view Lib/fftw-3.2.1/kernel/buffered.c @ 1:e86e9c111b29

Updates stuff that potentially fixes the memory leak and also makes it work on Windows and Linux (Need to test). Still have to fix fftw include for linux in Jucer.

/*
 * Copyright (c) 2003, 2007-8 Matteo Frigo
 * Copyright (c) 2003, 2007-8 Massachusetts Institute of Technology
 *
 * This program is free software; you can redistribute it and/or modify
 * it under the terms of the GNU General Public License as published by
 * the Free Software Foundation; either version 2 of the License, or
 * (at your option) any later version.
 *
 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 *
 * You should have received a copy of the GNU General Public License
 * along with this program; if not, write to the Free Software
 * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA  02111-1307  USA
 *
 */

/* routines shared by the various buffered solvers */

#include "ifftw.h"

#define DEFAULT_MAXNBUF ((INT)256)

/* approx. 512KB of buffers for complex data */
#define MAXBUFSZ (256 * 1024 / (INT)(sizeof(R)))

INT X(nbuf)(INT n, INT vl, INT maxnbuf)
{
     INT i, nbuf, lb; 

     if (!maxnbuf) 
	  maxnbuf = DEFAULT_MAXNBUF;

     nbuf = X(imin)(maxnbuf,
		    X(imin)(vl, X(imax)((INT)1, MAXBUFSZ / n)));

     /*
      * Look for a buffer number (not too small) that divides the
      * vector length, in order that we only need one child plan:
      */
     lb = X(imax)(1, nbuf / 4);
     for (i = nbuf; i >= lb; --i)
          if (vl % i == 0)
               return i;

     /* whatever... */
     return nbuf;
}

#define SKEW 6 /* need to be even for SIMD */
#define SKEWMOD 8 

INT X(bufdist)(INT n, INT vl)
{
     if (vl == 1)
	  return n;
     else 
	  /* return smallest X such that X >= N and X == SKEW (mod SKEWMOD) */
	  return n + X(modulo)(SKEW - n, SKEWMOD);
}

int X(toobig)(INT n)
{
     return n > MAXBUFSZ;
}

/* TRUE if there exists i < which such that maxnbuf[i] and
   maxnbuf[which] yield the same value, in which case we canonicalize
   on the minimum value */
int X(nbuf_redundant)(INT n, INT vl, int which, 
		      const INT *maxnbuf, int nmaxnbuf)
{
     int i;
     for (i = 0; i < which; ++i)
	  if (X(nbuf)(n, vl, maxnbuf[i]) == X(nbuf)(n, vl, maxnbuf[which]))
	       return 1;
     return 0;
}