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https://github.com/NGSolve/netgen.git
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336 lines
9.2 KiB
C
336 lines
9.2 KiB
C
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/* $XConsortium: CmapAlloc.c,v 1.9 94/04/17 20:15:52 rws Exp $ */
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/*
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Copyright (c) 1989, 1994 X Consortium
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Permission is hereby granted, free of charge, to any person obtaining a copy
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of this software and associated documentation files (the "Software"), to deal
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in the Software without restriction, including without limitation the rights
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to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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copies of the Software, and to permit persons to whom the Software is
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furnished to do so, subject to the following conditions:
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The above copyright notice and this permission notice shall be included in
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all copies or substantial portions of the Software.
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THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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X CONSORTIUM BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN
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AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
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CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
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Except as contained in this notice, the name of the X Consortium shall not be
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used in advertising or otherwise to promote the sale, use or other dealings
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in this Software without prior written authorization from the X Consortium.
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*/
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/*
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* Author: Donna Converse, MIT X Consortium
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*/
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#include <X11/Xlib.h>
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#include <X11/Xatom.h>
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#include <X11/Xutil.h>
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#include <stdio.h>
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#define lowbit(x) ((x) & (~(x) + 1))
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static int default_allocation();
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static void best_allocation();
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static void gray_allocation();
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static int icbrt();
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static int icbrt_with_bits();
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static int icbrt_with_guess();
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/* To determine the best allocation of reds, greens, and blues in a
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* standard colormap, use XmuGetColormapAllocation.
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* vinfo specifies visual information for a chosen visual
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* property specifies one of the standard colormap property names
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* red_max returns maximum red value
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* green_max returns maximum green value
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* blue_max returns maximum blue value
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*
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* XmuGetColormapAllocation returns 0 on failure, non-zero on success.
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* It is assumed that the visual is appropriate for the colormap property.
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*/
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Status XmuGetColormapAllocation(vinfo, property, red_max, green_max, blue_max)
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XVisualInfo *vinfo;
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Atom property;
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unsigned long *red_max, *green_max, *blue_max;
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{
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Status status = 1;
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if (vinfo->colormap_size <= 2)
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return 0;
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switch (property)
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{
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case XA_RGB_DEFAULT_MAP:
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status = default_allocation(vinfo, red_max, green_max, blue_max);
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break;
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case XA_RGB_BEST_MAP:
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best_allocation(vinfo, red_max, green_max, blue_max);
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break;
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case XA_RGB_GRAY_MAP:
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gray_allocation(vinfo->colormap_size, red_max, green_max, blue_max);
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break;
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case XA_RGB_RED_MAP:
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*red_max = vinfo->colormap_size - 1;
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*green_max = *blue_max = 0;
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break;
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case XA_RGB_GREEN_MAP:
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*green_max = vinfo->colormap_size - 1;
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*red_max = *blue_max = 0;
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break;
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case XA_RGB_BLUE_MAP:
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*blue_max = vinfo->colormap_size - 1;
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*red_max = *green_max = 0;
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break;
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default:
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status = 0;
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}
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return status;
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}
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/****************************************************************************/
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/* Determine the appropriate color allocations of a gray scale.
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*
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* Keith Packard, MIT X Consortium
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*/
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static void gray_allocation(n, red_max, green_max, blue_max)
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int n; /* the number of cells of the gray scale */
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unsigned long *red_max, *green_max, *blue_max;
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{
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*red_max = (n * 30) / 100;
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*green_max = (n * 59) / 100;
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*blue_max = (n * 11) / 100;
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*green_max += ((n - 1) - (*red_max + *green_max + *blue_max));
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}
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/****************************************************************************/
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/* Determine an appropriate color allocation for the RGB_DEFAULT_MAP.
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* If a map has less than a minimum number of definable entries, we do not
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* produce an allocation for an RGB_DEFAULT_MAP.
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*
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* For 16 planes, the default colormap will have 27 each RGB; for 12 planes,
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* 12 each. For 8 planes, let n = the number of colormap entries, which may
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* be 256 or 254. Then, maximum red value = floor(cube_root(n - 125)) - 1.
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* Maximum green and maximum blue values are identical to maximum red.
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* This leaves at least 125 cells which clients can allocate.
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*
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* Return 0 if an allocation has been determined, non-zero otherwise.
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*/
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static int default_allocation(vinfo, red, green, blue)
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XVisualInfo *vinfo;
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unsigned long *red, *green, *blue;
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{
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int ngrays; /* number of gray cells */
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switch (vinfo->class) {
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case PseudoColor:
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if (vinfo->colormap_size > 65000)
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/* intended for displays with 16 planes */
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*red = *green = *blue = (unsigned long) 27;
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else if (vinfo->colormap_size > 4000)
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/* intended for displays with 12 planes */
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*red = *green = *blue = (unsigned long) 12;
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else if (vinfo->colormap_size < 250)
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return 0;
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else
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/* intended for displays with 8 planes */
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*red = *green = *blue = (unsigned long)
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(icbrt(vinfo->colormap_size - 125) - 1);
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break;
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case DirectColor:
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if (vinfo->colormap_size < 10)
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return 0;
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*red = *green = *blue = vinfo->colormap_size / 2 - 1;
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break;
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case TrueColor:
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*red = vinfo->red_mask / lowbit(vinfo->red_mask);
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*green = vinfo->green_mask / lowbit(vinfo->green_mask);
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*blue = vinfo->blue_mask / lowbit(vinfo->blue_mask);
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break;
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case GrayScale:
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if (vinfo->colormap_size > 65000)
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ngrays = 4096;
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else if (vinfo->colormap_size > 4000)
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ngrays = 512;
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else if (vinfo->colormap_size < 250)
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return 0;
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else
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ngrays = 12;
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gray_allocation(ngrays, red, green, blue);
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break;
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default:
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return 0;
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}
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return 1;
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}
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/****************************************************************************/
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/* Determine an appropriate color allocation for the RGB_BEST_MAP.
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*
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* For a DirectColor or TrueColor visual, the allocation is determined
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* by the red_mask, green_mask, and blue_mask members of the visual info.
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*
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* Otherwise, if the colormap size is an integral power of 2, determine
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* the allocation according to the number of bits given to each color,
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* with green getting more than red, and red more than blue, if there
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* are to be inequities in the distribution. If the colormap size is
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* not an integral power of 2, let n = the number of colormap entries.
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* Then maximum red value = floor(cube_root(n)) - 1;
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* maximum blue value = floor(cube_root(n)) - 1;
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* maximum green value = n / ((# red values) * (# blue values)) - 1;
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* Which, on a GPX, allows for 252 entries in the best map, out of 254
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* defineable colormap entries.
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*/
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static void best_allocation(vinfo, red, green, blue)
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XVisualInfo *vinfo;
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unsigned long *red, *green, *blue;
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{
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if (vinfo->class == DirectColor || vinfo->class == TrueColor)
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{
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*red = vinfo->red_mask;
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while ((*red & 01) == 0)
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*red >>= 1;
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*green = vinfo->green_mask;
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while ((*green & 01) == 0)
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*green >>=1;
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*blue = vinfo->blue_mask;
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while ((*blue & 01) == 0)
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*blue >>= 1;
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}
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else
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{
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register int bits, n;
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/* Determine n such that n is the least integral power of 2 which is
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* greater than or equal to the number of entries in the colormap.
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*/
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n = 1;
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bits = 0;
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while (vinfo->colormap_size > n)
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{
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n = n << 1;
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bits++;
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}
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/* If the number of entries in the colormap is a power of 2, determine
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* the allocation by "dealing" the bits, first to green, then red, then
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* blue. If not, find the maximum integral red, green, and blue values
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* which, when multiplied together, do not exceed the number of
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* colormap entries.
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*/
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if (n == vinfo->colormap_size)
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{
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register int r, g, b;
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b = bits / 3;
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g = b + ((bits % 3) ? 1 : 0);
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r = b + (((bits % 3) == 2) ? 1 : 0);
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*red = 1 << r;
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*green = 1 << g;
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*blue = 1 << b;
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}
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else
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{
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*red = icbrt_with_bits(vinfo->colormap_size, bits);
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*blue = *red;
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*green = (vinfo->colormap_size / ((*red) * (*blue)));
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}
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(*red)--;
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(*green)--;
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(*blue)--;
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}
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return;
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}
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/*
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* integer cube roots by Newton's method
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*
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* Stephen Gildea, MIT X Consortium, July 1991
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*/
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static int icbrt(a) /* integer cube root */
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int a;
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{
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register int bits = 0;
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register unsigned n = a;
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while (n)
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{
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bits++;
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n >>= 1;
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}
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return icbrt_with_bits(a, bits);
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}
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static int icbrt_with_bits(a, bits)
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int a;
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int bits; /* log 2 of a */
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{
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return icbrt_with_guess(a, a>>2*bits/3);
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}
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#ifdef _X_ROOT_STATS
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int icbrt_loopcount;
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#endif
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/* Newton's Method: x_n+1 = x_n - ( f(x_n) / f'(x_n) ) */
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/* for cube roots, x^3 - a = 0, x_new = x - 1/3 (x - a/x^2) */
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/*
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* Quick and dirty cube roots. Nothing fancy here, just Newton's method.
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* Only works for positive integers (since that's all we need).
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* We actually return floor(cbrt(a)) because that's what we need here, too.
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*/
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static int icbrt_with_guess(a, guess)
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int a, guess;
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{
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register int delta;
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#ifdef _X_ROOT_STATS
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icbrt_loopcount = 0;
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#endif
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if (a <= 0)
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return 0;
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if (guess < 1)
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guess = 1;
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do {
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#ifdef _X_ROOT_STATS
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icbrt_loopcount++;
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#endif
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delta = (guess - a/(guess*guess))/3;
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#ifdef DEBUG
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printf("pass %d: guess=%d, delta=%d\n", icbrt_loopcount, guess, delta);
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#endif
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guess -= delta;
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} while (delta != 0);
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if (guess*guess*guess > a)
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guess--;
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return guess;
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}
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