如何使用 MPI 传递带有动态数组的自定义结构?
How to use MPI to transfer a custom struct with a dynamic array?
有一个简单的例子可以描述我的问题。
我有一个包含动态数组的自定义结构
struct my_data_type {
int c;
int d[];
};
并且根进程(进程 0)有一个这样的结构数组 nums[4]
。
我想通过 MPI_Scatter
将数组的块发送到不同的进程(例如,2 个进程)。这里的主要问题是我希望这个数组 d[]
是动态的。
主要代码如下:
int main(int argc, char* argv[]) {
MPI_Init(NULL, NULL);
int my_size; MPI_Comm_size(MPI_COMM_WORLD, &my_size);
int my_rank; MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
int len = 2; //example: the dynamic array d contains len=2 elements
my_data_type *nums //nums[4]
= (my_data_type*)malloc((sizeof(my_data_type) + sizeof(int) * len) * 4);
my_data_type *sub_nums //sub_nums[2]
= (my_data_type*)malloc((sizeof(my_data_type) + sizeof(int) * len) * 2);
if (my_rank == 0) { //just some examples
nums[0].c = 0; nums[1].c = 1; nums[2].c = 2; nums[3].c = 3;
nums[0].d[0] = 10; nums[1].d[0] = 11; nums[2].d[0] = 12; nums[3].d[0] = 13;
nums[0].d[1] = 14; nums[1].d[1] = 15; nums[2].d[1] = 16; nums[3].d[1] = 17;
}
MPI_Datatype mpi_data_type; //new datatype
int blocklens[2];
MPI_Datatype old_types[2];
MPI_Aint indices[2];
blocklens[0] = 1; blocklens[1] = len;
old_types[0] = MPI_INT; old_types[1] = MPI_INT;
MPI_Address(&nums[0].c, &indices[0]);
MPI_Address(&nums[0].d[0], &indices[1]);
indices[1] = indices[1] - indices[0];
indices[0] = 0;
MPI_Type_create_struct(2, blocklens, indices, old_types, &mpi_data_type);
MPI_Type_commit(&mpi_data_type);
MPI_Scatter(nums, 2, mpi_data_type,
sub_nums, 2, mpi_data_type,
0, MPI_COMM_WORLD);
cout << "rank " << my_rank << ": " << endl;
cout << "c: " << sub_nums[0].c << ", " << sub_nums[1].c << endl;
cout << "d: " << sub_nums[0].d[0] << ", " << sub_nums[0].d[1] << ", ";
cout << sub_nums[1].d[0] << ", " << sub_nums[1].d[1] << endl;
MPI_Finalize();
return 0;
}
如果我在struct my_data_type
的定义中将int d[];
改成int d[2];
,我肯定会得到像
这样的预期结果
rank 0:
c: 0, 1
d: 10, 14, 11, 15
rank 1:
c: 2, 3
d: 12, 16, 13, 17
但如果没有,结果如下:
rank 0:
c: 0, 10
d: 10, 14, 14, 15
rank 1:
c: 33, 0
d: 0, 0, 0, 1
如您所见,我知道问题出在动态数组上,但我不能在我的项目中使用静态数组。那么我怎样才能更改上面的代码以获得预期的结果呢?
我在这方面可能是错的,但我认为你想要的是保存一个指向数组的指针(即 int** myArrPointer
),因为我认为你需要做的,因为我不认为可以在C中分配一个数组(即myArr = myOtherArr
),就是:
- 计算新数组的大小
- 为新数组分配内存
- 在您的结构中存储指向该新数组的指针
您的结构可能最终需要看起来像这样:
struct my_data_type
{
int ArrSize;
int** PointerToAnArray;
};
void SomeFunForSwappingArrays(my_data_type* instance, int newArrSize)
{
int* newArr = (int*)malloc(newArrSize*sizeof(int));
//free the memory of the old array. if you don't need the data anymore, i would
//consider doing this.
free(*(instance->PointerToAnArray));
//save the memory address of the new array
instance->PointerToAnArray = &newArr;
instance->ArrSize = newArrSize;
}
希望对您有所帮助。
您的根本问题不是 mpi,而是使用具有灵活数组成员的结构数组。这是一个示例程序来说明问题
#include <assert.h>
#include <stdlib.h>
#include <stdint.h>
typedef struct s s;
struct s
{
int c;
int d[];
};
int main(int argc, char* argv[])
{
assert(sizeof(s) == sizeof(int));
int len = 4;
s* okay = malloc(sizeof(*okay) + sizeof(int)*len);
intptr_t true_size = (intptr_t)&okay->d[len] -(intptr_t)(okay);
assert(true_size == ((len+1)*sizeof(int)));
int nbad = 6;
s* bad = malloc((sizeof(*bad) + sizeof(int)*len)*nbad);
intptr_t bad_size = (intptr_t)&bad[1] -(intptr_t)&bad[0];
/* this size mismatch means arrays of `s` do not do what you think they do */
assert(bad_size != true_size);
assert(bad_size == sizeof(int));
assert((char*)&bad[1] == (char*)&bad[0].d[0]);
assert((char*)&bad[2] == (char*)&bad[0].d[1]);
assert((char*)&bad[3] == (char*)&bad[0].d[2]);
assert((char*)&bad[1].d[0] == (char*)&bad[0].d[1]);
assert((char*)&bad[2].d[0] == (char*)&bad[0].d[2]);
assert((char*)&bad[3].d[0] == (char*)&bad[0].d[3]);
}
要处理具有灵活数组成员的结构数组,您需要手动计算索引的内存偏移量,而不是依赖编译器。所以你可以像这样定义一个辅助函数:
s* s_index(const s* a, int len, int index)
{
uintptr_t true_size = sizeof(*a) + len*sizeof(int);
return (s*)((char*)a + index * true_size);
}
然后使用 s_index
访问所需的数组成员,而不是 bad[0]
、bad[1]
构造:
s* first = s_index(bad, len, 0);
s* second = s_index(bad, len, 1);
assert((char*)&first->d[len] == (char *)second);
有一个简单的例子可以描述我的问题。 我有一个包含动态数组的自定义结构
struct my_data_type {
int c;
int d[];
};
并且根进程(进程 0)有一个这样的结构数组 nums[4]
。
我想通过 MPI_Scatter
将数组的块发送到不同的进程(例如,2 个进程)。这里的主要问题是我希望这个数组 d[]
是动态的。
主要代码如下:
int main(int argc, char* argv[]) {
MPI_Init(NULL, NULL);
int my_size; MPI_Comm_size(MPI_COMM_WORLD, &my_size);
int my_rank; MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
int len = 2; //example: the dynamic array d contains len=2 elements
my_data_type *nums //nums[4]
= (my_data_type*)malloc((sizeof(my_data_type) + sizeof(int) * len) * 4);
my_data_type *sub_nums //sub_nums[2]
= (my_data_type*)malloc((sizeof(my_data_type) + sizeof(int) * len) * 2);
if (my_rank == 0) { //just some examples
nums[0].c = 0; nums[1].c = 1; nums[2].c = 2; nums[3].c = 3;
nums[0].d[0] = 10; nums[1].d[0] = 11; nums[2].d[0] = 12; nums[3].d[0] = 13;
nums[0].d[1] = 14; nums[1].d[1] = 15; nums[2].d[1] = 16; nums[3].d[1] = 17;
}
MPI_Datatype mpi_data_type; //new datatype
int blocklens[2];
MPI_Datatype old_types[2];
MPI_Aint indices[2];
blocklens[0] = 1; blocklens[1] = len;
old_types[0] = MPI_INT; old_types[1] = MPI_INT;
MPI_Address(&nums[0].c, &indices[0]);
MPI_Address(&nums[0].d[0], &indices[1]);
indices[1] = indices[1] - indices[0];
indices[0] = 0;
MPI_Type_create_struct(2, blocklens, indices, old_types, &mpi_data_type);
MPI_Type_commit(&mpi_data_type);
MPI_Scatter(nums, 2, mpi_data_type,
sub_nums, 2, mpi_data_type,
0, MPI_COMM_WORLD);
cout << "rank " << my_rank << ": " << endl;
cout << "c: " << sub_nums[0].c << ", " << sub_nums[1].c << endl;
cout << "d: " << sub_nums[0].d[0] << ", " << sub_nums[0].d[1] << ", ";
cout << sub_nums[1].d[0] << ", " << sub_nums[1].d[1] << endl;
MPI_Finalize();
return 0;
}
如果我在struct my_data_type
的定义中将int d[];
改成int d[2];
,我肯定会得到像
rank 0:
c: 0, 1
d: 10, 14, 11, 15
rank 1:
c: 2, 3
d: 12, 16, 13, 17
但如果没有,结果如下:
rank 0:
c: 0, 10
d: 10, 14, 14, 15
rank 1:
c: 33, 0
d: 0, 0, 0, 1
如您所见,我知道问题出在动态数组上,但我不能在我的项目中使用静态数组。那么我怎样才能更改上面的代码以获得预期的结果呢?
我在这方面可能是错的,但我认为你想要的是保存一个指向数组的指针(即 int** myArrPointer
),因为我认为你需要做的,因为我不认为可以在C中分配一个数组(即myArr = myOtherArr
),就是:
- 计算新数组的大小
- 为新数组分配内存
- 在您的结构中存储指向该新数组的指针
您的结构可能最终需要看起来像这样:
struct my_data_type
{
int ArrSize;
int** PointerToAnArray;
};
void SomeFunForSwappingArrays(my_data_type* instance, int newArrSize)
{
int* newArr = (int*)malloc(newArrSize*sizeof(int));
//free the memory of the old array. if you don't need the data anymore, i would
//consider doing this.
free(*(instance->PointerToAnArray));
//save the memory address of the new array
instance->PointerToAnArray = &newArr;
instance->ArrSize = newArrSize;
}
希望对您有所帮助。
您的根本问题不是 mpi,而是使用具有灵活数组成员的结构数组。这是一个示例程序来说明问题
#include <assert.h>
#include <stdlib.h>
#include <stdint.h>
typedef struct s s;
struct s
{
int c;
int d[];
};
int main(int argc, char* argv[])
{
assert(sizeof(s) == sizeof(int));
int len = 4;
s* okay = malloc(sizeof(*okay) + sizeof(int)*len);
intptr_t true_size = (intptr_t)&okay->d[len] -(intptr_t)(okay);
assert(true_size == ((len+1)*sizeof(int)));
int nbad = 6;
s* bad = malloc((sizeof(*bad) + sizeof(int)*len)*nbad);
intptr_t bad_size = (intptr_t)&bad[1] -(intptr_t)&bad[0];
/* this size mismatch means arrays of `s` do not do what you think they do */
assert(bad_size != true_size);
assert(bad_size == sizeof(int));
assert((char*)&bad[1] == (char*)&bad[0].d[0]);
assert((char*)&bad[2] == (char*)&bad[0].d[1]);
assert((char*)&bad[3] == (char*)&bad[0].d[2]);
assert((char*)&bad[1].d[0] == (char*)&bad[0].d[1]);
assert((char*)&bad[2].d[0] == (char*)&bad[0].d[2]);
assert((char*)&bad[3].d[0] == (char*)&bad[0].d[3]);
}
要处理具有灵活数组成员的结构数组,您需要手动计算索引的内存偏移量,而不是依赖编译器。所以你可以像这样定义一个辅助函数:
s* s_index(const s* a, int len, int index)
{
uintptr_t true_size = sizeof(*a) + len*sizeof(int);
return (s*)((char*)a + index * true_size);
}
然后使用 s_index
访问所需的数组成员,而不是 bad[0]
、bad[1]
构造:
s* first = s_index(bad, len, 0);
s* second = s_index(bad, len, 1);
assert((char*)&first->d[len] == (char *)second);