我正在C ++中探索MPI,我想并行化Mandelbrot集图片的创建。我正在使用ppm
格式。每个处理器构建其部分,并将其发送回作为MPI_CHAR接收它的主进程。这是代码:
#include "mpi.h"
#include <iostream>
#include <string>
#include <fstream>
#include <complex>
using namespace std;
int mandelbrot(int x, int y, int width, int height, int max) {
complex<float> point((float) (y - height/2.0) * 4.0/width, (float) (x - width/2.0) * 4.0/width);
complex<float> z(0, 0);
unsigned int iteration = 0;
while (abs(z) < 4 && iteration < max) {
z = z * z + point;
iteration++;
}
return iteration;
}
int main(int argc, char **argv) {
int numprocs;
int myid;
int buff_size = 404270; // 200x200
char buff[buff_size];
int i;
MPI_Status stat;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&numprocs);
MPI_Comm_rank(MPI_COMM_WORLD,&myid);
int width = 200, height = 200, max_iter = 1000;
if (myid == 0) {
ofstream image("mandel.ppm");
image << "P3\n" << width << " " << height << " 255\n";
for(i=1; i < numprocs; i++) {
MPI_Probe(i, 0, MPI_COMM_WORLD, &stat);
int length;
MPI_Get_count(&stat, MPI_CHAR, &length);
MPI_Recv(buff, length, MPI_CHAR, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
image << buff;
}
} else {
stringstream ss;
// proc rank: 1, 2, ..., n
int part = height/(numprocs-1), start = (myid - 1) * part, end = part * myid;
printf("%d -> %d\n", start, end);
for (int row = start; row < end; row++) {
for (int col = 0; col < width; col++) {
int iteration = mandelbrot(row, col, width, height, max_iter);
if (row == start) ss << 255 << ' ' << 255 << ' ' << 255 << "\n";
else if (iteration < max_iter) ss << iteration * 255 << ' ' << iteration * 20 << ' ' << iteration * 5 << "\n";
else ss << 0 << ' ' << 0 << ' ' << 0 << "\n";
}
}
printf("\n sizeof = %d\n", ss.str().length());
MPI_Send(ss.str().c_str(), ss.str().length(), MPI_CHAR, 0, 0, MPI_COMM_WORLD);
}
MPI_Finalize();
return 0;
}
代码编译:
$ mpic++ -std=c++0x mpi.mandel.cpp -o mpi.mandel
运行3个流程(流程主+流程等级1和2)
$ mpirun -np 3 ./mpi.mandel
以3、4和5进程运行时生成的ppm
图片:
当3个以上的进程尝试将MPI_CHAR元素发送到主进程时,似乎发送-接收的点对点通信正在混合结果。如何避免这种行为?
答案 0 :(得分:0)
在创建与接收消息长度相同的缓冲区buff
时有效:
.
.
for (int i=1; i < numprocs; i++) {
MPI_Probe(i, 0, MPI_COMM_WORLD, &stat);
int length;
MPI_Get_count(&stat, MPI_CHAR, &length);
printf("\nfrom %d <<-- %d (stat.source=%d) Receiving %d chars\n", myid, i, stat.MPI_SOURCE, length);
char buff[length + 1];
MPI_Recv(buff, length, MPI_CHAR, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
buff[length] = '\0';
image << buff;
}
.
.
因此,我们不再需要int buff_size = 404270;
开头的声明,也不需要char buff[buff_size];