Behaviour of CRgn class

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Hi All,

C++

CPoint ptArray[5];
ptArray[0] = CPoint(657, 79);
ptArray[1] = CPoint(675, 71);
ptArray[2] = CPoint(694, 66);
ptArray[3] = CPoint(912, 64);
ptArray[4] = CPoint(792, 66);
CRgn rgnObject;
rgnObject.CreatePolygonRgn(ptArray, 5, ALTERNATE);
CRect rectBounding;
rgnObject.GetRgnBox(rectBounding);

For bounding rect, I was expecting left = 657, right = 912, top = 64 and bottom 79 i.e. minimum and maximum X and Y coordinates.
Actual results are left = 660, right = 852, top = 65 and bottom 79

Can anyone explain this?

Thanks in advance.

Posted 4-Jul-11 1:04am

Roger Gonsalves

Updated 4-Jul-11 2:21am

Ramalinga Koushik

v3

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Richard MacCutchan 4-Jul-11 8:21am

Please use <pre> tags around your code.

Roger Gonsalves 4-Jul-11 22:47pm

I was in little bit hurry, so forgot to do that.
Anyway, thanks for advice.

Richard MacCutchan 5-Jul-11 5:54am

It takes less than a second, and only one mouse click!

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CPallini · Answer 1 · 2011-07-04T02:24:00

Solution 1

The bounding rectangle actually bounds the region, however, some points (the first, fourth and fifth) defining the region actually aren't inside the region itself (you may verify this using the PtInRegion method). I think this is related to roundings happening in region calculation.

Posted 4-Jul-11 2:24am

CPallini

Comments

Sergey Alexandrovich Kryukov 4-Jul-11 19:49pm

If this is rounding error, it would be amazingly low accuracy, don't you think so? I could hardly accept this explanation. Did you try it in code? maybe OP's report is not accurate? This is the second report of incorrect point set operation I see in CodeProject. Lousy piece of Win32 code?
--SA

CPallini 5-Jul-11 5:50am

Of course I tried. I don't know if it is acceptable or not it depends on the region algo definition (and Microsoft doesn't provide it in the documentation).
Suppose part of the region is a triangle: if one of the angles is small enough then, near the corrensponding vertex, the arc between the sides approaches zero.

Sergey Alexandrovich Kryukov 5-Jul-11 6:32am

I think the real problem is the algorithm for non-convex regions, and not sure how and if is can be formulated correctly for self-intersecting boundary (which effectively stop being a "boundary"). Will it work?
--SA

CPallini 5-Jul-11 6:54am

You are wrong: try it with a(n appropriate) triangle.

Sergey Alexandrovich Kryukov 6-Jul-11 17:32pm

A sequence of connected segment is not a set of triangles. You can represent this picture with triangles, but some of those triangles will be "empty" and some -- the subsets of the region. Do you see my point? The sub-set is not a union of those rectangle. I am sure the right (and probably pretty simple) algorithm does exists, I just doubt it is that simple. I only asked you: will it work? I did not even make is my statement. I could easily check your references algorithm on a simple example with self-intersecting chain of segments and "empty" triangles inside, but I wanted first to ask you: did you check it up?

Also, did you check up if "my solution" (actually, a Windows solution, not my; I only posted it) works?

Thank you.
--SA

CPallini 7-Jul-11 1:20am

You didn't get me: try ONE triangle.

Sergey Alexandrovich Kryukov 8-Jul-11 1:15am

And?...
The problem is general, a chain of segments. Why taking one triangle?
--SA

CPallini 8-Jul-11 3:13am

If the algo 'fails' (it doesn't fullfil our expectations) with a single triangle (i.e. a convex polygon) then my hypothesys is good.
Did you try (just choose one very small angle at one vertex)?

Sergey Alexandrovich Kryukov 12-Jul-11 15:00pm

I say no, but this issue looks disturbing; what is MS code is lousy here? When and if I have time I'll check it up. I know how it works in good vector editors -- all quite logical.
--SA

Sergey Alexandrovich Kryukov 6-Jul-11 17:34pm

This is, however, depends on implementation. You can consider a region as a union by definition. But the question is what a Windows region really does. I think it is traditionally considers a self-intersecting chain of segment as a set of "empty" triangles and subset triangles -- every vector editor demonstrates this.
--SA

Sergey Alexandrovich Kryukov 5-Jul-11 6:37am

I found available method in this class, it would be good to try. .NET region also has such method called Visible...
Please see my solution.
--SA

Sergey Alexandrovich Kryukov · Answer 2 · 2011-07-05T00:36:00

Solution 2

The solution is supposed to be implemented by the available method CRgn::PtInRegion, see http://msdn.microsoft.com/en-us/library/07zzbzfh(v=VS.71).aspx[^].

—SA

Posted 5-Jul-11 0:36am

Sergey Alexandrovich Kryukov

Comments

CPallini 5-Jul-11 6:50am

Did you try it? I did. It works accordingly with their region algo definition (hence, for instance, it returns FALSE for the first OP point).

Sergey Alexandrovich Kryukov 6-Jul-11 17:37pm

Sorry, I did not try it. That's why I said only "The solution is supposed to be implemented". Are you saying it does not work as expected? This disturbs me. I already saw two earlier reports of incorrect point-set operations with region in Windows.

A lousy Windows code? What do you think?

Thank you.
--SA