compiz - Feb 2008 - [PATCH] Make outputDeviceForGeometry behave smarter when dealing with overlapping outputs. Currently, the current output is returned if some part of the rectangle is on it; otherwise the output device the rectangle center is on is returned. This works

This makes scenarios like laptops with external monitors on Xrandr 1.2 work.
Those e.g. have a 1280x800 output and a 1280x1024 output, completely overlapping
each other. With this patch, outputDeviceForGeometry will return the smaller
head for a rectangle that's completely inside the smaller head and the
larger head otherwise.
---
 src/screen.c |   88 ++++++++++++++++++++++++++++++++++++++++++++++++---------
 1 files changed, 74 insertions(+), 14 deletions(-)

diff --git a/src/screen.c b/src/screen.c
index ca308df..701a57f 100644
--- a/src/screen.c
+++ b/src/screen.c
@@ -36,6 +36,7 @@
 #include <sys/types.h>
 #include <unistd.h>
 #include <assert.h>
+#include <limits.h>
 
 #include <X11/Xlib.h>
 #include <X11/Xatom.h>
@@ -3953,6 +3954,27 @@ viewportForGeometry (CompScreen *s,
     }
 }
 
+static int
+rectangleOverlapArea (BOX *rect1,
+		      BOX *rect2)
+{
+    int left, right, top, bottom;
+
+    /* extents of overlapping rectangle */
+    left = MAX (rect1->x1, rect2->x1);
+    right = MIN (rect1->x2, rect2->x2);
+    top = MAX (rect1->y1, rect2->y1);
+    bottom = MIN (rect1->y2, rect2->y2);
+
+    if (left > right || top > bottom)
+    {
+	/* no overlap */
+	return 0;
+    }
+
+    return (right - left) * (bottom - top);
+}
+
 int
 outputDeviceForGeometry (CompScreen *s,
 			 int	    x,
@@ -3961,26 +3983,64 @@ outputDeviceForGeometry (CompScreen *s,
 			 int	    height,
 			 int	    borderWidth)
 {
-    int output = s->currentOutputDev;
-    int x1, y1, x2, y2;
+    int        *overlapAreas;
+    int        i, highest, seen, highestScore;
+    BOX        geomRect;
+ 
+    if (s->nOutputDev == 1)
+	return 0;
 
-    width  += borderWidth * 2;
-    height += borderWidth * 2;
+    overlapAreas = malloc (s->nOutputDev * sizeof (int));
+    if (!overlapAreas)
+	return 0;
 
-    x1 = s->outputDev[output].region.extents.x1;
-    y1 = s->outputDev[output].region.extents.y1;
-    x2 = s->outputDev[output].region.extents.x2;
-    y2 = s->outputDev[output].region.extents.y2;
+    geomRect.x1 = x;
+    geomRect.y1 = y;
+    geomRect.x2 = x + width + 2 * borderWidth;
+    geomRect.y2 = y + height + 2 * borderWidth;
 
-    if (x1 >= x + width  ||
-	y1 >= y + height ||
-	x2 <= x		 ||
-	y2 <= y)
+    /* get amount of overlap on all output devices */
+    for (i = 0; i < s->nOutputDev; i++)
+	overlapAreas[i] = rectangleOverlapArea
(&s->outputDev[i].region.extents,
+						&geomRect);
+
+    /* find head with largest overlap */
+    for (i = 0, highest = 0, highestScore = 0; i < s->nOutputDev; i++)
+	if (overlapAreas[i] > highestScore)
+	{
+	    highest = i;
+	    highestScore = overlapAreas[i];
+	}
+
+    /* look if the highest score is unique */
+    for (i = 0, seen = 0; i < s->nOutputDev; i++)
+	if (overlapAreas[i] == highestScore)
+	    seen++;
+
+    if (seen > 1)
     {
-	output = outputDeviceForPoint (s, x + width  / 2, y + height / 2);
+	/* it's not unique, so find the head which is more overlapped by
+	   the window, which in turns means finding the smallest of the
+	   given heads */
+	unsigned int currentSize, smallestHead = UINT_MAX;
+
+	for (i = 0, highest = 0; i < s->nOutputDev; i++)
+	    if (overlapAreas[i] == highestScore)
+	    {
+		BOX *box = &s->outputDev[i].region.extents;
+
+		currentSize = (box->x2 - box->x1) * (box->y2 - box->y1);
+		if (currentSize < smallestHead)
+		{
+		    highest = i;
+		    smallestHead = currentSize;
+		}
+	    }
     }
+    
+    free (overlapAreas);
 
-    return output;
+    return highest;
 }
 
 Bool
-- 
1.5.3.8


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