Table des matières

PART 14 : Glenzing and faster Polygons

Hello there. Exams are just around the corner (again :( ), so I thought I better get round to doing the next trainer. As usual, there seems to have been a big delay between this one and the last one… sorry about that ;-)

Well, this trainer is mainly on four things : Glenzing, faster polys, fixed point and assembler. The sample program is basically tut 9 rewritten to include the above.

I'll go through them in order, and hopefully you won't have any hassles grasping the concepts. By the way, do any of you read the text files? I find myself answering questions via E-Mail etc. that were discussed in the text sections of the trainers … oh well, I'll just ramble along anyway ;-)

Please dont send any mail to smith9@batis.bis.und.ac.za anymore … I don't know for how much longer the account will be valid (How can a non-BIS person get onto the BIS UNIX machine in the BIS2 directory? If his name is Denthor I suppose ;-) Oh well, I got about 8 months use out of it. The account expires on Christmas day anyway…) So anyway, please leave all messages to denthor@beastie.cs.und.ac.za

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What is glenzing?

This is an easy one. Imagine, in a 3D object, that all the sides are made out of colored glass. That means that every time you look through that side, everything behind it is tinged in a certain color.

In ascii …

          +---------+
          |      <--|---Light blue
          |         |
     +--------+     |
     |    | <-|-----|---Dark blue
     |    +---|-----+
     |     <--|---------Light blue
     +--------+

So where the two sides overlap, the color values of the two sides are added. Easy huh? It is also easy to code. This is how you do it :

Set up your pallette to be a nice run of colors.

Draw your first poly.

While drawing poly 1, instead of plonking down a set pixel color, grab the backgrond pixel, add 1 to it, then put the result down.

Draw your second poly.

While drawing poly 2, instead of plonking down a set pixel color, grab the backgrond pixel, add 2 to it, then put the result down.

and so forth.

So if the color behind poly 1 was 5, you would place pixel 6 down instead.

If you do this for every single pixel of every single side of your 3d object, you then have glenzing going. This is obviously slightly slower then just drawing an item straight, but in the sample program it goes quite quickly … this is because of the following sections…

Faster Polygons

In Tut 9, you probably noticed that we were using a multiply for every single line of the poly that we drew. This is not good. Let's find out how to speed it up, shall we…

With the multiply method, we went through every line, to find out the minimum x and maximum x value for that line.

                           +
                   ------/---\------- Find min x and max x, draw a line
                       /       \        between them.
                     +           +
                       \       /
                         \   /
                           +

How about if we found out all the min and max x's for every line first, then just went through an array drawing them. We could do it by “scanning” each side in turn. Here is how we do it :

                          + 1
                        /
                      /
                  2 +

We go from point one to point two. For every single y we go down, we move a constant x value. This value is found like this :

           xchange := (x1-x2)/(y1-y2)

Remember gradients? This is how you calulated the slope of a line waaay back in school. You never thought it would be any use, didn't you ;-)

Anyway, with this value, we can do the following :

    For loop1:=y1 to y2 do BEGIN
      [ Put clever stuff here ]
      x:=x+xchange;
    END;

and we will go through all the x-values we need for that line. Clever, huh?

Now for the clever bit. You have an array, from 0 to 199 (which is all the possible y-values your onscreen poly can have). Inside this is two values, which will be your min x and your max x. You start off with the min x being a huge number, and the max x being a low number. Then you scan a side. For each y, check to see if one of the following has happened :

   If the x value is smaller then the xmin value in your
     array, make the xmin value equal to the x value
   If the x value is larger then the xmax value in your
     array, make the xmax value equal to the x value

The loop now looks like this :

    For loop1:=y1 to y2 do BEGIN
      if x>poly[loop1,1] then poly[loop1,1]:=x;
      if x<poly[loop1,1] then poly[loop1,1]:=x;
      x:=x+xchange;
    END;

Easy? Do this for all four sides (you can change this for polys with different numbers of sides), and then you have all the x min and x max values you need to draw your polygon.

In the sample program, if you replaced the Hline procedure with one that draws solid lines, you could use the given drawpoly for solids.

Even this procedure is sped up by the next section, on fixed point.

What is fixed point?

Have you ever noticed how slow reals are? I mean slooooow. You can get a massive speed increase in most programs by replacing your reals with integers, words etc. But, I hear you cry, what happens to the much needed fraction bit after the decimal point? The answer? You keep it.

Here's how.

Let us say you have a word, which is 16 bits. If you want to use it as a fixed point value, you can separate it into 2 sections, one of which holds the whole value, and one which holds the fraction.

             00000000  00000000   <-Bits
             Whole     Fraction

The number 6.5 would therefore be shown as follows :

 
             Top byte    :  6
             Bottom byte :  128

128 is half (or .5) of 256, and in the case of the fraction section, 256 would equal one whole number.

So let us say we had 6.5 * 2. Using reals this would be a slow mul, but with fixed point …

            Top Byte    :  6     
            Bottom Byte :  128
            Value       :  1664   <-This is the true value of the word ie. (top byte*256)+bottom byte).
                                     this is how the computer sees the word.
             1664 shl 1 = 3328    <-shl 1 is the same as *2, just faster.
            Top byte    :  13
            Bottom byte :  0

As you can see, we got the correct result! And in a fraction of the time that a multiplication of a real would have taken us. You can add and subtract fixed point values with no hassles, and multiply and divide them by normal values too. When you need the whole value section, you can just read the high byte, or do the following

             whole = word shr 8
         eg  1664 shr 8 = 6

As you can see, the fraction is truncated. Obviously, the more bits you set aside for the fraction section, the more accurate your calculation is, but the lesser the maximum whole number you can have. For example, in the above numbers, the maximum value of your whole number was 256, a far cry from the 65535 a normal (non fixed point) word's maximum.

There are a lot of hassles using fixed point (go on, try shift a negative value), most of which have to do with the fact that you have severely decreased the maximum number you may have, but trust me, the speed increase is worth it (With longintegers, and/or extended 386 registers, you can even have 16×16 fixed point, which means high accuracy and high maximum values)

Try write a program using fixed point. It is not difficult and you will get it perfect easily. Trust me, I'm a democoder ;-)

Assembler

In the sample program I used one or two assembler commands that I havent discussed with you … here they are …

imul  value
This is the same as mul, but for integer values. It multiplies ax by the value. If the value is a word, it returns the result in DX:AX
sal  register,value
This is the same as shl, but it is arithmetic, in other words it works on integers. If you had to shl a negative value, the result would mean nothing to you.
rcl  register,value
This is the same as shl, but after you have shifted, the value in the carry flag is placed in the now-vacated rightmost bit. The carry flag is set when you do an operation where the result is greater then the upmost possible value of the variable (usually 65535 or 32767)
mov ax,64000
shl ax,1     {<- Carry flag now = 1}

The sample program is basically Tut 9 rewritten. To see how the assembler stuff is working, do the following … Go into 50 line mode (-Much- easier to debug), then hit [Alt - D] then R. A little box with all your registers, segments etc and their values will pop up. Move it down to where you want it, then go back to the program screen (Hit Alt and it's number together), and resize it so that you have both it and the register box onscreen at once (Alt - 5 to resize) … then use F4, F7 and F8 to trace though the program (you know how). The current value of the registers will always be in that box.

In closing

Well, that is about it. The sample program may start as being a little intimidating to some when they first look at it, just remember to read it with tut 9, very little is different, it's just with fixed point and a bit of assembler.

Before I forget, with Tut 13, the program crashes if you have error checking on. This is how you sort it out :

1) Turn off error checking or

2) Make logo a pointer, and get and free the memory or

3) Read the logo directly to screen or

4) Use the {$M ……} command with various values at the top of the program till it works.

I prefer options 3 or 2, but hey … the problem was that the logo was rather large (16k), and Pascal likes complaining ;-)

I am in doubt as to weather to continue doing quotes … here is a conversation I had with Pipsy, after the group conversation got around to weather we were normal or not …

Me : I'm normal.

Pipsy : No your not.

Me : Prove it.

Pipsy : Just look at your quotes in your trainers.

Me : What? You think those are weird?

Pipsy : Too weird.

Me : You mean that there is a weirdness line, and I crossed it?

Pipsy : Yes.

Bit of a conversation killer that, so we stopped there.

Anyway, this trainer won't have a quote in it … how about a disclaimer instead? Feel free to use it in your messages …


The views expressed above are mine and not Novells. In fact, I've never worked for them in my life!

Byeeee…. Denthor 18:57 9-9-94

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