83 lines
		
	
	
		
			2.3 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			83 lines
		
	
	
		
			2.3 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
/*
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 * Copyright (c) 2003, 2007-14 Matteo Frigo
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 * Copyright (c) 2003, 2007-14 Massachusetts Institute of Technology
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 *
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 * This program is free software; you can redistribute it and/or modify
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 * it under the terms of the GNU General Public License as published by
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 * the Free Software Foundation; either version 2 of the License, or
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 * (at your option) any later version.
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 *
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 * This program is distributed in the hope that it will be useful,
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 * but WITHOUT ANY WARRANTY; without even the implied warranty of
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 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
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 * GNU General Public License for more details.
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 *
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 * You should have received a copy of the GNU General Public License
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 * along with this program; if not, write to the Free Software
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 * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301  USA
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 *
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 */
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/* routines shared by the various buffered solvers */
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#include "kernel/ifftw.h"
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#define DEFAULT_MAXNBUF ((INT)256)
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/* approx. 512KB of buffers for complex data */
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#define MAXBUFSZ (256 * 1024 / (INT)(sizeof(R)))
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INT X(nbuf)(INT n, INT vl, INT maxnbuf)
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{
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     INT i, nbuf, lb; 
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     if (!maxnbuf) 
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	  maxnbuf = DEFAULT_MAXNBUF;
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     nbuf = X(imin)(maxnbuf,
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		    X(imin)(vl, X(imax)((INT)1, MAXBUFSZ / n)));
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     /*
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      * Look for a buffer number (not too small) that divides the
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      * vector length, in order that we only need one child plan:
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      */
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     lb = X(imax)(1, nbuf / 4);
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     for (i = nbuf; i >= lb; --i)
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          if (vl % i == 0)
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               return i;
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     /* whatever... */
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     return nbuf;
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}
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#define SKEW 6 /* need to be even for SIMD */
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#define SKEWMOD 8 
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INT X(bufdist)(INT n, INT vl)
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{
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     if (vl == 1)
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	  return n;
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     else 
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	  /* return smallest X such that X >= N and X == SKEW (mod SKEWMOD) */
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	  return n + X(modulo)(SKEW - n, SKEWMOD);
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}
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int X(toobig)(INT n)
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{
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     return n > MAXBUFSZ;
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}
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/* TRUE if there exists i < which such that maxnbuf[i] and
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   maxnbuf[which] yield the same value, in which case we canonicalize
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   on the minimum value */
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int X(nbuf_redundant)(INT n, INT vl, size_t which, 
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		      const INT *maxnbuf, size_t nmaxnbuf)
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{
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     size_t i;
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     (void)nmaxnbuf; /* UNUSED */
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     for (i = 0; i < which; ++i)
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	  if (X(nbuf)(n, vl, maxnbuf[i]) == X(nbuf)(n, vl, maxnbuf[which]))
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	       return 1;
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     return 0;
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}
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