87 lines
1.9 KiB
C
87 lines
1.9 KiB
C
#include <stdio.h>
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#include <stdlib.h>
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#include <sys/time.h>
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#include <string.h>
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#include <math.h>
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#include "common.h"
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int main(int argc, char **argv){
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int N, NB, B, NN;
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double flops, nrm2, nrmf;
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double *Acpy1, *Acpy2, *x, *b;
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Matrix A;
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info_type info;
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long t_start,t_end;
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int err, nth, i, j;
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int ISEED[4] = {0,0,0,1};
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int IONE=1;
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char NoTran = 'N', *nt;
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double DONE=1.0, DMONE=-1.0;
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if(argc != 3){
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printf("Usage:\n\n./main B NB\n\nwhere B is the size of block-columns and \n\
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NB is the number of block-columns the matrix is made of.\n");
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return 1;
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}
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B = atoi(argv[1]); /* block size */
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NB = atoi(argv[2]); /* dimension in blocks */
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N = B*NB;
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NN = N*N;
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nt = getenv("OMP_NUM_THREADS");
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flops = ((double)2.0*(double) N)*((double) N)*((double) N)/3.0;
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Acpy1 = (double *)malloc(NN*sizeof(double));
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Acpy2 = (double *)malloc(NN*sizeof(double));
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info.ipiv = (int *)malloc(N*sizeof(int));
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dlarnv_(&IONE, ISEED, &NN, Acpy1);
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info.B = B;
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info.NB = NB;
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info.N = N;
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printf("Matrix size: %d\n",N);
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A = (double **)malloc(info.NB*sizeof(double*));
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/* get pointers to block-columns */
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for (j=0; j<info.NB; j+=1)
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A[j] = Acpy2+j*info.B*info.N;
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printf("\n========== Sequential (1 threads) ==========\n" );
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memcpy( Acpy2, Acpy1, N*N*sizeof(double) );
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t_start = usecs();
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lu_seq(A, info);
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t_end = usecs();
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printf("Time (msec.) : %7.1f\n",(t_end-t_start)/1e3);
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printf("Gflop/s : %7.1f\n",flops/(t_end-t_start)/1e3);
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checkres(Acpy2, Acpy1, info);
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printf("\n========== Tasks Parallel (%1s threads) ==========\n",nt);
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memcpy( Acpy2, Acpy1, N*N*sizeof(double) );
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t_start = usecs();
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lu_par_tasks(A, info);
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t_end = usecs();
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printf("Time (msec.) : %7.1f\n",(t_end-t_start)/1e3);
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printf("Gflop/s : %7.1f\n",flops/(t_end-t_start)/1e3);
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checkres(Acpy2, Acpy1, info);
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return 0;
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}
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