On 3/30/11, Mathieu Bouchard matju@artengine.ca wrote:
I have no use for that. The code I write nowadays does not work on anything else than the GNU compilers. I also had good knowledge of 16-bit DOS stuff back then, but chose to try to forget it.
The only thing I want to do relative to MS compilers, is know how to call GEM functions from GridFlow, where GEM is compiled by VC++, and GridFlow is compiled by MinGW. This is (probably) required to get [#from_pix], [#to_pix] and [gemdead] to work again on Windows.
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
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...
Indeed the GEM exports are declared like this # define GEM_EXPORT __declspec(dllexport)
The thing to do is use vc++ to compile to assembly language an export function. Then get gcc to compile the corresponding import function.
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
I am unsufficiently sophisticated for that kind of technology. Instead, I settled for using Linux.
Ha Ha likewise on the unsufficiently.
What do you mean by « not correct » ?
newx = ((x * x) - (y * y) - (z *z)) +k;
for newx=0, x²-y²-z² = -k, hyberboloïd equation (revolution of hyperbola around x axis)
newy = ((y *x) + (x *y)) +l;
y*x = x*y, so for newy=0, 2*x*y = -l, another hyperbola formula (diagonally), but this one has translation symmetry instead, along z axis (because z is not used in this formula)
newz = ((z * x) + (x * z)) +m;
similar thing. but those * above are not products of floats, tell me right away.
Hopefully the compiler understood.
I have also used the quaternion with W removed.
Here are my notes on the matter.
**********************START OF HISTORY 1996 Formula:
newx=((x*x)-(y*y)-(z*z))+k; newy=((x*y)+(x*y))+l; newz=(x*z)+(x*z)+m;
From my green folder that has ENG 111 on the front
(guess this explains my bad grammar)
A*B=C where A:[x1,y1,z1] B:[x2,y2,z2] C:[x,y,z]
[x1,y1,z1] * [x2,y2,z2] = [x,y,z]
x = (x1*x2) - (y1*y2) - (z1*z2) y = (x1*y2) + (y1*x2) z = (x1*z2) + (z1*x2)
Correct only if A and B are the same Or if x1,y1,z1 are the same and x2,y2,z2 are the same and a few other cases otherwize magnitude of x,y,z is slightly different than A * B ********************************************END OF HISTORY
START OF CODE SNIPET********************************************************************* void mandelbrot::iterate(void) {
int vtxi;
long int p,q,r; int n; double k,l,x,y,z,newx,newy,newz; double m;
vtxi=0;
for(r=1; r<depth; r++) { for (p=1; p<width; p++) { for (q=1; q<height; q++) { if(julia==0){ k=(double)x1+(x2-x1)*p /(width); l=(double)y1+(y2-y1)*q /(height); m=(double)z1+(z2-z1)*r /(depth);
x=realp;
y=imagp;
z=kapap;
}else{
x=(double)x1+(x2-x1)*p /(width);
y=(double)y1+(y2-y1)*q /(height);
z=(double)z1+(z2-z1)*r /(depth);
k=realp;
l=imagp;
m=kapap;
}
for (n=1; n<numits; n++)
{
//newx=x*x-y*y+k;
//newy=2*x*y+l;
//6-9-2002 equation
newx=((y*z)+(z*y))+k;
newy=((x*z)+(z*x))+l;
newz=((x*y)+(y*x))+m;
/*
* 2000 equation
*/
/*
* newx=((x*x)-(y*y)-(z*z))+k;
* newy=((y*x)-(z*y)+(x*z))+l;
* newz=((z*x)+(x*y)-(y*z))+m;
*/
/*
* 1996 equation - see history
*/
/*
* newx=((x*x)-(y*y)-(z*z))+k;
* newy=((x*y)+(x*y))+l;
* newz=(x*z)+(x*z)+m;
*/
x=newx;
y=newy;
z=newz;
//putpixel(surface, p+startx-1,q+starty-1,n);
//putpixel(surface,(int)(x*10+160),(int)(y*10+100),(unsigned char)n);
if (x*x+y*y+z*z>magnitude)
{
if((n>lowcolor)&&(n<highcolor))
{
vtx[vtxi].x=(p/scaledivisor);
vtx[vtxi].y=(q/scaledivisor);
vtx[vtxi].z=(r/scaledivisor);
vtx[vtxi].r=cMap[n].r;
vtx[vtxi].g=cMap[n].g;
vtx[vtxi].b=cMap[n].b;
if(alphamode==0)
{
vtx[vtxi].a=16;
}else{
vtx[vtxi].a=(unsigned
char)((0.9375*n)+16);//((256/(256-16)) * n)+16 } vtx[vtxi].n=n; vtxi++; }
// putpixel(surface, p+startx-1,q+starty-1,n);
//SDL_UpdateRect(screen, p+startx-1,
q+starty-1, 1, 1);
n=numits;
}//mag check
}//n
}//q
}//p
}//r nvertexes=vtxi; }
*******************************************************************END OF CODE SNIPPET
Complex numbers have 2 dimensions. The logic that originally led to finding them doesn't work for more dimensions. Looking at complex numbers in different ways (as modified vectors or as modified polynomials) leads to other structures that are interesting, in 2 or 4 or more dimensions, but fail to be as nice as complex numbers are. Complex numbers are very, very similar to real numbers.
I don't know of any 3-dimensional number system that is sufficiently similar to complex numbers to be comparable.
a: [x1,y1,z1] b: [x2,y2,z2] a*b=(y1 * z2 - z1 * y2)i + (x1 * z2 - z1 * x2)j + (x1 * y2 - y1 * x2)k
this is almost like cross-product, but the j part has the wrong sign. anyway. cross product is weird because a*a = 0, and this is also the case for your cross-product-like operator.
I have visited this perplexing search for the 3 dimensional equivalent of j or i sporadically. Elusive it is.
I get the following error message :
:-\ Cette vidéo est privée. Opération impossible
| Mathieu Bouchard ---- tél: +1.514.383.3801 ---- Villeray, Montréal, QC
Appologies. You should be able to view it now. I had it set to private instead of hidden.
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...
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@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.
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 ?
| Mathieu Bouchard ---- tél: +1.514.383.3801 ---- Villeray, Montréal, QC
-
Billy Stiltner -
Mathieu Bouchard