//*使用MPI_Sendrecv（）MPI_Send（）MPI_Recv（）實現MPI_Alltoall*//
#include?“mpi.h“
#include?“stdio.h“

#define??maxlen?10
//ALLtoAll?函數
int??My_Alltoall（int??*sendBufferint?sendcntMPI_Datatype?sendtype
?????????????????int?*receiveBufferint?recvcntMPI_Datatype?recvtypeMPI_Comm?commint?rankint?size）
{
??int?i;
??int?j;
??MPI_Status?status;

??if（size!=sendcnt||sendtype!=recvtype）
??????return?0;
??for（i=0;i??{
?????if（rank==i）
?????{
???????MPI_Sendrecv（&sendBuffer[i]1sendtypei99&receiveBuffer[i]1recvtypei99comm&status）;
?????}
?????else
?????{
???????MPI_Send（&sendBuffer[i]1sendtypeiicomm）;
???????MPI_Recv（&receiveBuffer[i]1recvtypeirankcomm&status）;?
?????}
?????
?}
return?1;??

}

int?main（int?argcchar?*argv[]）
{
???int?ranksize;
???MPI_Status?status;
???

???int?sendBuffer[maxlen]receiveBuffer[maxlen];
???int?ij;
???int?count;
??
???MPI_Init（&argc&argv）;
???MPI_Comm_rank（MPI_COMM_WORLD&rank）;
???MPI_Comm_size（MPI_COMM_WORLD&size）;
	//判斷進程數是否合法
????if（?size??10?）???
?????{?if（?rank?==?0?）?printf（“Please?input?a??number?between?1-10\n“）;
???????MPI_Finalize（）;
???????return?0;
??????}
???count=size;

???for（i=0;i???{
???????sendBuffer[i]=（rank+1）*（i+1）;?//初始化發送緩沖區
???????receiveBuffer[i]=0;					?//初始化接收緩沖區
???}
??????

???My_Alltoall（sendBuffercountMPI_INTreceiveBuffercountMPI_INTMPI_COMM_WORLDranksize）;
???
???if（rank==0）
???{
?????for（i=0;i?????{
???????printf（“%d\t“receiveBuffer[i]）;
?????}
??}

???MPI_Finalize（）;
???return?（0）;
}

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????10587??2010-06-06?16:39??高性能并行計算\exams.rar
?????文件?????5024074??2010-06-05?09:31??高性能并行計算\MPI.rar
?????文件??????943249??2010-06-01?22:44??高性能并行計算\mpi并行編程技術郁志輝.pdf
?????目錄???????????0??2010-06-20?16:29??高性能并行計算\并行計算復習提綱與答案\
?????文件????????1578??2010-06-09?13:49??高性能并行計算\并行計算復習提綱與答案\1.MyAlltoall.c
?????文件????????1266??2010-06-09?14:13??高性能并行計算\并行計算復習提綱與答案\2.Lower.c
?????文件????????1409??2010-06-08?16:33??高性能并行計算\并行計算復習提綱與答案\3.myfox.c
?????文件?????????755??2010-06-09?15:34??高性能并行計算\并行計算復習提綱與答案\4.spit.c
?????文件?????????247??2010-06-09?13:46??高性能并行計算\并行計算復習提綱與答案\并行計算復習提綱.txt
?????文件????????2960??2009-06-03?10:23??高性能并行計算\并行計算復習提綱與答案\課后習題?Page119?crecomm.c
?????目錄???????????0??2010-06-20?16:28??高性能并行計算\并行計算復習提綱與答案\另一版本\
?????文件????????2051??2010-06-08?16:36??高性能并行計算\并行計算復習提綱與答案\另一版本\myall.c
?????文件????????1409??2010-06-08?16:33??高性能并行計算\并行計算復習提綱與答案\另一版本\myfox.c
?????文件?????????799??2010-06-08?16:30??高性能并行計算\并行計算復習提綱與答案\另一版本\mygroup.c
?????文件????????1045??2010-06-08?16:27??高性能并行計算\并行計算復習提綱與答案\另一版本\mysum.c
?????文件????????2514??2010-06-08?16:34??高性能并行計算\并行計算復習提綱與答案\另一版本\xiti5_2.c
?????文件?????2217139??2010-05-19?19:46??高性能并行計算\高性能并行計算.pdf