tk3d.c

linux系统下的音频通信

 */

void
Tk_Draw3DPolygon(tkwin, drawable, border, pointPtr, numPoints,
	borderWidth, leftRelief)
    Tk_Window tkwin;		/* Window for which border was allocated. */
    Drawable drawable;		/* X window or pixmap in which to draw. */
    Tk_3DBorder border;		/* Token for border to draw. */
    XPoint *pointPtr;		/* Array of points describing
				 * polygon.  All points must be
				 * absolute (CoordModeOrigin). */
    int numPoints;		/* Number of points at *pointPtr. */
    int borderWidth;		/* Width of border, measured in
				 * pixels to the left of the polygon's
				 * trajectory.   May be negative. */
    int leftRelief;		/* TK_RELIEF_RAISED or
				 * TK_RELIEF_SUNKEN: indicates how
				 * stuff to left of trajectory looks
				 * relative to stuff on right. */
{
    XPoint poly[4], b1, b2, newB1, newB2;
    XPoint perp, c, shift1, shift2;	/* Used for handling parallel lines. */
    register XPoint *p1Ptr, *p2Ptr;
    TkBorder *borderPtr = (TkBorder *) border;
    GC gc;
    int i, lightOnLeft, dx, dy, parallel, pointsSeen;
    Display *display = Tk_Display(tkwin);

    if (borderPtr->lightGC == None) {
	TkpGetShadows(borderPtr, tkwin);
    }

    /*
     * Handle grooves and ridges with recursive calls.
     */

    if ((leftRelief == TK_RELIEF_GROOVE) || (leftRelief == TK_RELIEF_RIDGE)) {
	int halfWidth;

	halfWidth = borderWidth/2;
	Tk_Draw3DPolygon(tkwin, drawable, border, pointPtr, numPoints,
		halfWidth, (leftRelief == TK_RELIEF_GROOVE) ? TK_RELIEF_RAISED
		: TK_RELIEF_SUNKEN);
	Tk_Draw3DPolygon(tkwin, drawable, border, pointPtr, numPoints,
		-halfWidth, (leftRelief == TK_RELIEF_GROOVE) ? TK_RELIEF_SUNKEN
		: TK_RELIEF_RAISED);
	return;
    }

    /*
     * If the polygon is already closed, drop the last point from it
     * (we'll close it automatically).
     */

    p1Ptr = &pointPtr[numPoints-1];
    p2Ptr = &pointPtr[0];
    if ((p1Ptr->x == p2Ptr->x) && (p1Ptr->y == p2Ptr->y)) {
	numPoints--;
    }

    /*
     * The loop below is executed once for each vertex in the polgon.
     * At the beginning of each iteration things look like this:
     *
     *          poly[1]       /
     *             *        /
     *             |      /
     *             b1   * poly[0] (pointPtr[i-1])
     *             |    |
     *             |    |
     *             |    |
     *             |    |
     *             |    |
     *             |    | *p1Ptr            *p2Ptr
     *             b2   *--------------------*
     *             |
     *             |
     *             x-------------------------
     *
     * The job of this iteration is to do the following:
     * (a) Compute x (the border corner corresponding to
     *     pointPtr[i]) and put it in poly[2].  As part of
     *	   this, compute a new b1 and b2 value for the next
     *	   side of the polygon.
     * (b) Put pointPtr[i] into poly[3].
     * (c) Draw the polygon given by poly[0..3].
     * (d) Advance poly[0], poly[1], b1, and b2 for the
     *     next side of the polygon.
     */

    /*
     * The above situation doesn't first come into existence until
     * two points have been processed;  the first two points are
     * used to "prime the pump", so some parts of the processing
     * are ommitted for these points.  The variable "pointsSeen"
     * keeps track of the priming process;  it has to be separate
     * from i in order to be able to ignore duplicate points in the
     * polygon.
     */

    pointsSeen = 0;
    for (i = -2, p1Ptr = &pointPtr[numPoints-2], p2Ptr = p1Ptr+1;
	    i < numPoints; i++, p1Ptr = p2Ptr, p2Ptr++) {
	if ((i == -1) || (i == numPoints-1)) {
	    p2Ptr = pointPtr;
	}
	if ((p2Ptr->x == p1Ptr->x) && (p2Ptr->y == p1Ptr->y)) {
	    /*
	     * Ignore duplicate points (they'd cause core dumps in
	     * ShiftLine calls below).
	     */
	    continue;
	}
	ShiftLine(p1Ptr, p2Ptr, borderWidth, &newB1);
	newB2.x = newB1.x + (p2Ptr->x - p1Ptr->x);
	newB2.y = newB1.y + (p2Ptr->y - p1Ptr->y);
	poly[3] = *p1Ptr;
	parallel = 0;
	if (pointsSeen >= 1) {
	    parallel = Intersect(&newB1, &newB2, &b1, &b2, &poly[2]);

	    /*
	     * If two consecutive segments of the polygon are parallel,
	     * then things get more complex.  Consider the following
	     * diagram:
	     *
	     * poly[1]
	     *    *----b1-----------b2------a
	     *                                \
	     *                                  \
	     *         *---------*----------*    b
	     *        poly[0]  *p2Ptr   *p1Ptr  /
	     *                                /
	     *              --*--------*----c
	     *              newB1    newB2
	     *
	     * Instead of using x and *p1Ptr for poly[2] and poly[3], as
	     * in the original diagram, use a and b as above.  Then instead
	     * of using x and *p1Ptr for the new poly[0] and poly[1], use
	     * b and c as above.
	     *
	     * Do the computation in three stages:
	     * 1. Compute a point "perp" such that the line p1Ptr-perp
	     *    is perpendicular to p1Ptr-p2Ptr.
	     * 2. Compute the points a and c by intersecting the lines
	     *    b1-b2 and newB1-newB2 with p1Ptr-perp.
	     * 3. Compute b by shifting p1Ptr-perp to the right and
	     *    intersecting it with p1Ptr-p2Ptr.
	     */

	    if (parallel) {
		perp.x = p1Ptr->x + (p2Ptr->y - p1Ptr->y);
		perp.y = p1Ptr->y - (p2Ptr->x - p1Ptr->x);
		(void) Intersect(p1Ptr, &perp, &b1, &b2, &poly[2]);
		(void) Intersect(p1Ptr, &perp, &newB1, &newB2, &c);
		ShiftLine(p1Ptr, &perp, borderWidth, &shift1);
		shift2.x = shift1.x + (perp.x - p1Ptr->x);
		shift2.y = shift1.y + (perp.y - p1Ptr->y);
		(void) Intersect(p1Ptr, p2Ptr, &shift1, &shift2, &poly[3]);
	    }
	}
	if (pointsSeen >= 2) {
	    dx = poly[3].x - poly[0].x;
	    dy = poly[3].y - poly[0].y;
	    if (dx > 0) {
		lightOnLeft = (dy <= dx);
	    } else {
		lightOnLeft = (dy < dx);
	    }
	    if (lightOnLeft ^ (leftRelief == TK_RELIEF_RAISED)) {
		gc = borderPtr->lightGC;
	    } else {
		gc = borderPtr->darkGC;
	    }
	    XFillPolygon(display, drawable, gc, poly, 4, Convex,
		    CoordModeOrigin);
	}
	b1.x = newB1.x;
	b1.y = newB1.y;
	b2.x = newB2.x;
	b2.y = newB2.y;
	poly[0].x = poly[3].x;
	poly[0].y = poly[3].y;
	if (parallel) {
	    poly[1].x = c.x;
	    poly[1].y = c.y;
	} else if (pointsSeen >= 1) {
	    poly[1].x = poly[2].x;
	    poly[1].y = poly[2].y;
	}
	pointsSeen++;
    }
}

/*
 *----------------------------------------------------------------------
 *
 * Tk_Fill3DRectangle --
 *
 *	Fill a rectangular area, supplying a 3D border if desired.
 *
 * Results:
 *	None.
 *
 * Side effects:
 *	Information gets drawn on the screen.
 *
 *----------------------------------------------------------------------
 */

void
Tk_Fill3DRectangle(tkwin, drawable, border, x, y, width,
	height, borderWidth, relief)
    Tk_Window tkwin;		/* Window for which border was allocated. */
    Drawable drawable;		/* X window or pixmap in which to draw. */
    Tk_3DBorder border;		/* Token for border to draw. */
    int x, y, width, height;	/* Outside area of rectangular region. */
    int borderWidth;		/* Desired width for border, in
				 * pixels. Border will be *inside* region. */
    int relief;			/* Indicates 3D effect: TK_RELIEF_FLAT,
				 * TK_RELIEF_RAISED, or TK_RELIEF_SUNKEN. */
{
    register TkBorder *borderPtr = (TkBorder *) border;
    int doubleBorder;

    /*
     * This code is slightly tricky because it only draws the background
     * in areas not covered by the 3D border. This avoids flashing
     * effects on the screen for the border region.
     */
  
    if (relief == TK_RELIEF_FLAT) {
	borderWidth = 0;
    }
    doubleBorder = 2*borderWidth;

    if ((width > doubleBorder) && (height > doubleBorder)) {
	XFillRectangle(Tk_Display(tkwin), drawable, borderPtr->bgGC,
		x + borderWidth, y + borderWidth,
		(unsigned int) (width - doubleBorder),
		(unsigned int) (height - doubleBorder));
    }
    if (borderWidth) {
	Tk_Draw3DRectangle(tkwin, drawable, border, x, y, width,
		height, borderWidth, relief);
    }
}

/*
 *----------------------------------------------------------------------
 *
 * Tk_Fill3DPolygon --
 *
 *	Fill a polygonal area, supplying a 3D border if desired.
 *
 * Results:
 *	None.
 *
 * Side effects:
 *	Information gets drawn on the screen.
 *
 *----------------------------------------------------------------------
 */

void
Tk_Fill3DPolygon(tkwin, drawable, border, pointPtr, numPoints,
	borderWidth, leftRelief)
    Tk_Window tkwin;		/* Window for which border was allocated. */
    Drawable drawable;		/* X window or pixmap in which to draw. */
    Tk_3DBorder border;		/* Token for border to draw. */
    XPoint *pointPtr;		/* Array of points describing
				 * polygon.  All points must be
				 * absolute (CoordModeOrigin). */
    int numPoints;		/* Number of points at *pointPtr. */
    int borderWidth;		/* Width of border, measured in
				 * pixels to the left of the polygon's
				 * trajectory.   May be negative. */
    int leftRelief;			/* Indicates 3D effect of left side of
				 * trajectory relative to right:
				 * TK_RELIEF_FLAT, TK_RELIEF_RAISED,
				 * or TK_RELIEF_SUNKEN. */
{
    register TkBorder *borderPtr = (TkBorder *) border;

    XFillPolygon(Tk_Display(tkwin), drawable, borderPtr->bgGC,
	    pointPtr, numPoints, Complex, CoordModeOrigin);
    if (leftRelief != TK_RELIEF_FLAT) {
	Tk_Draw3DPolygon(tkwin, drawable, border, pointPtr, numPoints,
		borderWidth, leftRelief);
    }
}

/*
 *--------------------------------------------------------------
 *
 * BorderInit --
 *
 *	Initialize the structures used for border management.
 *
 * Results:
 *	None.
 *
 * Side effects:
 *	Read the code.
 *
 *-------------------------------------------------------------
 */

static void
BorderInit()
{
    initialized = 1;
    Tcl_InitHashTable(&borderTable, sizeof(BorderKey)/sizeof(int));
}

/*
 *--------------------------------------------------------------
 *
 * ShiftLine --
 *
 *	Given two points on a line, compute a point on a
 *	new line that is parallel to the given line and
 *	a given distance away from it.
 *
 * Results:
 *	None.
 *
 * Side effects:
 *	None.
 *
 *--------------------------------------------------------------
 */

static void
ShiftLine(p1Ptr, p2Ptr, distance, p3Ptr)
    XPoint *p1Ptr;		/* First point on line. */
    XPoint *p2Ptr;		/* Second point on line. */
    int distance;		/* New line is to be this many
				 * units to the left of original
				 * line, when looking from p1 to
				 * p2.  May be negative. */
    XPoint *p3Ptr;		/* Store coords of point on new
				 * line here. */
{
    int dx, dy, dxNeg, dyNeg;

    /*
     * The table below is used for a quick approximation in
     * computing the new point.  An index into the table
     * is 128 times the slope of the original line (the slope
     * must always be between 0 and 1).  The value of the table
     * entry is 128 times the amount to displace the new line
     * in y for each unit of perpendicular distance.  In other
     * words, the table maps from the tangent of an angle to
     * the inverse of its cosine.  If the slope of the original
     * line is greater than 1, then the displacement is done in
     * x rather than in y.
     */

    static int shiftTable[129];

    /*
     * Initialize the table if this is the first time it is
     * used.
     */

    if (shiftTable[0] == 0) {
	int i;
	double tangent, cosine;

	for (i = 0; i <= 128; i++) {
	    tangent = i/128.0;
	    cosine = 128/cos(atan(tangent)) + .5;
	    shiftTable[i] = (int) cosine;
	}
    }

    *p3Ptr = *p1Ptr;
    dx = p2Ptr->x - p1Ptr->x;
    dy = p2Ptr->y - p1Ptr->y;
    if (dy < 0) {
	dyNeg = 1;
	dy = -dy;
    } else {
	dyNeg = 0;
    }
    if (dx < 0) {
	dxNeg = 1;
	dx = -dx;
    } else {
	dxNeg = 0;
    }
    if (dy <= dx) {
	dy = ((distance * shiftTable[(dy<<7)/dx]) + 64) >> 7;
	if (!dxNeg) {
	    dy = -dy;
	}
	p3Ptr->y += dy;
    } else {
	dx = ((distance * shiftTable[(dx<<7)/dy]) + 64) >> 7;
	if (dyNeg) {
	    dx = -dx;
	}
	p3Ptr->x += dx;
    }
}

/*
 *--------------------------------------------------------------
 *
 * Intersect --
 *
 *	Find the intersection point between two lines.
 *
 * Results:
 *	Under normal conditions 0 is returned and the point
 *	at *iPtr is filled in with the intersection between
 *	the two lines.  If the two lines are parallel, then
 *	-1 is returned and *iPtr isn't modified.
 *
 * Side effects:
 *	None.
 *
 *--------------------------------------------------------------
 */

static int
Intersect(a1Ptr, a2Ptr, b1Ptr, b2Ptr, iPtr)
    XPoint *a1Ptr;		/* First point of first line. */
    XPoint *a2Ptr;		/* Second point of first line. */
    XPoint *b1Ptr;		/* First point of second line. */
    XPoint *b2Ptr;		/* Second point of second line. */
    XPoint *iPtr;		/* Filled in with intersection point. */
{
    int dxadyb, dxbdya, dxadxb, dyadyb, p, q;

    /*
     * The code below is just a straightforward manipulation of two
     * equations of the form y = (x-x1)*(y2-y1)/(x2-x1) + y1 to solve
     * for the x-coordinate of intersection, then the y-coordinate.
     */

    dxadyb = (a2Ptr->x - a1Ptr->x)*(b2Ptr->y - b1Ptr->y);
    dxbdya = (b2Ptr->x - b1Ptr->x)*(a2Ptr->y - a1Ptr->y);
    dxadxb = (a2Ptr->x - a1Ptr->x)*(b2Ptr->x - b1Ptr->x);
    dyadyb = (a2Ptr->y - a1Ptr->y)*(b2Ptr->y - b1Ptr->y);

    if (dxadyb == dxbdya) {
	return -1;
    }
    p = (a1Ptr->x*dxbdya - b1Ptr->x*dxadyb + (b1Ptr->y - a1Ptr->y)*dxadxb);
    q = dxbdya - dxadyb;
    if (q < 0) {
	p = -p;
	q = -q;
    }
    if (p < 0) {
	iPtr->x = - ((-p + q/2)/q);
    } else {
	iPtr->x = (p + q/2)/q;
    }
    p = (a1Ptr->y*dxadyb - b1Ptr->y*dxbdya + (b1Ptr->x - a1Ptr->x)*dyadyb);
    q = dxadyb - dxbdya;
    if (q < 0) {
	p = -p;
	q = -q;
    }
    if (p < 0) {
	iPtr->y = - ((-p + q/2)/q);
    } else {
	iPtr->y = (p + q/2)/q;
    }
    return 0;
}