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    <p>Probably the most efficient way would be computing a rectangle-rectangle subtraction that may seem difficult and with a lot of cases, but indeed it's not that hard:</p> <pre><code>struct Rect{ int x0, y0, x1, y1; Rect(int x0, int y0, int x1, int y1) : x0(x0), y0(y0), x1(x1), y1(y1) {} }; std::vector&lt;Rect&gt; subtract(const Rect&amp; a, const Rect&amp; b) { std::vector&lt;Rect&gt; result; if (a.y1 &lt;= b.y0 || a.y0 &gt;= b.y1 || a.x1 &lt;= b.x0 || a.x0 &gt;= b.x1) { // Trivial case: rectangles are not overlapping result.push_back(a); } else { int ystart = a.y0, yend = a.y1; if (ystart &lt; b.y0) { // Something visible above result.push_back(Rect(a.x0, ystart, a.x1, b.y0)); ystart = b.y0; } if (yend &gt; b.y1) { // Something visible below result.push_back(Rect(a.x0, b.y1, a.x1, yend)); yend = b.y1; } if (a.x0 &lt; b.x0) { // Something visible on the left result.push_back(Rect(a.x0, ystart, b.x0, yend)); } if (a.x1 &gt; b.x1) { // Something visible on the right result.push_back(Rect(b.x1, ystart, a.x1, yend)); } } return result; } </code></pre> <p>The above function given two rectangles <code>A</code> and <code>B</code> returns a vector of rectangles with the result of <code>A-B</code>. This vector may be empty (<code>B</code> covers <code>A</code>) or may have from one to four rectangles (four is when <code>B</code> is stricly contained in <code>A</code>, thus the result will be a rectangle with a rectangular hole in it).</p> <p>Using this function you can easily compute <code>new-old</code> and <code>old-new</code> areas.</p> <p>Note that the coordinate schema used in the above code assumes the point-base coordinate system (not a pixel-based coordinate system):</p> <p><img src="https://i.stack.imgur.com/LENHg.png" alt="enter image description here"></p> <p>In the above picture note that horizontal X coordinates of rectangles go from 0 to W (not W-1) and vertical Y coordinates go from 0 to H (and not to H-1).</p> <p>Pixels are just rectangles of area 1 with coordinates <code>(x, y)-(x+1, y+1)</code>; the center of this pixel is <code>(x+0.5, y+0.5)</code>. A rectangle with <code>x0==x1</code> or <code>y0==y1</code> is empty.</p> <p>Note also that the code assumes (and returns) non-empty oriented rectangles, i.e. <code>x0&lt;x1 &amp;&amp; y0&lt;y1</code>.</p> <p>This approach of separating the concept of pixel coordinate from the concept of point coordinate simplifies a lot of the pixel math: for example rectangle area is <code>width*height</code> and not <code>(width-1)*(height-1)</code>.</p> <p>A little program to test with your input case is the following</p> <pre><code>void print_result(const char *name, const std::vector&lt;Rect&gt;&amp; rects) { printf("Result '%s' (%i rects):\n", name, int(rects.size())); for (int i=0,n=rects.size(); i&lt;n; i++) { printf(" %i) (%i, %i) - (%i, %i)\n", i+1, rects[i].x0, rects[i].y0, rects[i].x1, rects[i].y1); } } int main() { Rect A(1, 1, 6, 6); Rect B(3, 2, 8, 7); print_result("A-B", subtract(A, B)); print_result("B-A", subtract(B, A)); return 0; } </code></pre> <p>and the output of this program is</p> <pre><code>Result 'A-B' (2 rects): 1) (1, 1) - (6, 2) 2) (1, 2) - (3, 6) Result 'B-A' (2 rects): 1) (3, 6) - (8, 7) 2) (6, 2) - (8, 6) </code></pre>
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    1. COFYI this isn't entirely correct - Your trivial case subtraction will find that there are new and old points if the 2 boxes are right on top of each other due to the <= instead of <.
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    2. CO@S.Richmond: No. The code is correct; simply it's based on the point coordinate system and not on the pixel coordinate system. I've added an explanation. The point coordinate system is much nicer when dealing with pixel-accurate rendering and allows to avoid both special cases and the need of `-1` and `+1` in the code.
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    3. COAh! Well that makes a lot more sense! I've been having a lot of trouble with my code due to the fact that I was working with the pixel coord system. The problem I'm facing is how I loop over the returned coords properly. For example, I get (4,4)(5,8) returned. Which is 1 row of 5 'points'. But I can't figure out a way to loop over that without getting 2 rows instead. For loop: `for(int x = block->x0; x <= block->x1; x++) { for(int y = block->y0; y <= block->y1; y++) { } }`
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