[2/3] gdiplus: fix GdipPathIterNextMarker behaviour on path without markers. fix tests.

[Michael Karcher]

Am Sonntag, den 13.07.2008, 13:51 +0400 schrieb Nikolay Sivov:
> Reece Dunn wrote:
> > Is this the way it works on Windows? You have 4 basic cases:
> > a.  exactly equal;
> > b.  equal + epsilon for rounding/representation errors;
> > c.  equal - epsilon for rounding/representation errors;
> > d.  not equal.
I don't see a difference between b and c. A matrix has four elements.
Some might be slightly too big and others slightly too small. Would this
make b or c? It's just one case: equal +/- epsilon.

> Exactly, so I plan to test this deeper on native. As I see now after 
> some basic tests, most probably native call does a bitwise comparison 
> cause the result is affected by a smallest matrix element change (e.g. 
> 1.0000000f -> 1.0000001f). But I'll test it more.
1+FLT_EPSILON is the value to use for the "smallest matrix element
change". But sour test already is a very strong indication that it
really is strict equality test. FLT_EPSILON is in <float.h>

> Reece, do you have any 
> suggestions to do some bounds test (I mean test is call affected by 
> representation or round errors or not)?
Another thing you could test is: Is the Matrix (1,FLT_EPSILON/2,0,1)
equal to (1,0,0,1). A matrix like te former can easily be the result of
multiplying a matrix by its inverse matrix (if represented as floating
point numbers). One might expect that the product of a matrix and its
inverse always is the identity matrix.

> >> 2. GdipIsMatrixInvertible which contains a check for 'not above zero'
> >> determinant.
> >>     
> > a.  det(A) > 0
> > b.  det(A) == 0
> > c.  det(A) < 0
> >
> > So does Windows match your assumption for the answer to these (a ==>
> > yes; b, c ==> no).   
> What do you mean here by (c => no)? Native allows negative det and I do so.
He implied it form your "not above zero" quote. A hint how I would
implement that test (if it turns out that a strict comparison to 0 is
not the right result): Do not calculate the sum of a*d and -b*c, but
compare these numbers in a sensible way.

Regards,
  Michael Karcher