MPI_Reduce 和 MPI_MIN 如何工作? [英] How does MPI_Reduce with MPI_MIN work?

问题描述

如果我有这个代码:

int main(void) {
    int result=0;
    int num[6] = {1, 2, 4, 3, 7, 1};
    if (my_rank != 0) {
        MPI_Reduce(num, &result, 6, MPI_INT, MPI_MIN, 0, MPI_COMM_WORLD);
    } else {
        MPI_Reduce(num, &result, 6, MPI_INT, MPI_MIN, 0, MPI_COMM_WORLD)
        printf("result = %d\n", result);
    }
}

结果打印为 1 ;

但是如果 num[0]=9;那么结果是9

But if the num[0]=9; then the result is 9

为了解决这个问题，我必须将变量 num 定义为数组.我无法理解函数 MPI_Reduce 如何与 MPI_MIN 一起工作.为什么，如果num[0] 不等于最小数，那么我必须将变量num 定义为数组?

I read to solve this problem I must to define the variable num as array. I can't understand how the function MPI_Reduce works with MPI_MIN. Why, if the num[0] is not equal to the smallest number, then I must to define the variable num as array?

推荐答案

MPI_Reduce 对通信器的成员执行归约 - 而不是本地数组的成员.sendbuf 和 recvbuf 必须具有相同的 size.

MPI_Reduce performs a reduction over the members of the communicator - not the members of the local array. sendbuf and recvbuf must both be of the same size.

我认为标准说得最好:

因此，所有进程都提供相同长度的输入缓冲区和输出缓冲区，具有相同类型的元素.每个进程可以提供一个元素或元素序列，在这种情况下，组合操作在序列的每个条目上按元素执行.

Thus, all processes provide input buffers and output buffers of the same length, with elements of the same type. Each process can provide one element, or a sequence of elements, in which case the combine operation is executed element-wise on each entry of the sequence.

MPI 不会获取数组中所有元素的最小值，您必须手动获取.

MPI does not get the minimum of all elements in the array, you have to do that manually.

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