i是通过从限制0到1的曲线4/1 (x*x(下的曲线下的区域来计算Pi(3.14(的值所以。
但是,当有一个过程时,它给出了正确的值。如果我给出一个以上的过程,那么只有等级0的过程给出一些值,而其他过程将为0.0作为本地计算的值
以下代码中有什么错误?
#include<mpi.h>
#include<math.h>
#include<stdio.h>
#define MAX_NAME 80
int main(int argc,char *argv[])
{
MPI_INIT(&argc,&argv);
int rank,nprocs,len;
double i=0.0;
double n=1000000000.0;
double PI25DT =3.141592653589793238462643;
double mypi,pi,step,sum,x;
char name[MAX_NAME];
double start_time,end_time,computation_time;
MPI_Comm_size(MPI_COMM_WORLD,&nprocs);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);
MPI_Get_processor_name(name,&len);
start_time=MPI_Wtime();
sum=0.0;
step=1.0/(double)n;
x=0.0;
x=(double)rank*(n/nprocs);
x=x+step;
double temp=x;
for(i=temp;i<(temp+(n/nprocs));i=i+1.0)
{
sum+=step*(4.0/(1.0+(double)(x*x)));
x=x+step;
}
mypi=sum;
printf("nProcessor: %d Name: %s Sum: %.16f n",rank,name,mypi);
MPI_Barrier(MPI_COMM_WORLD);
MPI_Reduce(&mypi,&pi,1,MPI_DOUBLE,MPI_SUM,0,MPI_COMM_WORLD);
if(rank==0)
{
printf("nValue of Pi is %.16f approximately .Error is %.16f n",pi,fabs(pi-PI25DT));
end_time=MPI_Wtime();
computation_time=end_time-start_time;
printf("nComputation time is: %f seconds.n",computation_time);
}
MPI_Finalize();
}
一个过程的上述代码的输出和多个过程为:
输出
代码中的整个逻辑似乎有缺陷 - 您对"循环"中的整数和双打混淆了,这意味着循环限制都是错误的。
调试简单程序的最简单方法只是打印到屏幕上 - 我永远对那些让人有多么令人惊讶的是让他们的程序告诉他们实际在做什么。
如果我设置n = 5.0并将循环更改为:
x=x+step;
printf("rank %d has x = %lfn", rank, x);
然后在两个过程中我得到:
rank 0 has x = 0.400000
rank 0 has x = 0.600000
rank 0 has x = 0.800000
Processor: 0 Name: starless Sum: 2.0471212357622095
Value of Pi is 2.3040395511642187 approximately .Error is 0.8375531024255745
Computation time is: 0.000067 seconds.
rank 1 has x = 2.900000
rank 1 has x = 3.100000
rank 1 has x = 3.300000
Processor: 1 Name: starless Sum: 0.2569183154020093
显示循环限制是错误的。
我认为它们在一个过程中甚至都不正确:
rank 0 has x = 0.400000
rank 0 has x = 0.600000
rank 0 has x = 0.800000
rank 0 has x = 1.000000
rank 0 has x = 1.200000
肯定的积分在0.0和1.0之间?
之间问:
大卫