Re: [hlcoders] Checking to see if an object/world vector is in a
> visible leaf?
>
> The only reliable way to do this is to find out exactly what leaves the
> projector projects into and then see if any of them is in the PVS. After
> searching the code, I found that there are som
The only reliable way to do this is to find out exactly what leaves the
projector projects into and then see if any of them is in the PVS. After
searching the code, I found that there are some lists of all shadows in
the game, and that each leaf has a handle per shadow (projectors?).
Try search
Problem with this: the light won't draw if it is out of PVS, even if
surfaces it shines on are in.
Testing the PVS of the entity and the end of a traceline down AbsAngles
to a maximum of m_flFarZ /should/ overcome that...most of the time
Kyle Gospodnetich wrote:
> HLCoders wrote:
>
>> After
HLCoders wrote:
> After a very quick search in some header files, I think these two
> functions might interest you.
>
> bool IVRenderView::AreAnyLeavesVisible( int *leafList, int nLeaves );
> int IEngineTrace::GetLeafContainingPoint( const Vector &ptTest ) = 0;
>
> I haven't tested them, but if th
HLCoders wrote:
> After a very quick search in some header files, I think these two
> functions might interest you.
>
> bool IVRenderView::AreAnyLeavesVisible( int *leafList, int nLeaves );
> int IEngineTrace::GetLeafContainingPoint( const Vector &ptTest ) = 0;
>
> I haven't tested them, but if th
After a very quick search in some header files, I think these two
functions might interest you.
bool IVRenderView::AreAnyLeavesVisible( int *leafList, int nLeaves );
int IEngineTrace::GetLeafContainingPoint( const Vector &ptTest ) = 0;
I haven't tested them, but if they work like I think they do
Is it possible to check if an object or vector is within a visible
(currently being rendered) visleaf? I need to occlude an entity that
isn't normally occluded, and this is the best way I can think of doing
it. Any help is much appreciated. I can't find any examples of this in
the code.
-Kyle
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