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Program of expression Splitting using multiple barriers

Posted By: Easy Tutor     Category: C Programming     Views: 1908

Write a program of expression Splitting using multiple barriers.

Code for Program of expression Splitting using multiple barriers in C Programming

# include <stdio.h>
# include "/usr/include/sys/types.h"
# include "/usr/include/sys/shm.h"
# include "/usr/include/sys/ipc.h"
# include "/usr/include/sys/sem.h"
# include "forkjoin.h"
# include "sharedlib.h"
# include "spinlock.h"
# include "barrier.h"int main()
{
    int iCount;
    int nProc=10;
    intconst n=19;
    floatconst r=1,t=1;        
    int id,k,**bar_ary,**shmid;    
    float *x,y[20],u[27],z[20],*sum;
    
        
    shmid=(int **)malloc(sizeof(int)* ((nProc-1)/2 +3) );
    
    for(iCount=0;iCount< ((nProc-1)/2+3); iCount++)
    {
        shmid[iCount]=(int *)malloc(sizeof(int));
    }

    sum=(float *)malloc(sizeof(float)*20);
    x=(float *)malloc(sizeof(float)*20);

    sum=(float *)sshared(sizeof(float)*20,shmid[0]);
    x=(float *)sshared(sizeof(float)*20,shmid[1]);

    bar_ary=(int**)malloc(sizeof(int)*((nProc-1)/2 +2));
    
    for(iCount=0; iCount <= ((nProc-1)/2); iCount++)
    {
        bar_ary[iCount]=(int *)malloc(sizeof(int)*5);
        bar_ary[iCount]=(int *)sshared(sizeof(int)*5,shmid[iCount+2]);
    }

    for(iCount=0;iCount<27;iCount++)
    {
        u[iCount]=iCount+3;
        if(iCount<20)
        {
            y[iCount]=iCount+1;
            z[iCount]=iCount+2;
            x[iCount]=0;
            sum[iCount]=0;
        }
    }

    for(iCount=0;iCount <= ((nProc-1)/2);iCount++)
    {
        barrier_init(bar_ary[1,iCount],2);
    }
    
    id=process_fork(nProc);

    for(k=1+(id/2); k<n; k+=nProc/2)
    {        
        if((id%2)==0)
        {
        sum[id]=u[k] + r * ( z[k] + r * y[k] ) + t * ( u[k+3] + r * u[k+2] );
        }
        else
        {
            sum[id]=t * r * r * u[k+1] + t * t * ( u[k+6] + r * ( u[k+5] + r * u[k+4] ));
        }

        barrier(bar_ary[1,id/2]);

        if((id%2) == 0)
        {
        x[k]=sum[id]+sum[id+1];
        }         
        barrier(bar_ary[1,id/2]);
    }

    process_join(nProc,id);

    printf(" Results \n");
    for(iCount=0; iCount<n ;iCount++)
    {
        printf(" x[%d]=%.2f \n",iCount,x[iCount]);
    }

    iCount=0;
    while(iCount <  ((nProc-1)/2 + 3))
    {
        cleanup_memory(shmid[iCount]);
        iCount++;
    }
    return 0;
}
  
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Easy Tutor
Easy Tutor author of Program of expression Splitting using multiple barriers is from United States. Easy Tutor says

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