I believe there is something wrong with Matrix.RotateAt

The following example shows the problem.

Form needs a picture box and a context menu.

PictureBox1

ContextMenuStrip1

ToolStripMenuItemPlus20

ToolStripMenuItemMinus20

ToolStripMenuItemNewImage

When the image is rotated it loses the rightmost pixels



I'd be interested in your comments





Imports System.Drawing.Drawing2D

Public Class Form1

Private Const NumbOfPixels As Integer = 99

Private Sub Rotate(ByVal angle As Single)

Dim TmpBitmap As New Drawing.Bitmap(CInt(PictureBox1.Image.Width),
CInt(PictureBox1.Image.Height))

Dim TmpBitmapGraphics As Graphics = Graphics.FromImage(TmpBitmap)

Dim mx As Matrix = New Matrix()

mx.RotateAt(angle, New PointF(CSng((PictureBox1.Image.Width - 1) / 2),
CSng((PictureBox1.Image.Height - 1) / 2))) 'Rotates CW about the specified
point

TmpBitmapGraphics.Transform = mx

TmpBitmapGraphics.DrawImage(PictureBox1.Image, 0, 0,
PictureBox1.Image.Width, PictureBox1.Image.Height)

TmpBitmap.SetPixel(PictureBox1.Image.Width - 1, PictureBox1.Image.Height \
2, Color.Green)

TmpBitmapGraphics.Dispose()

PictureBox1.Image.Dispose()

PictureBox1.Image = TmpBitmap

End Sub

Private Sub ToolStripMenuItemPlus20_Click(ByVal sender As Object, ByVal e As
System.EventArgs) Handles ToolStripMenuItemPlus20.Click

Rotate(90)

End Sub

Private Sub ToolStripMenuItemMinus20_Click(ByVal sender As Object, ByVal e
As System.EventArgs) Handles ToolStripMenuItemMinus20.Click

Rotate(-90)

End Sub

Private Sub ToolStripMenuItemNewImage_Click(ByVal sender As Object, ByVal e
As System.EventArgs) Handles ToolStripMenuItemNewImage.Click

If PictureBox1.Image IsNot Nothing Then PictureBox1.Image.Dispose()

PictureBox1.Image = New Bitmap(NumbOfPixels, NumbOfPixels)

Dim g As Graphics = Graphics.FromImage(PictureBox1.Image)

g.Clear(Color.Yellow)

g.DrawRectangle(New Pen(Color.Black, 1), 0, 0, NumbOfPixels - 1,
NumbOfPixels - 1) 'NumbOfPixels pixels wide rectangle even tho width and
height equal NumbOfPixels-1

g.DrawRectangle(New Pen(Color.Red, 1), 1, 1, NumbOfPixels - 3,
NumbOfPixels - 3)

g.Dispose()

End Sub

End Class

" active" <activeNOSPAM (AT) a-znet (DOT) com> wrote

Why are you subtracting 1?

Michael

Doesn't affect this problem. I tried it without first.

Just want to see if it changed anything.

I shouldn't have left it in.

I also tried different widths. 100 works the same as 99.

Thanks



"Michael C" <nospam (AT) nospam (DOT) com> wrote

" active" <activeNOSPAM (AT) a-znet (DOT) com> wrote

It worked perfectly on mine. For a 100x100 bitmap I need to rotate around
50,50. For a 101x101 bitmap I can rotate around 50.5,50.5 to get the correct
result. Try putting the PointF into a var and checking to make sure it has
the correct values.

Michael

"Michael C" <nospam (AT) nospam (DOT) com> wrote

I used QuickWatch and PointF does have the values you mentioned above. Tried
both 100 ,101.

Did you notice that in the picture I sent you the black line on the right
side is missing?

When you say it worked perfectly, do you mean the right side has both the
red and the black lines?


Thanks

" active" <activeNOSPAM (AT) a-znet (DOT) com> wrote

I missed the problem the first few times I read this. First I should say
that using the extra set of brackets around Image.Width is unnecessary and
makes the code more confusing (it confused me for sure!). Anyway, the
problem is you are dividing by 2 and then converting to single. This will do
an integer divide, hence dropping the decimal part. Convert to width to
single first and then divide by 2.

Also, are you using PixelOffsetMode.Half?

Michael

"Michael C" <nospam (AT) nospam (DOT) com> wrote

You do C#, riight?
In VB we have "/" which results in a Double and "\" which does integer
devision.
The extra brackets are probably left over from when I tried subtracting 1.



No I was not, and when I set it the problem went away.
Isn't it strange that it affected the right side but not the botton.
Since the example was completely simmetrical one might expect an operation
would not treat X any differently then Y.

Is PixelOffsetMode.Half something that I should use regulary?

What about SmoothingMode. Should I be setting this before operations like
"Rotate"


Thanks a lot for sticking throught this for me!
I really appreciate it.

" active" <activeNOSPAM (AT) a-znet (DOT) com> wrote

That's true although I came from Vb6.

That was true for VB6 but not for vb.net. In vb.net if both sides of the
division are integers then it will do an integer divide (hence it is
irrelevant whether you use / or \). If you try it with 11 and 2 the result
will be 5 when you really want 5.5.

Possibly a combination of PixelOffsetMode and rounding error?

I'd have to go back through my code and check where I have and haven't used
it but I would say that it is something that should regularly be used for
painting bitmaps to the screen. I don't see any reason myself for not using
it.

This doesn't appear to apply to painting bitmaps.

No problems, I've learn a couple things myself. :-)

Michael

"Michael C" <nospam (AT) nospam (DOT) com> wrote

I believe
5 \ 2 will return type integer equal to 2 but

Dim n As Double = 5

Dim d As Double = 2

n / d will return a type double equal to 2.5

at least in VB.Net
========================

Just checked it out and the above is correct.
Dim a As Double = 5 \ 2 ' returns 2 but

Dim n As Double = 5

Dim d As Double = 2

Dim q As Double = n / d 'returns 2.5


also
Dim qq As Double = n \ d

If Option Strict is On will produce a compiler error since n and d are
converted to integers before the division.

I always have to look it up when it's important but if rounding vs.
truncation is important cint(), fix() and \ do not all behave it the same.


Thanks again for the help

" active" <activeNOSPAM (AT) a-znet (DOT) com> wrote

That's true but my point was if you try this

dim n as integer =5
dim d as integer = 2
n / d will return 2 as an integer.
this is the same result if you use n \ d which is interesting because it
means with option strict turned on there is no real use for the \.