我已遵循 this question 提供的指导和 this link它处理将指针数组传递给设备并返回的概念,但当指针指向一个对象时,我似乎在为我的特定情况而苦苦挣扎。请参阅下面的示例代码,为简洁起见,我删除了错误检查。
// Kernel
__global__ void myKernel(Obj** d_array_of_objs)
{
// Change the scalar of each object to 5
// by dereferencing device array to get
// appropriate object pointer.
*d_array_of_objs->changeToFive(); <--------- SEE QUESTION 4
}
// Entry point
int main()
{
/********************************/
/* INITIALISE OBJ ARRAY ON HOST */
/********************************/
// Array of 3 pointers to Objs
Obj* h_obj[3];
for (int i = 0; i < 3; i++) {
h_obj[i] = new Obj(); // Create
h_obj[i]->scalar = i * 10; // Initialise
}
// Write out
for (int i = 0; i < 3; i++) {
std::cout << h_obj[i]->scalar << std::endl;
}
/**************************************************/
/* CREATE DEVICE VERSIONS AND STORE IN HOST ARRAY */
/**************************************************/
// Create host pointer to array-like storage of device pointers
Obj** h_d_obj = (Obj**)malloc(sizeof(Obj*) * 3); <--------- SEE QUESTION 1
for (int i = 0; i < 3; i++) {
// Allocate space for an Obj and assign
cudaMalloc((void**)&h_d_obj[i], sizeof(Obj));
// Copy the object to the device (only has single scalar field to keep it simple)
cudaMemcpy(h_d_obj[i], &(h_obj[i]), sizeof(Obj), cudaMemcpyHostToDevice);
}
/**************************************************/
/* CREATE DEVICE ARRAY TO PASS POINTERS TO KERNEL */
/**************************************************/
// Create a pointer which will point to device memory
Obj** d_d_obj = nullptr;
// Allocate space for 3 pointers on device at above location
cudaMalloc((void**)&d_d_obj, sizeof(Obj*) * 3);
// Copy the pointers from the host memory to the device array
cudaMemcpy(d_d_obj, h_d_obj, sizeof(Obj*) * 3, cudaMemcpyHostToDevice);
/**********
* After the above, VS2013 shows the memory pointed to by d_d_obj
* to be NULL <------- SEE QUESTION 2.
**********/
// Launch Kernel
myKernel <<<1, 3>>>(d_d_obj);
// Synchronise and pass back to host
cudaDeviceSynchronize();
for (int i = 0; i < 3; i++) {
cudaMemcpy(&(h_obj[i]), h_d_obj[i], sizeof(Obj), cudaMemcpyDeviceToHost); <--------- SEE QUESTION 3
}
// Write out
for (int i = 0; i < 3; i++) {
std::cout << h_obj[i]->scalar << std::endl;
}
return 0;
}
所以问题是:
如果上面
SEE QUESTION 1
指示的行为指针分配主机内存,一旦我在后续循环中使用了cudaMalloc
来分配设备内存,h_d_obj 指向的指针被设备地址覆盖,这是否意味着我已经为 3 个Obj*
分配了主机内存,现在没有指针指向它?为什么当我测试返回的状态时
cudaMemcpy
成功但显然没有正确复制地址?我期望h_d_obj
和d_d_obj
的内存地址“数组”相同,因为它们应该指向相同的Obj
设备地址空间。在
SEE QUESTION 3
行,假设我在问题 2 中是正确的。我还希望能够使用h_d_obj
或d_d_obj
从设备中检索Obj
对象,因为区别仅在于我是取消引用主机指针以访问指向Obj
的设备指针还是设备指针这两个我都可以在cudaMemcpy
方法中完成,对吗?如果我使用写入的内容,复制成功,但h_obj[0]
处的指针已损坏,我无法写出数据。在
SEE QUESTION 4
行,为什么我不能取消引用Obj**
来获取Obj*
然后使用->
运算符调用 device 方法?编译器提示说它不是指向类类型的指针,事实上它是一个Obj*
告诉我它是。
最佳答案
首先,如果您提供完整的代码(包括Obj
类的定义)会很方便。我根据对您的代码的检查和一些猜测提供了一个。
其次,您在这里的大部分困惑似乎是 C(或 C++)中带有指针的不太清晰的工具。在主机和设备之间使用具有双指针结构 (**
) 的 CUDA API 需要清晰的理解和可视化正在发生的事情的能力。
If the line indicated by SEE QUESTION 1 above allocates host memory for the pointers, and once I have used
cudaMalloc
in the subsequent loop to allocate device memory, the pointer pointed to byh_d_obj
get overwritten with device addresses, does that mean I have allocated host memory for 3 Obj* that now has no pointer pointing to it?
没有。 h_d_obj
由 malloc
操作建立(即赋予有意义的值)。您之后所做的任何事情都不会修改 h_d_obj
的值。
Why is the cudaMemcpy succeeding when I test the status returned but clearly does not copy the addresses correctly? I was expecting the "arrays" of memory address of both
h_d_obj
andd_d_obj
to be the same since they should point to the same Obj in the device address space.
到目前为止,我没有发现您的代码有任何问题。 h_d_obj
的值是(之前)由malloc
建立的,它的数值是主机内存中的一个地址。 d_d_obj
的值是由cudaMalloc
建立的,它的数值是设备内存中的一个地址。在数字上,我希望它们有所不同。
At the line SEE QUESTION 3, assuming I'm correct in question 2. I also expect to be able to use either
h_d_obj
ord_d_obj
to retrieve the Obj objects from the device since the difference would be only whether I dereference a host pointer to access a device pointer to Obj or a device pointer both of which I can do in a cudaMemcpy method right? If I use what is written, the copy succeeds but the pointer at h_obj[0] is corrupted and I cannot write out the data.
没有。如果它是 cudaMemcpy
中的参数,则您不能取消引用主机代码中的设备指针,即使。这作为 cudaMemcpy
操作中的源或目标是合法的:
h_d_obj[i]
这是不合法的:
d_d_obj[i]
原因是为了获得实际的目标地址,我必须在第一种情况下取消引用主机指针(即访问主机上的内存位置),但在第二种情况下必须取消引用设备指针。从主机代码中,我可以检索 h_d_obj[i]
的内容。我不允许尝试在主机代码中检索 d_d_obj[i]
的内容(cudaMemcpy
的参数操作是 host 代码) . d_d_obj
的值可以用作主机代码的目标。 d_d_obj[i]
不能。
At the line SEE QUESTION 4, why can I not dereference an Obj** to get a Obj* then use the -> operator to call a device method? The compiler moans that it is not a pointer to class type which the fact that it is a Obj* tells me it is.
编译器对你咆哮是因为你不明白你正在使用的各种运算符(*
、->
)之间的操作顺序。如果您添加括号来标识正确的顺序:
(*d_array_of_objs)->changeToFive();
然后编译器不会反对(尽管我会像下面那样稍微不同地做)。
这是您的代码的修改版本,其中添加了 Obj
定义,对内核进行了轻微更改,以便独立线程在独立对象上工作,并进行了一些其他修复。您的代码大部分是正确的:
$ cat t1231.cu
#include <iostream>
class Obj{
public:
int scalar;
__host__ __device__
void changeToFive() {scalar = 5;}
};
// Kernel
__global__ void myKernel(Obj** d_array_of_objs)
{
// Change the scalar of each object to 5
// by dereferencing device array to get
// appropriate object pointer.
int idx = threadIdx.x+blockDim.x*blockIdx.x;
// (*d_array_of_objs)->changeToFive(); // <--------- SEE QUESTION 4 (add parenthesis)
d_array_of_objs[idx]->changeToFive();
}
// Entry point
int main()
{
/********************************/
/* INITIALISE OBJ ARRAY ON HOST */
/********************************/
// Array of 3 pointers to Objs
Obj* h_obj[3];
for (int i = 0; i < 3; i++) {
h_obj[i] = new Obj(); // Create
h_obj[i]->scalar = i * 10; // Initialise
}
// Write out
for (int i = 0; i < 3; i++) {
std::cout << h_obj[i]->scalar << std::endl;
}
/**************************************************/
/* CREATE DEVICE VERSIONS AND STORE IN HOST ARRAY */
/**************************************************/
// Create host pointer to array-like storage of device pointers
Obj** h_d_obj = (Obj**)malloc(sizeof(Obj*) * 3); // <--------- SEE QUESTION 1
for (int i = 0; i < 3; i++) {
// Allocate space for an Obj and assign
cudaMalloc((void**)&h_d_obj[i], sizeof(Obj));
// Copy the object to the device (only has single scalar field to keep it simple)
cudaMemcpy(h_d_obj[i], &(h_obj[i]), sizeof(Obj), cudaMemcpyHostToDevice);
}
/**************************************************/
/* CREATE DEVICE ARRAY TO PASS POINTERS TO KERNEL */
/**************************************************/
// Create a pointer which will point to device memory
Obj** d_d_obj = NULL;
// Allocate space for 3 pointers on device at above location
cudaMalloc((void**)&d_d_obj, sizeof(Obj*) * 3);
// Copy the pointers from the host memory to the device array
cudaMemcpy(d_d_obj, h_d_obj, sizeof(Obj*) * 3, cudaMemcpyHostToDevice);
/**********
* After the above, VS2013 shows the memory pointed to by d_d_obj
* to be NULL <------- SEE QUESTION 2.
**********/
// Launch Kernel
myKernel <<<1, 3>>>(d_d_obj);
// Synchronise and pass back to host
cudaDeviceSynchronize();
for (int i = 0; i < 3; i++) {
cudaMemcpy(h_obj[i], h_d_obj[i], sizeof(Obj), cudaMemcpyDeviceToHost); // <--------- SEE QUESTION 3 remove parenthesis
}
// Write out
for (int i = 0; i < 3; i++) {
std::cout << h_obj[i]->scalar << std::endl;
}
return 0;
}
$ nvcc -o t1231 t1231.cu
$ cuda-memcheck ./t1231
========= CUDA-MEMCHECK
0
10
20
5
5
5
========= ERROR SUMMARY: 0 errors
$
h_d_obj
和 d_d_obj
的图表可能会有所帮助:
HOST | DEVICE
h_d_obj-->(Obj *)-------------------------->Obj0<---(Obj *)<----|
(Obj *)-------------------------->Obj1<---(Obj *) |
(Obj *)-------------------------->Obj2<---(Obj *) |
| |
d_d_obj---------------------------------------------------------|
HOST | DEVICE
您可以在主机代码或 cudaMemcpy
操作中访问上图左侧 (HOST) 的任何数量(位置)。您不能访问主机代码右侧的任何数量(位置)。
关于c++ - 指向 CUDA 中对象指针数组的指针,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38978297/