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Parallel Processing in C Using Unix Commands

Posted on 2003-12-04
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Last Modified: 2013-12-26
I'm trying to write a program in C using Unix commands that will perform parallel processing of matrices for subtraction, multiplication, and transpose purposes. It will look like this when called:
matrix_proc rp cp <pathname1> <a/s/m/T> [pathname2] <pathename3>
pathname 2 is only optional if the operation is transposing. The operation to be performed will end as such:
a: m3=m1+m2
s: m3=m1-m2
m: m3=m1*m2
T: m3= transpose of m1

The program should spawn rp*cp processes to calculate the elements of the resulting matrix. rcp and cp indicate the number of processes working in parallel on the rows and columns of the resulting matrix respecitvely. The elements of the resulting matrix should be evenly split among the rp*cp processes with no less than one element being assigned to each process. (Assume that the matrices will have valid sizes for the operation being preformed, as I already made the error checking for this part.)

Also, passing of information from parents to child processes should be done through shared memory variables, whereas passing from child to parents processes should be done through pipes.
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Question by:JeebusFnChrist
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by:sunnycoder
ID: 9880080
homework ?
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by:JeebusFnChrist
ID: 9880151
No, it's related to a class I'm taking but it's an outside project more for my own benefit since, clearly, I haven't been able to learn much from the teacher of that course.
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ozo earned 500 total points
ID: 9880336
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#define rp 3
#define cp 3
main(){
  int m1[rp][cp]={{1,2,3},{4,5,6},{7,8,9}};
  int m2[rp][cp]={{10,20,30},{40,50,60},{70,80,90}};
  int m0[rp][cp];
  int p[rp][cp][3];
  int i,j;
  for( i=0;i<rp;i++ ){
    for(j=0;j<cp;j++ ){
      int shared1,shared2,f;
      if( pipe(p[i][j]) ){ perror("pipe"); exit(1); }
      shared1 = m1[i][j];
      shared2 = m2[i][j];
      while( (f = fork()) < 0 ){  perror("fork"); sleep(1); }
      if( f==0 ){
         close(p[i][j][0]);
         shared1 *= shared2;
         write(p[i][j][1], &shared1,sizeof(shared1));
         exit(0);
      }else{
        close(p[i][j][1]);
        p[i][j][2] = f;
      }
    }
  }
  for( i=0;i<rp;i++ ){
    for(j=0;j<cp;j++ ){
      int status;
      read(p[i][j][0],&m0[i][j],sizeof(m0[i][j]));
      close(p[i][j][0]);
      waitpid(p[i][j][2],&status,WNOHANG);
     }
  }
  for( i=0;i<rp;i++ ){
    for(j=0;j<cp;j++ ){
      printf("%d ",m0[i][j]);
    }
    printf("\n");
  }
}
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by:liddler
ID: 10241382
No comment has been added lately, so it's time to clean up this TA.
I will leave a recommendation in the Cleanup topic area that this question is:

Answered by ozo

Please leave any comments here within the next seven days.

PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER!

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