Question

我正在尝试使用mpicc编译MPI程序。编译器仅抱怨没有对MPI_RECIVE和MPI_SEND的引用，并结束了编译错误。我在.c文件中有#include。

有人可以告诉我如何解决此问题吗？

此处提供了代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <assert.h>
#include <mpi.h>

#include "random.h"
#include "md5tool.h"

/* horizontal size of the configuration */
#define XSIZE 1024

/* "ADT" State and line of states (plus border) */
typedef char State;
typedef State Line[XSIZE + 2];

/* determine random integer between 0 and n-1 */
#define randInt(n) ((int)(nextRandomLEcuyer() * n))

/* random starting configuration */
static void initConfig(Line *buf, int lines){
    int x, y;

    initRandomLEcuyer(424243);
    for (y = 1;  y <= lines;  y++) {
        for (x = 1;  x <= XSIZE;  x++) {
            buf[y][x] = randInt(100) >= 50;
        }
    }
}

/* annealing rule from ChoDro96 page 34 
 * the table is used to map the number of nonzero
 * states in the neighborhood to the new state
 */
static State anneal[10] = {0, 0, 0, 0, 1, 0, 1, 1, 1, 1};

/* a: pointer to array; x,y: coordinates; result: n-th element of anneal,
      where n is the number of neighbors */
#define transition(a, x, y) \
   (anneal[(a)[(y)-1][(x)-1] + (a)[(y)][(x)-1] + (a)[(y)+1][(x)-1] +\
           (a)[(y)-1][(x)  ] + (a)[(y)][(x)  ] + (a)[(y)+1][(x)  ] +\
           (a)[(y)-1][(x)+1] + (a)[(y)][(x)+1] + (a)[(y)+1][(x)+1]])

/* treat torus like boundary conditions */
static void boundary(Line *buf, int lines){
    int x,y;
    for (y = 0;  y <= lines+1;  y++) {
        /* copy rightmost column to the buffer column 0 */
        buf[y][0      ] = buf[y][XSIZE];

        /* copy leftmost column to the buffer column XSIZE + 1 */
        buf[y][1+1] = buf[y][1    ];
    }

    for (x = 0;  x <= XSIZE+1;  x++) {
        /* copy bottommost row to buffer row 0 */
        buf[0][x      ] = buf[lines][x];

        /* copy topmost row to buffer row lines + 1 */
        buf[lines+1][x] = buf[1][x    ];
    }
}
/* make one simulation iteration with lines lines.
 * old configuration is in from, new one is written to to.
 */

 //umschreiben
 /**
static void simulate(Line *from, Line *to, int lines){


    boundary(from, lines); 
    for (y = 1;  y <= lines;  y++) {
        for (x = 1;  x <= XSIZE;  x++) {
            to[y][x  ] = transition(from, x  , y);
        }
    }
}
*/

/* --------------------- measurement ---------------------------------- */

int main(int argc, char** argv){
    int lines, its;
    int i;
    Line *from, *to, *temp, *next;
    char* hash;

    assert(argc == 3);

    lines = atoi(argv[1]);
    its   = atoi(argv[2]);
    from = malloc((lines + 2) * sizeof(Line));
    to   = malloc((lines + 2) * sizeof(Line));

    MPI_Init(NULL, NULL);
    // Get the number of processes
    int world_size;
    MPI_Comm_size(MPI_COMM_WORLD, &world_size);

    // Get the rank of the process
    int world_rank;
    MPI_Comm_rank(MPI_COMM_WORLD, &world_rank);

    if(world_rank == 0){ 
        int y;
        next = malloc((lines + 2) * sizeof(Line));
        initConfig(from, lines);
        for (i = 0;  i < its;  i++) {
            boundary(from, lines);
            int z;
            for(z = 0; z < world_size;z++){
                if(z !=world_rank ){
                    MPI_SEND(from,(lines + 2) * sizeof(Line),MPI_CHARACTER,z,0,MPI_COMM_WORLD);
                }
            }
            MPI_Status status;
            for(z = 0; z < world_size;z++){
                if(z !=world_rank ){
                    MPI_RECIVE(next,(lines + 2) * sizeof(Line),MPI_CHARACTER,z,1,&status);
                    if(status.MPI_ERROR){
                        //TODO
                        MPI_Abort(MPI_COMM_WORLD,1);
                    }

                    for (y = 1;  y <= (lines%world_size+lines/world_size);  y++) {
                        stpcpy(to[y*z],next[y*z]);
                    }

                }
            }

            temp = from;  
            from = to;  
            to = temp;
    }

    hash = getMD5DigestStr(from[1], sizeof(Line) * (lines));
    printf("hash: %s\n", hash);
        free(next);
    }else{
        int x,y;
        MPI_Status status;
        for(i = 0;  i < its;  i++){
            MPI_RECIVE(from,(lines + 2) * sizeof(Line),MPI_CHARACTER,0,0,&status);
            if(status.MPI_ERROR){
                MPI_Abort(MPI_COMM_WORLD,2);
            }
            for (y = 1;  y <= (lines%world_size+lines/world_size);  y++) {
                for (x = 1;  x <= XSIZE;  x++) {
                    to[y*world_rank][x  ] = transition(from, x  , y*world_rank);
                }
            }
            MPI_SEND(to,(lines + 2) * sizeof(Line),MPI_CHARACTER,0,1,MPI_COMM_WORLD);

        }


    }



    MPI_Finalize();

    free(from);
    free(to);
    free(hash);
    return 0;
}

这是我为大学编写的作为作业的C序列实现。

Answer 1

您是在谈论MPI_Send和MPI_Recv吗？

不知道任何MPI_SEND或MPI_RECIV函数...

我想你只是拼错了。

顺便说一句：这是一个很好的教程，介绍如何使用它们http://mpitutorial.com/tutorials/mpi-send-and-receive/

MPI_SEND和MPI_RECIEVE没有编译参考

1 个答案: