The Array Compaction Assignment. Programming Assignment 0.

Due: March 31, 2004


In the array compaction assignment (one of many versions) there are 3 arrays A,C & D with the same number of elements (N).



The goal is to compact the elements of array A into array D according to the elements of array C.



Specifically if C[i] (0<=i<=N) is equal to 1 then A[i] is copied into the next free element in D.



To get the next free element number in D, a prefix sum (ps) instruction is used.



Here is a program that implements the array compaction assignment:



#include <stdio.h>



int A[8]; 		// Global variables are initiated during the execution

int C[8];		// of the XMT simulator with an input data file.

int D[8];



main() 

{ 

	int i;

	register int LR,low,high; // The variables of 'spawn' & 'ps' must be held in 

	global register GR;       // registers. The global register of 'ps' is 

			          // defined here. Global registers must be the 
				  // last ones to be defined.



 	GR = 0; 	

	low=0; 	

	high=7; 	

	spawn low,high;            // A special instruction in a separate line. 

	{

		if ( c[$] == 1)    // '$' represents a running virtual thread ID.

		{

			LR = 1; 		

			ps LR , GR; 	// The left parameter of 'ps' is local and 		

			d[LR]= a[$]; 	// the right parameter is global.

		}

	}

	join;				// A special instruction in a separate line. 



	// For now it is recommended to print inside main(). 	

	// You are supposed to print in serial mode only.  



	for ( i=0 ; i<8 ; i++ )

		printf( "A[%]=%d  ",i,A[i]);	// At most 3 parameters can 
						// be passed to 'printf'.

	printf("\n");				// Do not use '"' inside a string.

	

	for ( i=0 ; i<8 ; i++ )

		printf( "C[%]=%d  ",i,C[i]);

	printf("\n");



	for ( i=0 ; i<8 ; i++ )

		printf( "D[%]=%d  ",i,D[i]);

	printf("\n");

}