int ks = max(1, level-(y + z));
int ke = min(x, level-2);
int js = max(1, level-(x + z));
int je = min(y, level-2);
#pragma acc region
{
#pragma acc loop independent
for(int k = ks; k <= ke; k++){
#pragma acc loop independent
for(int j = js; j <= je; j++){
int i = level - (k + j);
if(i > 0 && i <= z){
int index = i * NX * NY + j * NX + k;
p->distance[index] = solve(p, index);
}
}
}
}
}
}
void create_phi(Phi *p){
p->dx = 1;
p->dy = 1;
p->dz = 1;
p->distance = (double *) malloc(sizeof(double) * NX * NY * NZ);
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
p->distance[index] = (i*j*k == 0) ? 0 : 1;
}
}
}
}
int main()
{
printf("start \n");
Phi *p = (Phi *) malloc(sizeof(Phi));
create_phi(p);
printf("calling fast sweep \n");
fast_sweep(p);
printf(" print the results \n");
for(int i = 0; i < NZ; i++){
for(int j = 0; j < NY; j++){
for(int k = 0; k < NX; k++){
int index = i * NX * NY + j * NX + k;
printf("%f ", p->distance[index]);
}
printf("\n");
}
printf("\n");
}
return 0;
}
% pgcc -acc -ta=tesla:cc35 -Minfo=accel test1.c -V15.7 ; a.out
solve:
19, Generating acc routine seq
fast_sweep:
34, Generating create(p[:1])
Generating copy(p->distance[:1000])
45, Loop is parallelizable
47, Loop is parallelizable
Accelerator kernel generated
Generating Tesla code
45, #pragma acc loop gang /* blockIdx.y */
47, #pragma acc loop gang, vector(128) /* blockIdx.x threadIdx.x */
start
calling fast sweep
print the results
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
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