Two questions. 1) Am I transforming this rectangle by the matrix correctly?
2) Is there an easier way?
The Content Stream
q
1.1342565 0 0 1.199793 0 0 cm
41.906 201.086 0.936 -69.448 re
f
Q
float rectXPt = ((PdfNumber)operands[0]).FloatValue; //41.906
float rectYPt =
Yeah, found it. I’ll get right on it as soon as my machine has finished
rebooting (my entire machine froze, so I think I beat your app crashes :-) ).
From: FDnC Red fdnc...@yahoo.commailto:fdnc...@yahoo.com
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Sorry, just replied to the wrong mail.
From: MacGregor, Duncan MacGregor
duncan.macgre...@ge.commailto:duncan.macgre...@ge.com
Date: Thursday, 18 September 2014 14:48
To: FDnC Red fdnc...@yahoo.commailto:fdnc...@yahoo.com, Post all your
questions about iText here
Darren,
FDnC Red wrote
1) Am I transforming this rectangle by the matrix correctly?
Well, you are not transforming the rectangle but merely the two corner
points on one diagonal (rectXPt, rectYPt) and (tempEndXPt, tempEndYPt).
In case you merely want to consider transformations like the one in
Thanks Michael. I guess another way of saying what you said is that the points
returned from the content stream are Cartesian points. I keep forgetting PDF
puts 0,0 in the BL instead of TL.
So in order to calculate all four points for the rectangle I should be able to
do something like this,
Yes - all geometry in PDF is Cartesian. It makes more sense that way :).
Leonard
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