MPI 收集字符串数组

Question

我正在尝试将字典集合合并到根进程中。这是一个简短的例子：

#define MAX_CF_LENGTH 55

    map<string, int> dict;

    if (rank == 0)
    {
        dict = {
            {"Accelerator Defective", 33},
            {"Aggressive Driving/Road Rage", 27},
            {"Alcohol Involvement", 19},
            {"Animals Action", 30}};
    }
    if (rank == 1)
    {
        dict = {
            {"Driver Inexperience", 6},
            {"Driverless/Runaway Vehicle", 46},
            {"Drugs (Illegal)", 38},
            {"Failure to Keep Right", 24}};
    }
    if (rank == 2)
    {
        dict = {
            {"Lost Consciousness", 1},
            {"Obstruction/Debris", 8},
            {"Other Electronic Device", 25},
            {"Other Lighting Defects", 43},
            {"Other Vehicular", 7}};
    }

    Scatterer scatterer(rank, MPI_COMM_WORLD, num_workers);
    scatterer.gatherDictionary(dict, MAX_CF_LENGTH);

gatherDictionary() 内部的想法是将每个键放在每个进程的char 数组中（允许重复）。之后，将所有键收集到根中并在广播之前创建最终（合并）字典。代码如下：

void Scatterer::gatherDictionary(map<string,int> &dict, int maxKeyLength)
{
    // Calculate destination dictionary size
    int numKeys = dict.size();
    int totalLength = numKeys * maxKeyLength;
    int finalNumKeys = 0;
    MPI_Reduce(&numKeys, &finalNumKeys, 1, MPI_INT, MPI_SUM, 0, comm);

    // Computing number of elements that are received from each process
    int *recvcounts = NULL;
    if (rank == 0)
        recvcounts = new int[num_workers];

    MPI_Gather(&totalLength, 1, MPI_INT, recvcounts, 1, MPI_INT, 0, comm);

    // Computing displacement relative to recvbuf at which to place the incoming data from each process
    int *displs = NULL;
    if (rank == 0)
    {
        displs = new int[num_workers];

        displs[0] = 0;
        for (int i = 1; i < num_workers; i++)
            displs[i] = displs[i - 1] + recvcounts[i - 1] + 1;
    }

    char(*dictKeys)[maxKeyLength];
    char(*finalDictKeys)[maxKeyLength];
    dictKeys = (char(*)[maxKeyLength])malloc(numKeys * sizeof(*dictKeys));
    if (rank == 0)
        finalDictKeys = (char(*)[maxKeyLength])malloc(finalNumKeys * sizeof(*finalDictKeys));

    // Collect keys for each process
    int i = 0;
    for (auto pair : dict)
    {
        strncpy(dictKeys[i], pair.first.c_str(), maxKeyLength);
        i++;
    }

    MPI_Gatherv(dictKeys, totalLength, MPI_CHAR, finalDictKeys, recvcounts, displs, MPI_CHAR, 0, comm);

    // Create new dictionary and distribute it to all processes
    dict.clear();
    if (rank == 0)
    {
        for (int i = 0; i < finalNumKeys; i++)
            dict[finalDictKeys[i]] = dict.size();
    }

    delete[] dictKeys;
    if (rank == 0)
    {
        delete[] finalDictKeys;
        delete[] recvcounts;
        delete[] displs;
    }

    broadcastDictionary(dict, maxKeyLength);
}

我确信broadcastDicitonary() 的正确性，因为我已经对其进行了测试。调试收集功能，我得到以下部分结果：

Recvcounts:
220
220
275

Displacements:
0
221
442

FinalDictKeys:
Rank:0 Accelerator Defective
Rank:0 Aggressive Driving/Road Rage
Rank:0 Alcohol Involvement
Rank:0 Animals Action
Rank:0 
Rank:0 
Rank:0 
Rank:0 
Rank:0 
Rank:0 
Rank:0 
Rank:0 
Rank:0 

由于只收集根数据，我想知道这是否与字符分配有关，即使它应该是连续的。我不认为这与最后缺少空字符有关，因为每个字符串/键已经有很多填充。 提前感谢您指出任何缺失或改进，如果您需要任何额外信息，请发表评论。

如果您想自己测试它，我已将所有代码放在一个文件中，它已准备好编译和运行（当然这适用于 3 个 mpi 进程）。 Code Here

Answer 1

displs[i] = displs[i - 1] + recvcounts[i - 1] + 1;

最后的+ 1 是多余的。将其更改为：

displs[i] = displs[i - 1] + recvcounts[i - 1];