Fast affine texture mapping (fatmap.txt) ---------------------------------------- by Mats Byggmastar a.k.a. MRI / Doomsday mri@penti.sit.fi 8 Jul. 1996 Jakobstad, Finland 19 Jun. 1996 Espoo, Finland Read this today, it might be obsolete tomorrow. Feel free to upload this document to wherever you find appropriate, as long as you keep it in it's original, unmodified form. This is free information, you may not charge anything for it. Table of contents ----------------- 1. About this document 2. Disclaimer 3. Definition of terms 4. Assume the following 5. The slow method 6. A faster method 7. General structure of the texture mapping function 8. Equations for the constant deltas 9. Traditional inner loops 10. Memories from the past 11. Selfmodifying code 12. Unrolled and selfmodifying inner loops 13. Table lookup inner loops 14. Problems with precalculated runs 15. Pre-stepping texture coordinates 16. Special case code 17. Clipping 18. Clipping using matrices 19. Writing a byte, word, dword and other weird things 20. The data cache 21. The code cache 22. Some pairing rules 23. Pipeline delays 24. The time stamp counter 25. Branch prediction 26. Reference 27. Where to go from here 28. Credits and greetings 1. About this document ----------------------- This document tries to describe how to make fast affine texture mapping. The document describes both the general structure as well as the more critical parts such as inner loops. The information is aimed at both beginners and also at people who maybe already have a working texture mapper but are looking for more speed. The goal was to make a good document that would be useful today, not already be obsolete. So I'm giving you the best information I can possibly come up with. You don't get the information for free though. You will have to invest some of your own effort and actually learn what's going on in the inner loops and select the parts that will be most suitable for you. The information is based on my own work and findings but also on information found on the net, especially articles posted to the newsgroup comp.graphics.algorithms and on ideas given to me by other coders. IRC channel #coders is usually a good place to get new ideas and help. Many of the coders there are willing to share ideas and answer decent questions. I am not claiming that the methods described here are THE fastest methods of doing texture mapping. But these methods are what coders are using today and they ARE fast. To get the most out of this document you should have a good understanding of 386+ Assembly and C. The asm code and optimizations are aimed especially for the Intel Pentium CPU. Note that the C code given is only meant as some sort of pseudo code. C is most of the time much easier to read that asm. For your information I have the whole texture mapping function in asm. This is overkill, I know, but this way I get full control over the optimization. In C I can only _hope_ that the compiler makes the best code possible. I'm certain that a human brain still is better to optimize code than a compiler. I do not say this because I'm a true asm-freak. In fact, I had programmed C for a year before even considering learning asm. I should say that I do not have a masters degree in computer graphics. I'm merely a 24 year old computer and telecom engineer (B.Sc.) that found interest in this area. I have never taken any computer graphics related course in school so if you think I misuses some expressions or terms, or even leave out some expressions or terms where I should use them, you might very well be right and I wrong. Also I have to confess that I haven't read Chris Hecker's articles in Game Developer magazine (http://www.gdmag.com). People tell me that they are good. You should probably take a look at them also. 2. Disclaimer -------------- Some parts of the technical discussion at the end of the document might not be 100% accurate as of the actual hardware in the Pentium. But from a programmers point of view the guidelines given should apply anyway. When I state that a inner loop is e.g. 5 clock ticks per pixel, this don't mean that it will actually run at 5 clock ticks. This is just a theoretical minimum when I assume that the instructions pair as expected and there are no cache misses and no delays writing to RAM. 3. Definition of terms ----------------------- Just so there won't be any confusion: triangle side Each triangle has two sides, the left side and the right side. triangle edge These makes up the outline of the triangle. Usually one interpolates variables along the triangle edges, both on the left and the right side of the triangle. triangle section A triangle is always made up of 3 sections. These are straight lines which makes up the triangle edges. When interpolating along the triangle edges we must calculate deltas for each of the 3 sections. triangle x The current x value of a triangle edge on the screen. There are two triangle x, one on the left side and one on the right side of the triangle. Between these is the current scanline. u and v The x and y components in the bitmap. dudx and dvdx Our constant deltas for u and v, du/dx and dv/dx. (constant texture gradients for u and v) 4. Assume the following ------------------------ We are only drawing triangles. (This is no problem to me as 3D Studio only uses triangles anyway.) Well actually it doesn't have to be triangles, this also works on other types of polygons as long as the texture gradients are constant for the whole polygon surface. You agree that a fractional part of 8 bit is enough for interpolating u and v on the scanlines of the triangle. Well actually a 16 bit fractional part is better but inner loops are usually much simpler to do if we only use 8 bits. Bitmaps always has a width of 256 pixels and a maximum height of 256 pixels. In some of the inner loops we must also assume that the bitmaps are aligned on 64k. The CPU is a Pentium and we are in 32 bit protected mode and flat memory model (TASM+WATCOM+PMODE/W). 5. The slow method ------------------- The slow method of doing texture mapping is to interpolate u and v (and triangle x) on both the left and right side of the triangle and then calculate du/dx and dv/dx for each scanline. Then interpolate u and v when drawing the scanline. To make this even slower you could first interpolate the u and v (and triangle x) on both sides of the triangle and store the values in a edge buffer. Then pick the values from the edge buffer and draw each scanline. 6. A faster method ------------------- Don't use a edge buffer as described in the slow method above. Calculate the edge deltas when you need them and interpolate along the edges at the same time you draw each scanline. It's just as simple to do it this way and a lot faster. One important thing you should realize is that when texture mapping a triangle (or any type of polygon that has constant texture gradients), you are interpolating two variables, u and v, whose deltas are constant over the whole triangle. I repeat, the deltas are constant for the whole triangle. Make sure you understand this because this is the key to fast texture mapping (or any other type of linear shading for that matter). I guess that the correct term isn't constant deltas, rather constant gradients, but I like the term delta better. Because the deltas (delta u and delta v) are constant, we only need to calculate them once for the whole triangle. No need to calculate them for each scanline. Also when interpolating u and v along the edges of the triangle you only need to interpolate u and v on one side of the triangle. Triangle x must be interpolated on both sides. 7. General structure of the texture mapping function ----------------------------------------------------- Here is the general structure of my texture mapping function. If you have Watcom C/C++ you can compile it as is. Just initialize VGA mode 0x13 and call it. I didn't want to include the clipping code as it only would make it more difficult to read. No kind of pre-stepping or any other type of compensation is presented here, this is just the bare bones of the function. It might look big (?) but it is pretty damn simple and efficient if I may say so myself. You should call the function by passing a pointer to an array of 3 vertex structures and a pointer to the bitmap. extern char myimage[]; // 256x256 256 color bitmap vertex array[3]; // fill in the values for each vertex in the array here DrawTextureTriangle(array, myimage); Note that the function doesn't move the vertex data to some local variables, it uses pointers to each of the structures instead. This makes it extremely simple to later on add more variables in the vertex structure which you will be doing in the case of an environment-bump or Phong-texture-bump mapper. The same function structure can still be used, just add a few variables to the vertex structure, calculate 2 more deltas, interpolate 2 more variables along the left side and make a new inner loop. // This is the only Watcom C/C++ specific part of the function. These // instructions take a 26:6 bit fixed point number and converts it // to 32:32 bit. Then divides it with another 16:16 bit fixed point // number. The result is 16:16 bit. This must be done in asm where we // can do 64/32 bit divides. int shl10idiv(int x, int y); #pragma aux shl10idiv = \ " mov edx, eax "\ " shl eax, 10 "\ " sar edx, 22 "\ " idiv ebx "\ parm [eax] [ebx] \ modify exact [eax edx] \ value [eax] // sizeof(int) is 4 struct vertex { int x,y; // screen coordinates (integers) int u,v; // vertex u,v (26:6 bit fixed point) }; static vertex * left_array[3], * right_array[3]; static int left_section, right_section; static int left_section_height, right_section_height; static int dudx, dvdx; static int left_u, delta_left_u, left_v, delta_left_v; static int left_x, delta_left_x, right_x, delta_right_x; inline int RightSection(void) { vertex * v1 = right_array[ right_section ]; vertex * v2 = right_array[ right_section-1 ]; int height = v2->y - v1->y; if(height == 0) return 0; // Calculate the deltas along this section delta_right_x = ((v2->x - v1->x) << 16) / height; right_x = v1->x << 16; right_section_height = height; return height; // return the height of this section } inline int LeftSection(void) { vertex * v1 = left_array[ left_section ]; vertex * v2 = left_array[ left_section-1 ]; int height = v2->y - v1->y; if(height == 0) return 0; // Calculate the deltas along this section delta_left_x = ((v2->x - v1->x) << 16) / height; left_x = v1->x << 16; delta_left_u = ((v2->u - v1->u) << 10) / height; left_u = v1->u << 10; delta_left_v = ((v2->v - v1->v) << 10) / height; left_v = v1->v << 10; left_section_height = height; return height; // return the height of this section } void DrawTextureTriangle(vertex * vtx, char * bitmap) { vertex * v1 = vtx; vertex * v2 = vtx+1; vertex * v3 = vtx+2; // Sort the triangle so that v1 points to the topmost, v2 to the // middle and v3 to the bottom vertex. if(v1->y > v2->y) { vertex * v = v1; v1 = v2; v2 = v; } if(v1->y > v3->y) { vertex * v = v1; v1 = v3; v3 = v; } if(v2->y > v3->y) { vertex * v = v2; v2 = v3; v3 = v; } // We start out by calculating the length of the longest scanline. int height = v3->y - v1->y; if(height == 0) return; int temp = ((v2->y - v1->y) << 16) / height; int longest = temp * (v3->x - v1->x) + ((v1->x - v2->x) << 16); if(longest == 0) return; // Now that we have the length of the longest scanline we can use that // to tell us which is left and which is the right side of the triangle. if(longest < 0) { // If longest is neg. we have the middle vertex on the right side. // Store the pointers for the right and left edge of the triangle. right_array[0] = v3; right_array[1] = v2; right_array[2] = v1; right_section = 2; left_array[0] = v3; left_array[1] = v1; left_section = 1; // Calculate initial left and right parameters if(LeftSection() <= 0) return; if(RightSection() <= 0) { // The first right section had zero height. Use the next section. right_section--; if(RightSection() <= 0) return; } // Ugly compensation so that the dudx,dvdx divides won't overflow // if the longest scanline is very short. if(longest > -0x1000) longest = -0x1000; } else { // If longest is pos. we have the middle vertex on the left side. // Store the pointers for the left and right edge of the triangle. left_array[0] = v3; left_array[1] = v2; left_array[2] = v1; left_section = 2; right_array[0] = v3; right_array[1] = v1; right_section = 1; // Calculate initial right and left parameters if(RightSection() <= 0) return; if(LeftSection() <= 0) { // The first left section had zero height. Use the next section. left_section--; if(LeftSection() <= 0) return; } // Ugly compensation so that the dudx,dvdx divides won't overflow // if the longest scanline is very short. if(longest < 0x1000) longest = 0x1000; } // Now we calculate the constant deltas for u and v (dudx, dvdx) int dudx = shl10idiv(temp*(v3->u - v1->u)+((v1->u - v2->u)<<16),longest); int dvdx = shl10idiv(temp*(v3->v - v1->v)+((v1->v - v2->v)<<16),longest); char * destptr = (char *) (v1->y * 320 + 0xa0000); // If you are using a table lookup inner loop you should setup the // lookup table here. // Here starts the outer loop (for each scanline) for(;;) { int x1 = left_x >> 16; int width = (right_x >> 16) - x1; if(width > 0) { // This is the inner loop setup and the actual inner loop. // If you keep everything else in C that's up to you but at // least remove this inner loop in C and insert some of // the Assembly versions. char * dest = destptr + x1; int u = left_u >> 8; int v = left_v >> 8; int du = dudx >> 8; int dv = dvdx >> 8; // Watcom C/C++ 10.0 can't get this inner loop any tighter // than about 10-12 clock ticks. do { *dest++ = bitmap[ (v & 0xff00) + ((u & 0xff00) >> 8) ]; u += du; v += dv; } while(--width); } destptr += 320; // Interpolate along the left edge of the triangle if(--left_section_height <= 0) // At the bottom of this section? { if(--left_section <= 0) // All sections done? return; if(LeftSection() <= 0) // Nope, do the last section return; } else { left_x += delta_left_x; left_u += delta_left_u; left_v += delta_left_v; } // Interpolate along the right edge of the triangle if(--right_section_height <= 0) // At the bottom of this section? { if(--right_section <= 0) // All sections done? return; if(RightSection() <= 0) // Nope, do the last section return; } else { right_x += delta_right_x; } } } 8. Equations for the constant deltas ------------------------------------- Sort the vertices in the triangle so that the topmost vertex is known as x1,y1 and the bottom vertex is known as x3,y3. Like the drawing below. x1,y1 p1 / / / / / / / / / / / x2,y2 / / p2 /_____________/ \ width / \ / \ / \ / \/ x3,y3 p3 xn,yn - x,y screen coordinates at vertex n (integers) pn - Value of variable at vertex n to calculate the constant delta for. Note that this variable is assumed to have a 6 bit fractional part (26:6 bit fixed point). width - Width of the longest scanline in the triangle The reason why I have p as a 26:6 bit fixed point and not 16:16 or 24:8 bit fixed point is just for being able to store u and v with a little higher precision in the 3D structure and still use only words to save space. Sorting 3 vertices is no more that 3 compares. Another thing: Don't load all x,y,u and v values of the vertices into registers. Use pointers to the vertex structures instead. This will also make it easier when you later on implement your Phong-texture-bump mapper. Something like this: ; EDX -> vertex 1 ; ESI -> vertex 2 ; EDI -> vertex 3 mov EAX, [EDX+vertex_y] cmp EAX, [ESI+vertex_y] jle short @@sorta xchg EDX, ESI ; swap v1 - v2 @@sorta: mov EAX, [EDX+vertex_y] cmp EAX, [EDI+vertex_y] jle short @@sortb xchg EDX, EDI ; swap v1 - v3 @@sortb: mov EAX, [ESI+vertex_y] cmp EAX, [EDI+vertex_y] jle short @@sortc xchg ESI, EDI ; swap v2 - v3 @@sortc: ; EDX -> topmost vertex ; ESI -> middle vertex ; EDI -> bottom vertex The following two equations needs only be calculated once for all the constant deltas in the triangle. Skip the triangle if y3 == y1, i.e. if the triangle has zero height. The width can be either positive or negative depending on which side the x2,y2 vertex is. This will be useful information when sorting out which is left and which is the right side of the triangle. (y2-y1) << 16 temp = -------------- y3-y1 width = temp * (x3-x1) + ((x1-x2) << 16) This will give you temp and width as 16:16 bit fixed point. The equation below is used to calculate the delta for a variable that should be interpolated over the triangle, e.g. texture u. Beware of the denominator in this equation! Make sure it won't cause divide overflow in case the width is less than one pixel. (Remember that width is a 16:16 bit fixed point number.) Note that shift by 10 in the equation. This is because p1,p2,p3 has a 6 bit fractional part. The resulting delta p is a 16:16 bit number. Note that this divide should be done in asm where we can do 64/32 bit divides. ( temp * (p3-p1) + ((p1-p2) << 16) ) << 10 delta p = -------------------------------------------- width So for a texture mapper where we have 2 variables (u,v) to interpolate over the triangle, we have a total of 3 divs and 3 muls to calculate dudx and dvdx. Here is another equation that can be used to calculate the deltas with. It was posted to the newsgroup comp.graphic.algorithm by Mark Pursey. There is a cleaner way, which doesn't rely on finding the widest line: A-B-C: a triangle with screen x and y components, as well as t, a value which could represent lightning, texture coordinates etc. The following equation gives you the increment for t per horizontal pixel: (At-Ct)*(By-Cy) - (Bt-Ct)*(Ay-Cy) dt/dx = --------------------------------- (Ax-Cx)*(By-Cy) - (Bx-Cx)*(Ay-Cy) I've been told that this is the correct way to calculate the deltas (or constant texture gradients). This might very well be true but the other equations gives me good results and the length of the longest scanline for free. In this equation the denominator is reusable for both u and v. This makes a total of 6 muls and 2 divs. Remember to add the necessary shifts if you do this in fixed point. 9. Traditional inner loops --------------------------- So assuming you have come so far that you have the triangle sorted, the constant deltas calculated, the u and v deltas on the left side calculated, deltas for triangle x calculated for both sides, and you are actually interpolating those values for each scanline, we come to the very core of the texture mapper, the inner loop. I'll first present a few traditional inner loops that interpolates u and v while plotting the scanline. These loops are simple, fast and works very well. The loops assume the following: ebx = ptr to bitmap aligned on 64k. (the low 16 bits zero) edi = ptr to first destination pixel to plot in this scanline ebp = width of scanline (loop counter) left_u = current u on the left edge of the triangle (16:16 bit fixed point) left_v = current v on the left edge of the triangle (16:16 bit fixed point) du = our constant delta u (24:8 bit fixed point) dv = our constant delta v (24:8 bit fixed point) The first loop interpolates the u and v in two 32 bit registers (ecx, edx). We are one register short here so we use the dudx variable directly in the inner loop. This loop should run at 6 ticks per pixel. eax is not used for anything else than holding the pixel so we could unroll this loop to plot a word or dword at a time. mov ecx, [left_u] ; current u mov edx, [left_v] ; current u shr ecx, 8 ; make them 28:8 bit fixed point shr edx, 8 mov bl, ch ; make ebx point to the first textel mov bh, dh mov esi, [du] @@inner: add edx, [dv] ; update v add ecx, esi ; update u mov al, [ebx] ; get pixel from aligned texture map mov bl, ch mov [edi], al ; plot pixel mov bh, dh inc edi dec ebp jnz @@inner Just to show that it is also possible to directly interpolate u and v in ebx I'll present this one that uses the carry flag to add the "overflow" from the fractional part to the whole part of u and v. mov cl, byte ptr [left_u+1] ; fractional part of current u mov ch, byte ptr [left_v+1] ; fractional part of current v mov dl, byte ptr [du] ; fractional part of delta u mov dh, byte ptr [dv] ; fractional part of delta v mov bl, byte ptr [left_u+2] ; whole part of current u mov bh, byte ptr [left_v+2] ; whole part of current v @@inner: mov al, [ebx] ; get pixel from aligned texture map add cl, dl ; update fractional part of u adc bl, byte ptr [du+1] ; + whole part of dudx (+carry) add ch, dh ; update fractional part of v adc bh, byte ptr [dv+1] ; + whole part of dvdx (+carry) mov [edi], al ; plot pixel inc edi dec ebp jnz @@inner The following loop uses a combination of interpolation in one 32 bit register (ecx) and the carry overflow method. We have just enough registers in this loop that we don't need to use any memory variables. On the other hand this makes it impossible to unroll it and plot a word or dword at a time. Anyway, this version should run at 5 ticks per pixel. mov ecx, [left_u] shr ecx, 8 ; make it 28:8 bit fixed point mov esi, [du] mov dl, byte ptr [dv] ; fractional part of delta v mov dh, byte ptr [left_v+1] ; fractional part of current v mov ah, byte ptr [dv+1] ; whole part of delta v mov bh, byte ptr [left_v+2] ; whole part of current v mov bl, ch @@inner: add ecx, esi ; update u mov al, [ebx] ; get pixel from aligned texture map mov bl, ch add dh, dl ; update fractional part of v adc bh, ah ; + whole part of of delta v (+carry) mov [edi], al ; plot pixel inc edi dec ebp jnz @@inner The loop counter (ebp) in the above loop can be removed if we reorder the registers a bit and plot the scanline from right to left. @@inner: add ecx, ebp mov al, [ebx] mov bl, ch add dh, dl adc dh, ah mov [edi+esi], al dec esi jnz @@inner The loop should now run at 4 clock ticks. I'm sure there are other ways to make these kind of loops but this is what I could come up with. After I wrote the above sentence, there was a post in the newsgroup comp.graphics.algorithms by Sean L. Palmer where he presented the following 4 tick loop: Texture must be aligned on a 64K boundary. Must be 256x256. Only 8 bits used for fractions, means shaky textures. Start at right end of scanline T=texture adr D=dest adr+count (start) E=dest adr (end) X=tex X int (whole part of initial u) x=tex X frac (fractional part of initial u) Y=tex Y int (whole part of initial v) y=tex Y frac (fractional part of initial v) H=tex X step int (whole part of delta u) h=tex X step frac (fractional part of delta u) V=tex Y step int (whole part of delta v) v=tex Y step frac (fractional part of delta v) m=account for borrow for negative Y step, either 0 or 0FFh p=texture pixel edi=DDDD esi=EEEE edx=TTYX eax=000p ebx=x0Yy ecx=hmVv ebp=000H esp= mov dh,bh @@L: mov al,[edx] add ebx,ecx adc edx,ebp dec edi mov dh,bh cmp edi,esi mov [edi],al jne @@L It's not necessary to simulate the loop counter this way. esi is not really used in the loop so we might as well use it as a loop counter and draw the scanline from left to right (the way I like to draw my scanlines). Like this: @@inner: mov al, [edx] add ebx, ecx adc edx, ebp inc edi mov dh, bh dec esi mov [edi], al jnz @@inner Both of these loops uses eax only to hold the pixel so they can be unrolled to plot a word or dword at a time. In fact, by unrolling this loop to plot a dword per turn it might very well beat the table lookup inner loop presented below. By unrolling this loop we can remove 3 instructions, "inc edi", "dec esi" and "jnz @@inner". This will also mean that the loop will become too tight that will lead to AGI delays instead. 10. Memories from the past -------------------------- I as many others, started coding asm in real mode and later on moved to protected mode and flat model. The thing I miss about real mode was the ability to have a pointer in the low 16 bit and a variable in the high 16 bit of a 32 bit register. In flat model we need all 32 bits for the pointer. Sure, one can setup a selector and address the data with only the low 16 bits but all prefix bytes can be seen as a 1 clock tick, nonpairable instruction on the Pentium. So addressing with only 16 bit and using a segment override will give 2 prefix bytes or 2 ticks delay. The following loop in real mode was for a bitmap scaler I once used. We have 4 variables in only 2 registers (edi, ebx). ; ebx = neg(loop counter) : source ptr ; edi = decision variable : destination pointer ; ecx = frac. part of delta : 1 ; edx = 1 : whole part of delta ; the delta is 16:16 bit @@inner: mov al, [bx] mov es:[di], al add edi, ecx ; update fractional part : move dest. pointer adc ebx, edx ; update loop counter : whole step in bmp (+carry) jnc @@inner ; jump if loop counter didn't overflow OK, this loop is crap on a Pentium but ain't it pretty? Just two adds to move both pointers, update the decision variable and loop counter. If we only had 64 bit registers on the Pentium... 11. Selfmodifying code ----------------------- One way to get rid of the memory variables in inner loops is to use selfmodifying code. When you have calculated a constant delta and are about to store it in a memory variable, why don't you store it right into a instruction as a constant in the inner loop? It's just as simple. Just remember to not use CS as segment override as we are in protected mode. I must warn you about this way of coding, especially on the Pentium (read about the code cache at the end). It can actually make the loop slower even if you think you cut away a few ticks. Doing more complex shadings like environment-bump or Phong-texture-bump, selfmodifying code might be the only way to get it to run at all. I.e. not having to write to any memory variables from the inner loop. If you are about to make your loop selfmodifying, compare it with your old loop by actually timing a typical scene. Then you'll know if you gained anything. If your loop is faster with selfmodifying code and the environment your application is aimed for allows selfmodifying code, I'd definitely say go for it, use selfmodifying code. 12. Unrolled and selfmodifying inner loops ------------------------------------------ I don't really see these as an alternative to the traditional inner loops on the Pentium. I present them here just because they are interesting. The deltas are constant so the offsets for each pixel in each scanline into the bitmap will also be constant. I.e. we can precalculate a whole run and use that in the inner loop. The inner loops for these type of texture mappers can look very different. The most radical must be to unroll it all the way and to plug in the offsets right into the mov instructions, i.e. selfmodifying code. These completely unrolled loops will be pretty big also. The loop below is 14 byte per pixel which means over 4k code for a whole 320 pixel scanline. The loop will take up half of the code cache. Ouch! (read about the code cache at the end). Here is some code that shows the principle of this type of "inner loop": jmp ecx ; Jump to the right place in the "loop" mov al, [esi+12345] mov [edi+319], al mov al, [esi+12345] ; Get pixel mov [edi+318], al ; Plot pixel ...... mov al, [esi+12345] ; '12345' is the selfmodifying part mov [edi+2], al ; that will be modified once per triangle mov al, [esi+12345] mov [edi+1], al mov al, [esi+12345] mov [edi+0], al Note that we are doing it backwards, from right to left. This makes it easier to setup esi and edi. As the code for each pixel in this loop is 14 byte you will be doing a X*14 when calculating the jump offset. X*14 is (X<<4)-X-X. You should of coarse not plug in the offsets for the whole loop if you only have a small triangle. The length of the longest scanline is a byproduct from the constant delta calculations. So what about the 1.5 tick per pixel loop? Well the following peace of code is usually what people think of. I'm not really sure that this is actually 1.5 tick per pixel as the 'mov [edi+?],ax' has a operand size prefix byte. This code will need some work to make the instructions pair on the Pentium. Of coarse this loop also suffers from the same problems as the previous selfmodifying, unrolled loop. jmp ecx ...... mov al, [esi+12345] mov ah, [esi+12345] mov [edi+4], ax mov al, [esi+12345] mov ah, [esi+12345] mov [edi+2], ax mov al, [esi+12345] mov ah, [esi+12345] mov [edi], ax 13. Table lookup inner loops ---------------------------- Now to a cooler method that is not selfmodifying and don't need to be unrolled all the way. The idea is very similar to the unrolled loops above but in this loop we have the offsets stored in a lookup table instead. For each pixel we get the address of the next pixel from the lookup table. This method should be much more Pentium friendly. Also this inner loop don't need to have the bitmap aligned on 64k as the traditional inner loops. The loop assume the following: esi = ptr to bitmap (no alignment needed) edi = ptr to first destination pixel to plot in this scanline ebp = width of scanline (loop counter) left_u = current u on the left edge of the triangle (16:16 bit fixed point) left_v = current v on the left edge of the triangle (16:16 bit fixed point) lookup = ptr to the precalculated lookup table. The lookup table is an array of dwords. mov edx, [lookup] xor eax, eax mov al, byte ptr [left_u+2] mov ah, byte ptr [left_v+2] add esi, eax @@inner: mov al, [esi+ebx] ; Get pixel mov ebx, [edx] ; Get offset for next pixel mov [edi], al ; Plot pixel add edx, 4 inc edi dec ebp jnz @@inner The same loop could look like this in C: // destptr = ptr to screen + y*320 // bitmap = ptr to bitmap // lookup = ptr to lookup table // x1 = start screen x coordinate of scanline // width = width of scanline char * dest = destptr + x1; char * src = bitmap + (left_u>>16) + (left_v>>16)*256; for(; width--; ) { *(dest++) = src[ *(lookup++) ]; } The above loop in asm should be 4 clock ticks per pixel on a Pentium. This loop can be changed to plot 4 pixels at a time: @@inner: mov al, [esi+ebx] ; Get pixel #1 mov ebx, [edx] mov ah, [esi+ecx] ; Get pixel #2 mov ecx, [edx+4] shl eax, 16 ; Move pixels 1 and 2 to the high word add edi, 4 mov al, [esi+ebx] ; Get pixel #3 mov ebx, [edx+8] mov ah, [esi+ecx] ; Get pixel #4 mov ecx, [edx+12] rol eax, 16 ; Swap the high and low words add edx, 16 mov [edi], eax ; Plot all 4 pixels dec ebp jnz @@inner Now this loop is 9 (8 if we assume that shl and rol are pairable in the U pipeline) ticks per 4 pixel with the pixels written as a dword. Very good if we align the write on dword. Use the other loop for very short lines or to get this one aligned on dword and use this for the rest of the scanline. Calculate the lookup table with the following loop (this loop can also be used to calculate the offsets in the selfmodifying example): (dudx and dvdx are 16:16 bit fixed point. lookup is an array of dwords) int du = dudx >> 8; int dv = dvdx >> 8; int u = 0; int v = 0; for( width of longest scanline ) { *lookup++ = (u>>8) + (v & 0xffffff00); u += du; v += dv; } ; ebx = ecx = 0 ; esi = delta u (26:8 bit fixed point) ; edi = delta v (26:8 bit fixed point) ; edx = ptr to lookup table ; ebp = length of table (the width of the longest scanline) @@mklookup: mov eax, ecx add ecx, edi ; update v mov al, bh add ebx, esi ; update u mov [edx], eax ; lookup[edx] = u+256*v add edx, 4 dec ebp jnz @@mklookup 14. Problems with precalculated runs ------------------------------------ The more I play around with inner loops that uses the same precalculated run for each scanline, the more skeptic I get. This is because they all suffers from the same problem, no matter if we use a lookup table or if we have a unrolled selfmodified loop. In the case of the lookup table inner loop we always start at the beginning of the table when drawing a scanline. This is wrong and will give very bad distortion especially when the triangle is zoomed in close. Always starting at the beginning of the table is the same as ignoring the fractional parts of the initial u and v of the scanline. So to fix this we should start somewhere into the table depending on the initial fractional parts of u and v. But this is impossible because u and v are interpolated separately on the triangle edge but are fixed to each other in the lookup table. Wilco Dijkstra posted the following solution in comp.graphics.algorithms: The basic idea is correct. What you mean is using subpixel positioning with one or two bits precision. For example, for 2 bits subpixel positioning you have to create 4 * 4 tables of the longest scanline. The first table starts at u = v = 0, second u = 0, v = 0.25, third u 0, v = 0.50 fourth u = 0, v = 0.75, fifth u = 0.25, v = 0, etc. When stepping down the scanlines, select the table giving the 2 most significant fractional bits of u and v. The maximum error you get is 1/8 in each direction (when proper rounding is used!). Thus this is 64 times more precise than using no subpixel positioning. The problem is that it's only faster for very large triangles (eg. more than 32 scanlines deep), so it may be faster (and more accurate) to draw the texture in the standard way, without a table. This method will reduce the distortion. On the other hand the lookup tables will require much more memory that in turn will push out other cached data, not to mention the additional time it takes to setup the tables. 15. Pre-stepping texture coordinates ------------------------------------ When we interpolate u, v and triangle x along the left edge of the triangle we always truncates triangle x when drawing a scanline. This is natural because we can only draw whole pixels. When we truncates x we must also adjust the initial u and v of the scanline. Adjusting u and v will give much cleaner and stable textures. Note that this only applies if you use a traditional inner loop. Don't bother doing this if you are using a table lookup inner loop. Kevin Baca sent me the following explanation: No matter how you compute screen pixels, you need to "pre-step" your texture coordinates by the difference between actual screen coordinates and screen pixels. It looks like this: // sp = screen pixel, sc = screen coordinate. float sc, diff, u, v, dudx, dvdx; int sp; sp = (int) sc; diff = sc - (float) sp; u -= dudx * diff; v -= dvdx * diff; You can actually do this without multiplies (by calculating a dda for each edge that determines when to add an extra 1 to the texel coordinates). 16. Special case code --------------------- It often pays off to make special case code that takes care of the edge delta calculations when a triangle section is 1, 2 or 4 pixels high. Then you can skip the divs and use shifts instead. I once made a histogram of the length of each scanline in the very popular chrmface.3ds object. This object has about 850 triangles and was scaled up so it just touched the top and the bottom of a 320x200 pixel screen. The histogram showed that most scanlines was only 1 or 2 pixels wide. This proves that the outer loop is just as important as the inner loop and also that it might be a good idea to have special case code for those 1 or 2 pixel lines. width number of scanlines 1 ********************* 2 ****************** 3 ********** 4 ****** 5 *** 6 ** 7 ** 17. Clipping ------------ Clipping is most of the time a real pain in the ass implementing. It will always mess up a nice looking routine with extra junk. One possibility is to have two separate functions, one with clipping and one with no clipping. Then test the triangle if it needs clipping before calling any of the functions. The actual clipping code is not that difficult to implement really. Say if you need to clip a texture mapped scanline, you first have to get the number of pixels you need to skip at the end of the scanline and the number of pixels in the beginning of the scanline. Then subtract the number of pixels skipped from the original scanline width. If you skipped some pixels at the start of the scanline, the new starting u and v must be calculated. This is done by multiplying the pixels skipped by delta u and delta v respectively. And adding the original starting u and v of coarse. The following code is what I'm using to sort out the stuff: movsx EBP, word ptr [left_x+2] ; Get the integer part from the movsx ECX, word ptr [right_x+2] ; 16:16 bit numbers. mov EDX, EBP sub EDX, ECX ; EDX = width of scanline ; ECX = x1 ; EBP = x2 mov EBX, EDX sub EBP, [_RightClip] jle short @@rightok sub EDX, EBP ; skip pixels at end @@rightok: xor EAX, EAX cmp ECX, [_LeftClip] jge short @@leftok mov EAX, [_LeftClip] sub EAX, ECX mov ECX, [_LeftClip] @@leftok: sub EDX, EAX ; skip pixels at start jle @@notvisible ; EAX = pixels skipped at start ; ECX = clipped x1 ; EDX = clipped width of scanline So now you just have to multiply EAX by delta u and add the original u to get the clipped u. The same apply for v. 18. Clipping using matrices --------------------------- I've been told that clipping should not be done scanline by scanline in the texture mapping function. But I have yet to find a simple alternative solution to this. Don't confuse the clipping I'm referring to with removal of nonvisible polygons. When we arrive at the texture mapping function we should already have removed those triangles that are backface or outside the viewcone. Kevin Baca sent me the following explanation on how to decide if vertices should be clipped or not. If you use homogeneous matrices to do your transformations it's actually very simple to clip before you do the perspective divide to get screen coordinates. Using homogeneous coordinates, you get vertices of the form [X Y Z W] after doing the perspective projection. To get actual screen coordinates, you divide X and Y by W. If you are going to "Normalized Device Coordinates" the results of these divisions will be -1 < X' < 1 and -1 < Y' < 1. Therefore, to do clipping you need to perform the following comparison before the perspective divide: -W < X < W, -W < Y < W. To clip along the Z axis, you can do the same thing, but I usually use the following comparison instead: 0 < Z < W. To do a perspective projection, multiply the projection matrix, P, by the view matrix, V: M = P * V. The view matrix is the result of all your transformations (translations, rotations, scalings, etc.) of both the model and the camera. For the projection matrix, I use the following: 1 0 0 0 0 a 0 0 0 0 b c 0 0 f 0 where: a = the aspect ratio (width / height of screen) b = f * (yon / (yon - hither)) c = -f * (yon * hither / (yon - hither)) f = sin(aov / 2) / cos(aov / 2) aov = angle of view yon = distance to far clipping plane hither = distance to near clipping plane These values allow me to clip using: -W < X < W -W < Y < W 0 < Z < W After clipping, divide X and Y by W and multiply by the width and height of your screen to get final screen coordinates. 19. Writing a byte, word, dword and other weird things ------------------------------------------------------ Now to a weird thing on the Pentium. The Pentium has a so called Write-Back cache. Well, the fact that the Pentium has a Write-Back cache is not weird at all. It's how the Write-Back cache works in practice that is weird if you are used to a Write-Trough cache that is used on the 486. Write-Trough: When we write a byte to memory the byte is always written to RAM. If that same byte is also present in the cache, the byte in the cache is also updated. Write-Back: When we write a byte to memory the byte is only written to RAM if the same byte is not present in the cache. If the byte is present in the cache, only the cache will be updated. It is first when a cacheline is pushed out from the cache that the whole cacheline will be written to RAM. I have done tests on my system (Pentium 120, L1:8+8k, L2:256k) using the time stamp counter to see how it actually behaves. These are the results: Writing to a byte (or aligned word or dword) that is not present in the L1 cache takes 8 clock ticks (no matter if the byte is present in the L2 cache). If the byte is present in the L1 cache, the same "mov" instruction takes the theoretical 0.5 clock tick. This is very interesting and potentially useful. If we e.g. manage to keep the cacheline where we have our memory variables in the L1 cache, we can write to them at the same speed as writing to a register. This could be very useful in the case of a Phong-texture or Phong-texture-bump inner loop where we need to interpolate many variables and only have 7 registers. The problem is that our cacheline will be pushed out from the cache as soon as we start getting cache misses when reading the texture data. Then we are back at 8 clock tick per write. To fix this we must read a byte from our cacheline so that it won't be marked as old and thrown out. But this is usually what we do anyway. We read a variable, interpolates it, uses it and writes it back. Juan Carlos Arevalo Baeza presented in an article to comp.graphics.algorithms another way to make use of the Write-Back cache in a texture mapping inner loop. The idea is to ensure that the destination pixel written is always present in the cache. This is done by reading a byte from the destination cacheline first: ; edi = ptr to first destination pixel (+1) to plot ; esi = ptr to last destination pixel to plot ; The scanline is plotted from right to left push esi mov al,[edi-1] ; read the first byte into the cache. @@L1: lea esi,[edi-32] cmp esi,[esp] jae @@C mov esi,[esp] @@C: mov al,[esi] ; read the last byte of the 32-byte chunk. @@L: mov al,[edx] add ebx,ecx adc edx,ebp dec edi mov dh,bh cmp edi,esi mov [edi],al jne @@L cmp edi,[esp] jne @@L1 pop esi This ensures that whenever you write a pixel, that address is already in the cache, and that's a lot faster. A LOT. My P90 takes 20-40 cycles to read a cache line, so that's around 1 more cycle per pixel. Problems: when painting polys, rows of very few pixels (let's say 1-8 pixels) are the most common, and those don't feel so good about this loop. You can always have two loops for the different lengths. Another way to speed up writes (that also works on 486) is to collect 4 pixels in a 32 bit register and write all 4 pixels at a time as a aligned dword. This will split the 8 clock tick delay on all 4 pixels making the delay only 2 clock ticks per pixel. This method will almost always gain speed especially if the scanlines are long. 20. The data cache ------------------ Although it is fun optimizing inner loops there are other important factors that one should look at. With the Pentium processor the cache aspects are very important. Maybe more important than the speed of the inner loop. Don't know how long this is true though as newer processors seems to get bigger and bigger caches that probably will become smarter also. The general idea of the cache is: When the CPU has decoded an instruction that needs to get a variable from memory, the CPU first checks the cache to see if the variable is already in the cache. If it is there the CPU reads the variable from the cache. This is called a cache hit. If the variable is not in the cache the CPU first has to wait for the data to be read from RAM (or the secondary cache, L2) into the cache and first after that get the variable from the cache. The cache always loads a full cacheline at a time so this will take a few clock ticks. A cacheline is 16 byte on a 486 and 32 byte on Pentium. The advantage of this is when reading byte after byte from the memory, the data will most of the time already be loaded into the cache because we have accessed the same cacheline just before. Also a cacheline is always aligned on 16 byte on the 486 and on 32 byte on the Pentium. I did a few tests on my system (Pentium 120 MHz, L1 cache 8+8k, L2 cache 256k) using the time stamp counter to check the actual time for loading a cacheline. In the first test I flushed the L2 cache so that each cacheline must be read all the way from RAM. This was done by allocating a 256k memory chunk and read each byte of that first. This would cause the memory I did the test on to be pushed out of the L2 cache. The testloop looked like this: mov ecx, 1000 next: mov al, [esi] add esi, ofs dec ecx jnz next The overhead of the loop was first timed by replacing the "mov al, [esi]" by "mov al, cl". The loop ran at exactly 2 clock tick per turn. The "ofs" value was replaced for each run with 1, 2, 4, 8, 16, 32, 64, ... In the second test I first forced the L2 cache to load the memory by reading each byte of a 128k memory chunk and then run the testloop on the same memory. Here are the results of both tests: clock ticks * * | * * * * * 40 + * * * * | * 35 + from RAM * | * 30 + * | * 25 + * | * 20 + * + + + + + + + + + + + | * + 15 + * + | * + from L2 cache 10 + * + | * + 5 + * + | * + 0 + -----+-----+-----+-----+-----+-----+-----+-----+-----+----- ofs 1 2 4 8 16 32 64 128 256 512 So this tells me that it takes 40-45 clock ticks minimum to load a cacheline all the way from RAM and exactly 18 clock ticks from the L2 cache. When "ofs" was 1 the "mov al, [esi]" ran at 2.0 ticks when loading from RAM and 1.1 ticks from the L2 cache. 0.5+40/32=1.75 and 0.5+18/32=1.06 so this makes sense. This is pretty scary! 18 clock ticks to load a cacheline from the L2 cache. 18 clock ticks minimum for the inner loops if we assume that a cacheline must be filled for each byte read. Ouch! So in the case of a texture mapper where we might be reading texels in a vertical line in the bitmap, the inner loop will be accessing pixels that are >256 bytes apart. The CPU will then be busy filling cachelines for each texel. A 64k bitmap won't fit inside a 8k cache, you know. So what can we do? Well, we can wait on Intel to invent bigger caches or we might consider storing our bitmaps some other, more cache friendly way. I got an interesting tip from Otto Chrons on channel #coders the other day about this. He said that one should store the bitmap as a set of tiles, say 8 x 8 pixels instead of the usual 256 x 256 pixel. This makes perfect sense. It would mean that a small part of the bitmap (8 x 4 pixel) would fit in the same 32 byte cacheline. This way, new cachelines don't need to be loaded that often when reading pixels in a vertical line in the bitmap. The following was suggested in a mail to me by Dare Manojlovic: If you are saving bitmap as a set of tiles (8*4) the inner loop wouldn't have to be more complicated (this is my opinion - not yet tested). For example, let's say that we have u&v texture coordinates, we only have to reorder bits to get the correct address (before the inner loop): Normally for a bitmap of 256*256 the texel address would look like: EAX AH AL oooo oooo oooo oooo oooo oooo oooo oooo v coordinate u coordinate And now: EAX AH AL oooo oooo oooo oooo oooo oooo oooo oooo v(other 6 bits) u(other 5 bits) v(lower 2 bits) u(lower 3 bits) Adding a constant value,that is also converted, in the loop shouldn't be a problem. Now, as I understand cache loading procedure,it always loads 32 bytes of data (Pentium), so the whole bitmap tile of (8*4 pixels) will be in cache. Of course bitmap tile must be 32 bytes aligned. This would also work faster on 486 where cache is loaded with 16 bytes. There is a small problem to the above method. We can't just add a constant value to a number in this format (even if they both are converted). This is because there is a gap between the bits. We must make the bits jump over the gap to make the add correct. There is a simple solution to this problem though. Just fill the gap with 1:s before adding the constant value. This will cause the bit to jump over the gap. Filling the gap is done with a bitwise OR instruction. Converting u and v (16:16 bit) to this format can be done with the following code: int uc = (u & 0x0007ffff) | ((u<<2) & 0xffe00000); int vc = (v & 0x0003ffff) | ((v<<5) & 0xff800000); ; eax = u --------wwwwwwwwffffffffffffffff (w=whole, f=fractional) ; ebx = v --------wwwwwwwwffffffffffffffff ; ecx = scratch register mov ecx, eax shl eax, 2 and ecx, 00000000000001111111111111111111b and eax, 11111111111000000000000000000000b or eax, ecx mov ecx, ebx shl ebx, 5 and ecx, 00000000000000111111111111111111b and ebx, 11111111100000000000000000000000b or ebx, ecx ; eax = u ------wwwww--wwwffffffffffffffff ; ebx = v ---wwwwww-----wwffffffffffffffff Adding dudx and dvdx to u and v in this format can be done with the following code (all variables are in the converterd format): uc = (uc | 0x00180000) + dudx; vc = (vc | 0x007c0000) + dvdx; ; eax = u ------wwwww--wwwffffffffffffffff ; ebx = v ---wwwwww-----wwffffffffffffffff ; dudx, dvdx = 16:16 bit converted to this format or eax, 00000000000110000000000000000000b ; fill the bit-gap in u or ebx, 00000000011111000000000000000000b ; fill the bit-gap in v add eax, [dudx] add ebx, [dvdx] In a mail sent to me, Russel Simmons preresented the following method to reorder the bits to acheive a simpler inner loop by eliminating a bit-gap: In one post, someone suggested a bit structure to find the corect position in your tiled texture given u and v. He suggested something like: high bits of v | high bits of u | low bits of v | low bits of u This way the high bits of u and v determine which tile our texel is in, and the low bits of u and v determine where in our tile the texel is. If we store our tiles in a different manner, we can simplify this to: high bits of u | high bits of v | low bits of v | low bits of u which is in other words: high bits of u | all bits of v | low bits of u In order to facilitate this, instead of storing our tiles in this order: ------------- | 0| 1| 2| 3| ... (here i am showing the upper 4x4 tiles of a 256x256 ------------- texture store in 8x8 tiles) |32|33|34|35| ... ------------- Original Method |64|65|66|67| ... ------------- |96|97|98|99| ... ------------- | | | | | store them in this order: ------------- | 0|32|64|96| ... (here i am showing the upper 4x4 tiles of a 256x256 ------------- texture store in 8x8 tiles) | 1|33|65|97| ... ------------- New Method, in order to acheive a simpler inner loop | 2|34|66|98| ... ------------- | 3|35|67|99| ... ------------- | | | | | Also, if we are storing our bitmap in a tiled fashion, then it would greatly improve our cache performance if we can back and forth across scan lines.. in other words alternate the direction we scan across lines. Say we have just scanned forward across one scan line. If we start backwards across the next scan line, we are likely to be pulling texels from the same tiles as we were at the end of the previous scan line. The last part about alternating the drawing direction is definitely something to try out! I was hoping I would be able to present some code here that uses all these techniques and 16:16 bit interpolation in a slick inner loop but due to lack of time and the fact that I'm fed up with this document, I leave this to you. 21. The code cache ------------------ The cool thing about Pentiums is that it can execute two instructions in parallel. This is called instruction pairing. But there is a lot of rules that must be fulfilled for the pairing to take place. One rule is that both instructions must already be in the code cache. This means that the first time trough a inner loop, no instructions will pair. There is one exception to this rule. If the first instruction is a 1 byte instruction, e.g. inc eax, and the other is a simple instruction, then they will pair the first time. If by chance our inner loop happens to be in the code cache, by modifying an instruction in the inner loop (selfmodifying code) the cacheline where we did the modification will be marked as not up to date. So that cacheline must be loaded into the cache again before we can execute the inner loop again. Loading of code cachelines seems to be exceptionally slow also. In other words, we have found yet another source of delay. So to have a completely unrolled loop that almost fills up the whole code cache and also is selfmodifying is a pretty bad idea on the Pentium. On the other hand, we are not modifying the loop for each scanline so chances are that parts of it will be in the code cache from drawing the previous scanline. 22. Some pairing rules ---------------------- As mentioned above, the Pentium can execute two instructions in parallel. This is possible because the CPU has dual integer pipelines, they are called the U and V pipelines. The Pentium has a so called superscalar architecture. The U pipeline is fully equipped and can execute all integer instructions. The V pipeline on the other hand is a bit crippled and can only execute simple, RISC type instructions. Simple instructions are: mov, inc, dec, add, adc, sub, sbb, and, or, xor, cmp, test, push, pop, lea, jmp, call, jcc, nop, sar, sal, shl, shr, rol, ror, (rcl), (rcr) (What I've heard there are different opinions on if the shift/rotate instructions are pairable or not. The book I have here states that these instructions are pairable but can only execute in the U pipeline) The first pairing rule is that both instructions must be simple instructions. Also, no segment registers can be involved in the instructions. Another rule is that the two instructions must be completely independent of each other. Also they must not write to the same destination register/memory. They can read from the same register though. Here are some examples: add ecx, eax ; store result in ecx add edx, ecx ; get result from ecx. No pairing! mov ecx, eax mov edx, ecx ; No pairing! mov al, bh ; al and ah is in the same register mov ah, ch ; No pairing! mov ecx, eax ; read from the same register mov edx, eax ; Pairs ok. mov ecx, eax ; note eax in this example add eax, edx ; Pairs ok. There are two exception to this rule. Namely the flag register and the stack pointer. Intel has been kind enough to optimize these. dec ecx ; modifies the flag register jnz @@inner ; Pairs ok. push eax ; both instructions are accessing esp push ebx ; Pairs ok. So for example the loop we used to calculate the lookup table with, all instructions are simple and not dependent on the previous one. The 8 instructions should execute in 4 clock ticks. @@mklookup: mov eax, ecx add ecx, edi ; Pairs ok. mov al, bh add ebx, esi ; Pairs ok. mov [edx], eax add edx, 4 ; Pairs ok. dec ebp jnz @@mklookup ; Pairs ok. 23. Pipeline delays ------------------- There are a whole bunch of these that will delay the pipelines: - data cache memory bank conflict - address generation interlock, AGI - prefix byte delay - sequencing delay I personally think that the AGI is most important to consider in the case of tight inner loops. Because that is what's happening in a inner loop, where we are calculating an address and need it right away to access some data. There will be a AGI delay if a register used in a effective address calculation is modified in the previous clock cycle. So if we have our instructions nicely pairing we might have to put 3 instructions in between to avoid the AGI delay. add esi, ebx ; Move the array pointer. mov eax, [esi+8] ; AGI delay. You just modified esi. add esi, ebx ; Move the array pointer. add ebx, ecx ; Do something useful here inc edi ; " add ebp, 4 ; " mov eax, [esi+8] ; Now it's OK to access the data. No AGI delay. If you don't have any useful instructions to fill out the gap with you could try to swap the two instructions so that you access the data first and then modify the index register. mov eax, [esi+8] add esi, ebx ; Pairs ok. No AGI delay. There are a lot more rules one must follow so I suggest you buy a good book on the subject. I don't know of any free info about this on the net as of this writing. Maybe you'll find something at Intel's www-site (http://www.intel.com). Anyway, a book that got me started was: "Pentium Processor Optimization Tools" by Michael L. Schmit ISBN 0-12-627230-1 This book has a few minor errors and some of the explanations are a bit cryptic but it is a good starting point. The way to really learn is to get the basics from e.g. a book and then time actual code to see what is faster and what's not. 24. The time stamp counter -------------------------- The Pentium has a built in 64 bit counter called the Time Stamp Counter that is incremented by 1 for each clock tick. To read the counter you use the semi-undocumented instruction RDTSC (db 0fh,31h). This will load the low 32 bit of the counter into EAX and the high 32 bit into EDX. Perfect for timing code! ; First time the overhead of doing the RDTSC instruction db 0fh,31h ; hex opcode for RDTSC mov ebx, eax ; save low 32 bit in ebx db 0fh,31h sub eax, ebx ; overhead = end - start mov [oh_low], eax ; Now do the actual timing db 0fh,31h mov [co_low], eax mov [co_hi], edx ; Run some inner loop here of whatever you want to time db 0fh,31h sub eax, [co_low] ; ticks = end - start sbb edx, [co_hi] sub eax, [oh_low] ; subtract overhead sbb edx, 0 ; Number of clock ticks is now in edx:eax You'll notice that I first time the overhead of doing the RDTSC instruction. This might be a bit overkill but it's no harm in doing it. Note also that I ignore the high 32 bit. The overhead should not be more than 2^32 clock ticks anyway. The RDTSC can be a privileged instruction under some extenders (?) but still be available (under the control of the extender) so there might actually be a overhead to time. You can usually ignore the high 32 bit. Using only the low 32 bit will allow a maximum of 2^32 clock ticks which is 35 seconds on a Pentium 120 MHz. When you are timing your code e.g. when you have done some optimizations on your texture mapper, don't time just one triangle over and over. Time how long it takes to draw a complete object with hundreds (thousands) of triangles. Then you'll know if that optimization made any difference. 25. Branch prediction --------------------- The Pentium has some sort of lookup table called the Branch Target Buffer (BTB) in which it stores the last 256 branches. With this it tries to determine the destination for each jump or call. This is done by keeping a history of whether a jump was taken or not the last time it was executed. If the prediction is correct then a conditional jump takes only 1 clock tick to execute. Because the history mechanism only remembers the last time the jump was executed, the prediction will always fail if we jump different each time. There is a 4-5 clock tick delay if the prediction fails. The branch prediction takes place in the second stage of the instruction pipeline and predicts if whether a branch will be taken or not and its destination. Then it starts filling the other instruction prefetch queue with instructions from the branch destination. If the prediction was wrong, then both prefetch queue must be flushed and prefetching restarted. So to avoid this delay you should strive to use simple loops that always takes the jump or always not takes the jump. Not like the following that jumps different depending on the carry flag. jmp @@inner @@extra: .... ; Do something extra when we get carry overflow dec ebp jz @@done @@inner: .... ; Do something useful here add eax, ebx jc @@extra ; Jump on carry overflow dec ebp jnz @@inner @@done: In this loop it's the 'jc @@extra' instruction that will mess up the branch prediction. Sometimes the jump will be taken and sometimes not. The typical way of doing masking with compares and jumps has this problem also. 26. Reference ------------- Most of the Pentium specific information on optimization was found in the book: "Pentium Processor Optimization Tools" by Michael L. Schmit ISBN 0-12-627230-1 27. Where to go from here ------------------------- When you have implemented your texture mapper you automatically also have Phong shading and environment mapping. It's only a matter of making a suitable bitmap and to use the normal vectors at each triangle vertex to get the u and v values. From there the step is not far from combining Phong shading and texture mapping. And then adding bumps to all this. The only difficult part is that you need to interpolate 4 variables in the inner loop when you do Phong-texture, environment-bump or Phong-texture-bump and still have registers left for pointers and loop counter. These shadings can't really be called "fast" as the inner loops will become pretty ugly. They can definitely be called real time though. 28. Credits and greetings ------------------------- Juan Carlos Arevalo Baeza (JCAB/Iguana-VangeliSTeam) Wilco Dijkstra Kevin Baca Sean L. Palmer Tiziano Sardone Mark Pursey Dare Manojlovic Russel Simmons (Armitage/Beyond) Aatu Koskensilta (Zaphod.B) Otto Chrons (OCoke) (a legend) Nix/Logic Design (a cool coder) Phil Carmody (FatPhil) (The optimizing guru, why all this silence?) Jmagic/Complex (another legend) MacFeenix (you are young) BX (keep on coding) thefear (a cool swede) John Gronvall (MIPS R8000 rules!) LoZEr (when will PacMan for Linux be out?) Addict, Damac, Dice, Doom, Swallow, Wode / Doomsday (a bunch of finns ;) When I started out writing this document I didn't know half of what I now know about texture mapping. I've learned a lot and this is much because of those 12 first persons in the credits and greetings list. Thanks a lot for the help. I hope that the readers of this document also will learn something. If you truly find this document useful, you could consider giving me a small greeting in your production. That would be cool.