[PD] 3d fractals was Re: access to pd table from another application

[Mathieu Bouchard]

On Thu, 31 Mar 2011, Billy Stiltner wrote:
> Here is the MingW docs for how to do that but the link to reimp
> appears to be dead.
> http://www.mingw.org/wiki/MSVC_and_MinGW_DLLs

I think that I've read that already, and that I tried to do that, but it 
wouldn't handle C++-specific things, such as encoding the argument types 
into the symbols, encoding class name into symbols, namespaces, 
exceptions, and such.

> If you could reverse engineer this it might work. Appears the stack is
> reversed and the segments are in diferent chunks.
> http://www.mmowned.com/forums/world-of-warcraft/bots-programs/memory-editing/281008-gcc-thiscall-calling-convention-linux-win32-mingw.html

I can't even get the symbols to be found, thus it's not worth trying %ecx 
tricks and stuff.

> The thing to do is use vc++ to compile to assembly language an  export function.
> Then get gcc to compile the corresponding import function.

I always expected not to have to install MSVC.

> Look at the differences and make some inline assembly code to compensate.
> Here is a reference from the gcc vcc+ veiwpoint.
> http://wyw.dcweb.cn/stdcall.htm

It's much of the same : even though it says «C++» in the title, it doesn't 
say how to deal with C++-specific name mangling such as :

   ??4imageStruct@@QAEAAU0 at ABU0@@Z

and because they contain «?» and «@», they can't be used directly in C++, 
and stdcall didn't do the job of mangling them VC++-style, and don't know 
how to get gridflow/src/Gem.def's aliases to work.

> Hopefully the compiler understood.

I hadn't looked at the code, only at what you had pasted in the email.

> I have visited this perplexing search for the 3 dimensional equivalent 
> of j or i sporadically. Elusive it is.

I'm pretty sure that there are already proofs that the nicest number 
systems made using a cartesian product of real numbers, are made with two 
or four dimensions, and everything else is weird. I mean, if you expect 
that a*(b*c) = (a*b)*c where a,b,c are three vectors, that's quite a 
difficult property to satisfy, and if you expect that a*a=0 implies a=0, 
it's even harder to satisfy, etc.

You could read those pages in detail (I haven't) :
   http://en.wikipedia.org/wiki/Hypercomplex_number
   http://en.wikipedia.org/wiki/Algebra_over_a_field

> Appologies. You should be able to view it now. I had it set to private 
> instead of hidden.
>
> http://www.youtube.com/watch?v=xZUTn-rie8w

I don't understand any of it, but I can certify that a 5-dimensional cube 
has 80 faces. It's in the big table in this article :

   http://en.wikipedia.org/wiki/Hypercube

... which, incidentally, is a very close variant of Pascal's triangle.

so, where does your video come from ?

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| Mathieu Bouchard ---- tél: +1.514.383.3801 ---- Villeray, Montréal, QC