Version 1 (modified by wade, 16 years ago) (diff) |
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Source code
/* Program: * mpich_example 內建範例,計算 pi 。 * History: * 2008-04-11 BETA * 2008-06-19 增加可重複輸入欲計算之精準度 * 2008-06-23 加入 MPI_Barrier 以確保每個 node 在接受 n 後才執行 * / #include "mpi.h" #include <stdio.h> #include <math.h> #include <time.h> double f( double ); double f( double a ) { return (4.0 / (1.0 + a*a)); } int main( int argc, char *argv[]) { int done = 0, n, myid, numprocs, i=0; double PI25DT = 3.141592653589793238462643; double mypi, pi, h, sum, x; double startwtime = 0.0, endwtime; int namelen; char processor_name[MPI_MAX_PROCESSOR_NAME]; MPI_Init(&argc,&argv); MPI_Comm_size(MPI_COMM_WORLD,&numprocs); MPI_Comm_rank(MPI_COMM_WORLD,&myid); MPI_Get_processor_name(processor_name,&namelen); fprintf(stderr,"Process %d on %s\n", myid, processor_name); n = 0; while (!done) { /* 由 node 0 將使用者輸入的值送給其它的 node */ if (myid == 0) { printf("Enter the number of intervals: (0 quits) "); scanf("%d", &n); startwtime = MPI_Wtime(); } /* 這非常重要,所有的 node 必需在此同步,才可以收到使用者輸入的 n */ MPI_Barrier(MPI_COMM_WORLD); /* 將 n 送給其它的 node */ MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD); if (n == 0) done = 1; else { /* 此為計算 pi 的演算法 */ h = 1.0 / (double) n; sum = 0.0; for (i = myid + 1; i <= n; i += numprocs) { x = h * ((double)i - 0.5); sum += f(x); } mypi = h * sum; /* 將算完的結果傳給 node 0 加總 */ MPI_Reduce(&mypi, &pi, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD); if (myid == 0) { printf("pi is approximately %.16f, Error is %.16f\n", pi, fabs(pi - PI25DT)); endwtime = MPI_Wtime(); printf("wall clock time = %f\n", endwtime-startwtime); } } } MPI_Finalize(); return 0; }
Result
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