MPI_Scatter和MPI_Gather与std :: vector

时间:2018-10-12 18:03:26

标签: c++ mpi

我是MPI编程的新手。我之所以使用C ++,是因为我想使用向量而不是数组(因此用户可以动态选择游戏网格的大小(Conway的《人生游戏》),而不是硬编码的数组大小。

假设:共有4个过程(p = 4

这是我的代码:

MPI_Init(&argc, &argv);
MPI_Comm_size(MPI_COMM_WORLD, &p); 
MPI_Comm_rank(MPI_COMM_WORLD, &rank); 

vector<CELL> *gameBoard = NULL;

if (rank == 0)
{
    //create the gameboard on just the root process
    gameBoard = new vector<CELL>();
    gameBoard->resize(16);
}

//create a partialGrid on EACH process
vector<CELL> *partialGrid = new vector<CELL>();
partialGrid->resize(4);

int partialSize = 4;

MPI_Scatter(gameBoard, partialSize, mpi_cell, partialGrid, partialSize, mpi_cell, 0, MPI_COMM_WORLD);

//do something on each process
cout << "HI!!! I'm scattered process number " << rank << endl;
//ideally, do something to the sub vector here (once it works)

vector<CELL> *rbuf = NULL; //Just make null for all other processes. Valid receive buffer only needed for root (rank = 0)

if (rank == 0)
{
    //make sure the receiving buff is created (for just the root rank 0)
    vector<CELL> *rbuf = new vector<CELL>();
    rbuf->resize(16);
    cout << "RANK 0 TEST!" << endl;
}

MPI_Gather(&partialGrid, (pow(n, 2) / p), mpi_cell, &rbuf, (pow(n, 2) / p), mpi_cell, 0, MPI_COMM_WORLD);

目标是(目前为硬编码)将16个元素的向量拆分为4个子向量,每个子元素用于4个过程中的每个要执行的操作。

我认为我的等级0的逻辑是正确的。我仅在排名过程中为MPI_Gather创建一个接收缓冲区,并且在MPI_Scatter之前为EACH过程创建子矢量。我尝试了很多事情,MPI_Scatter总是失败。请帮忙!

注意:CELL是九个struct值中的intmpi_cell是我在MPI程序中创建的自定义类型。为简单起见,已排除了代码,但如果要测试,可以将CELL替换为int,将mpi_cell替换为MPI_INT

我收到细分错误:

enter image description here

1 个答案:

答案 0 :(得分:1)

如评论中助理所述,您的new错误。使用std::vector的正确方法如下:

std::vector<CELL> gameBoard;
if (rank == 0)
    gameBoard.resize(16);

std::vector<CELL> partialGrid;
partialGrid.resize(4);

MPI_Scatter(gameBoard.data(), 4, mpi_cell, partialGrid.data(), 4, mpi_cell, 0, MPI_COMM_WORLD);

// ...

std::vector<CELL> rbuf;
if (rank == 0)
    rbuf.resize(16);

MPI_Gather(partialGrid.data(), 4, mpi_cell, rbuf.data(), 4, mpi_cell, 0, MPI_COMM_WORLD);

为简单起见,我也对所有常量进行了硬编码。