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+// -*- C++ -*-
+/* Copyright (C) 1989, 1990, 1991, 1992, 2000, 2001, 2002
+ Free Software Foundation, Inc.
+ Written by Gaius Mulley <gaius@glam.ac.uk>
+ using adjust_arc_center() from printer.cc, written by James Clark.
+
+This file is part of groff.
+
+groff 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, or (at your option) any later
+version.
+
+groff 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 groff; see the file COPYING. If not, write to the Free Software
+Foundation, 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */
+
+
+#include <stdio.h>
+#include <math.h>
+
+#undef MAX
+#define MAX(a, b) (((a) > (b)) ? (a) : (b))
+
+#undef MIN
+#define MIN(a, b) (((a) < (b)) ? (a) : (b))
+
+
+// This utility function adjusts the specified center of the
+// arc so that it is equidistant between the specified start
+// and end points. (p[0], p[1]) is a vector from the current
+// point to the center; (p[2], p[3]) is a vector from the
+// center to the end point. If the center can be adjusted,
+// a vector from the current point to the adjusted center is
+// stored in c[0], c[1] and 1 is returned. Otherwise 0 is
+// returned.
+
+#if 1
+int adjust_arc_center(const int *p, double *c)
+{
+ // We move the center along a line parallel to the line between
+ // the specified start point and end point so that the center
+ // is equidistant between the start and end point.
+ // It can be proved (using Lagrange multipliers) that this will
+ // give the point nearest to the specified center that is equidistant
+ // between the start and end point.
+
+ double x = p[0] + p[2]; // (x, y) is the end point
+ double y = p[1] + p[3];
+ double n = x*x + y*y;
+ if (n != 0) {
+ c[0]= double(p[0]);
+ c[1] = double(p[1]);
+ double k = .5 - (c[0]*x + c[1]*y)/n;
+ c[0] += k*x;
+ c[1] += k*y;
+ return 1;
+ }
+ else
+ return 0;
+}
+#else
+int printer::adjust_arc_center(const int *p, double *c)
+{
+ int x = p[0] + p[2]; // (x, y) is the end point
+ int y = p[1] + p[3];
+ // Start at the current point; go in the direction of the specified
+ // center point until we reach a point that is equidistant between
+ // the specified starting point and the specified end point. Place
+ // the center of the arc there.
+ double n = p[0]*double(x) + p[1]*double(y);
+ if (n > 0) {
+ double k = (double(x)*x + double(y)*y)/(2.0*n);
+ // (cx, cy) is our chosen center
+ c[0] = k*p[0];
+ c[1] = k*p[1];
+ return 1;
+ }
+ else {
+ // We would never reach such a point. So instead start at the
+ // specified end point of the arc. Go towards the specified
+ // center point until we reach a point that is equidistant between
+ // the specified start point and specified end point. Place
+ // the center of the arc there.
+ n = p[2]*double(x) + p[3]*double(y);
+ if (n > 0) {
+ double k = 1 - (double(x)*x + double(y)*y)/(2.0*n);
+ // (c[0], c[1]) is our chosen center
+ c[0] = p[0] + k*p[2];
+ c[1] = p[1] + k*p[3];
+ return 1;
+ }
+ else
+ return 0;
+ }
+}
+#endif
+
+
+/*
+ * check_output_arc_limits - works out the smallest box that will encompass
+ * an arc defined by an origin (x, y) and two
+ * vectors (p0, p1) and (p2, p3).
+ * (x1, y1) -> start of arc
+ * (x1, y1) + (xv1, yv1) -> center of circle
+ * (x1, y1) + (xv1, yv1) + (xv2, yv2) -> end of arc
+ *
+ * Works out in which quadrant the arc starts and
+ * stops, and from this it determines the x, y
+ * max/min limits. The arc is drawn clockwise.
+ *
+ * [I'm sure there is a better way to do this, but
+ * I don't know how. Please can someone let me
+ * know or "improve" this function.]
+ */
+
+void check_output_arc_limits(int x1, int y1,
+ int xv1, int yv1,
+ int xv2, int yv2,
+ double c0, double c1,
+ int *minx, int *maxx,
+ int *miny, int *maxy)
+{
+ int radius = (int)sqrt(c0*c0 + c1*c1);
+ int x2 = x1 + xv1 + xv2; // end of arc is (x2, y2)
+ int y2 = y1 + yv1 + yv2;
+
+ // firstly lets use the `circle' limitation
+ *minx = x1 + xv1 - radius;
+ *maxx = x1 + xv1 + radius;
+ *miny = y1 + yv1 - radius;
+ *maxy = y1 + yv1 + radius;
+
+ /* now to see which min/max can be reduced and increased for the limits of
+ * the arc
+ *
+ * Q2 | Q1
+ * -----+-----
+ * Q3 | Q4
+ *
+ *
+ * NB. (x1+xv1, y1+yv1) is at the origin
+ *
+ * below we ask a nested question
+ * (i) from which quadrant does the first vector start?
+ * (ii) into which quadrant does the second vector go?
+ * from the 16 possible answers we determine the limits of the arc
+ */
+ if (xv1 > 0 && yv1 > 0) {
+ // first vector in Q3
+ if (xv2 >= 0 && yv2 >= 0 ) {
+ // second in Q1
+ *maxx = x2;
+ *miny = y1;
+ }
+ else if (xv2 < 0 && yv2 >= 0) {
+ // second in Q2
+ *maxx = x2;
+ *miny = y1;
+ }
+ else if (xv2 >= 0 && yv2 < 0) {
+ // second in Q4
+ *miny = MIN(y1, y2);
+ }
+ else if (xv2 < 0 && yv2 < 0) {
+ // second in Q3
+ if (x1 >= x2) {
+ *minx = x2;
+ *maxx = x1;
+ *miny = MIN(y1, y2);
+ *maxy = MAX(y1, y2);
+ }
+ else {
+ // xv2, yv2 could all be zero?
+ }
+ }
+ }
+ else if (xv1 > 0 && yv1 < 0) {
+ // first vector in Q2
+ if (xv2 >= 0 && yv2 >= 0) {
+ // second in Q1
+ *maxx = MAX(x1, x2);
+ *minx = MIN(x1, x2);
+ *miny = y1;
+ }
+ else if (xv2 < 0 && yv2 >= 0) {
+ // second in Q2
+ if (x1 < x2) {
+ *maxx = x2;
+ *minx = x1;
+ *miny = MIN(y1, y2);
+ *maxy = MAX(y1, y2);
+ }
+ else {
+ // otherwise almost full circle anyway
+ }
+ }
+ else if (xv2 >= 0 && yv2 < 0) {
+ // second in Q4
+ *miny = y2;
+ *minx = x1;
+ }
+ else if (xv2 < 0 && yv2 < 0) {
+ // second in Q3
+ *minx = MIN(x1, x2);
+ }
+ }
+ else if (xv1 <= 0 && yv1 <= 0) {
+ // first vector in Q1
+ if (xv2 >= 0 && yv2 >= 0) {
+ // second in Q1
+ if (x1 < x2) {
+ *minx = x1;
+ *maxx = x2;
+ *miny = MIN(y1, y2);
+ *maxy = MAX(y1, y2);
+ }
+ else {
+ // nearly full circle
+ }
+ }
+ else if (xv2 < 0 && yv2 >= 0) {
+ // second in Q2
+ *maxy = MAX(y1, y2);
+ }
+ else if (xv2 >= 0 && yv2 < 0) {
+ // second in Q4
+ *miny = MIN(y1, y2);
+ *maxy = MAX(y1, y2);
+ *minx = MIN(x1, x2);
+ }
+ else if (xv2 < 0 && yv2 < 0) {
+ // second in Q3
+ *minx = x2;
+ *maxy = y1;
+ }
+ }
+ else if (xv1 <= 0 && yv1 > 0) {
+ // first vector in Q4
+ if (xv2 >= 0 && yv2 >= 0) {
+ // second in Q1
+ *maxx = MAX(x1, x2);
+ }
+ else if (xv2 < 0 && yv2 >= 0) {
+ // second in Q2
+ *maxy = MAX(y1, y2);
+ *maxx = MAX(x1, x2);
+ }
+ else if (xv2 >= 0 && yv2 < 0) {
+ // second in Q4
+ if (x1 >= x2) {
+ *miny = MIN(y1, y2);
+ *maxy = MAX(y1, y2);
+ *minx = MIN(x1, x2);
+ *maxx = MAX(x2, x2);
+ }
+ else {
+ // nearly full circle
+ }
+ }
+ else if (xv2 < 0 && yv2 < 0) {
+ // second in Q3
+ *maxy = MAX(y1, y2);
+ *minx = MIN(x1, x2);
+ *maxx = MAX(x1, x2);
+ }
+ }
+
+ // this should *never* happen but if it does it means a case above is wrong
+ // this code is only present for safety sake
+ if (*maxx < *minx) {
+ fprintf(stderr, "assert failed *minx > *maxx\n");
+ fflush(stderr);
+ *maxx = *minx;
+ }
+ if (*maxy < *miny) {
+ fprintf(stderr, "assert failed *miny > *maxy\n");
+ fflush(stderr);
+ *maxy = *miny;
+ }
+}
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