this is a very good tutor for making circles and lines in code. its written
by denthor of asphyxia. at the end you can find two procedures i use for
circles and lines in the blockart editor :
W E L C O M E
To the VGA Trainer Program
By
DENTHOR of ASPHYXIA
updated by Snowman
-- PART 3 --
Note: things in brackets have been added by Snowman. The original text
has remained mostly unaltered except for the inclusion of C++ material
Introduction
Greetings! This is the third part of the VGA Trainer series! Sorry it
took so long to get out, but I had a running battle with the traffic
department for three days to get my car registered, and then the MailBox
went down. Ahh, well, life stinks. Anyway, today will do some things
vital to most programs : Lines and circles.
Watch out for next weeks part : Virtual screens. The easy way to
eliminate flicker, doubled sprites, and subjecting the user to watch
you building your screen. Almost every ASPHYXIA demo has used a virtual
screen with the exception of the SilkyDemo, so this is one to watch out
for. I will also show you how to put all of these loose procedures into
units.
If you would like to contact me, or the team, there are many ways you
can do it : 1 Write a message to Grant Smith in private mail here on
the Mailbox BBS.
2 Write a message here in the Programming conference here
on the Mailbox Preferred if you have a general
programming query or problem others would benefit from
3 Write to ASPHYXIA on the ASPHYXIA BBS.
4 Write to Denthor, Eze or Livewire on Connectix.
5 Write to : Grant Smith
P.O.Box 270 Kloof
3640
6 Call me Grant Smith at 73 2129 leave a message if you
call during varsity
NB : If you are a representative of a company or BBS, and want ASPHYXIA
to do you a demo, leave mail to me we can discuss it.
NNB : If you have done/attempted a demo, SEND IT TO ME! We are feeling
quite lonely and want to meet/help out/exchange code with other demo
groups. What do you have to lose? Leave a message here and we can work
out how to transfer it. We really want to hear from you!
Circle Algorithim
You all know what a circle looks like. But how do you draw one on the
computer?
You probably know circles drawn with the degrees at these points :
0
270 ----+---- 90
180
Sorry about my ASCII - ... anyway, Pascal and C++ dont work that way ...
they work with radians instead of degrees. You can convert radians to
degrees, but Im not going to go into that now. Note though that in Pascal
and C++, the circle goes like this :
270
180 ----+---- 0
90
Even so, we can still use the famous equations to draw our circle ...
You derive the following by using the theorem of our good friend
Pythagoras
Sin deg Y/R
Cos deg X/R
This is standard 8? maths ... if you havent reached that level yet,
take this to your dad, or if you get stuck leave me a message and Ill
do a bit of basic Trig with you. I aim to please -
Where Y your Y-coord
X your X-coord
R your radius the size of your circle
deg the degree
To simplify matters, we rewrite the equation to get our X and Y values :
Y R*Sindeg
X R*Cosdeg
This obviousy is perfect for us, because it gives us our X and Y co-ords to
put into our Putpixel routine see Part 1. Because the Sin and Cos
functions return a Real long value, we use a round function to transform
it into an Integer.
Pascal
Procedure Circle oX,oY,rad:integerCol:Byte
VAR deg:real
X,Y:integer
BEGIN
deg:0
repeat
X:roundrad*COS deg
Y:roundrad*sin deg
putpixel x+ox,y+oy,Col
deg:deg+0.005
until deg6.4
END
C++
void Circleint X, int Y, int rad, int col
float deg 0
do
X roundrad * cosdeg
Y roundrad * sindeg
Putpixel X+160, Y+100, col
deg + 0.005
while deg 6.4
In the above example, the smaller the amount that deg is increased by,
the closer the pixels in the circle will be, but the slower the procedure.
0.005 seem to be best for the 320x200 screen. NOTE : ASPHYXIA does not use
this particular circle algorithm, ours is in assembly language, but this
one should be fast enough for most. If it isnt, give us the stuff you are
using it for and well give you ours.
Line algorithms
There are many ways to draw a line on the computer. I will describe one
and give you two. The second one you can figure out for yourselves it
is based on the first one but is faster
The first thing you need to do is pass what you want the line to look
like to your line procedure. What I have done is said that x1,y1 is the
first point on the screen, and x2,y2 is the second point. We also pass the
color to the procedure. Remember the screens top left hand corner is 0,0
see Part 1
Ie. o X1,Y1
ooooooooo
ooooooooo
oooooooo X2,Y2
Again, sorry about my drawings -
To find the length of the line, we say the following :
Pascal
XLength ABS x1-x2
YLength ABS y1-y2
C++
xlength absx1-x2
ylength absy1-y2
The ABS function means that whatever the result, it will give you an
absolute, or posotive, answer. At this stage I set a variable stating
wheter the difference between the two xs are negative, zero or posotive.
I do the same for the ys If the difference is zero, I just use a loop
keeping the two with the zero difference posotive, then exit.
If neither the xs or ys have a zero difference, I calculate the X and Y
slopes, using the following two equations :
Pascal
Xslope Xlength / Ylength
Yslope Ylength / Xlength
C++
Note: C++ is significantly different here. I had to add special divide
by zero checking. C++ doesnt like x/0. I also added in type-casting
which might not have been necessary. :
if xlength ! 0 ylength ! 0
xslope floatxlength/floatylength
yslope floatylength/floatxlength
else
xslope 0.0
yslope 0.0
As you can see, the slopes are real numbers.
NOTE : XSlope 1 / YSlope
Now, there are two ways of drawing the lines :
X XSlope * Y
Y YSlope * X
The question is, which one to use? if you use the wrong one, your line
will look like this :
o
o
o
Instead of this :
ooo
ooo
ooo
Well, the solution is as follows :
If the slope angle is in the area of the stars * then use the first
equation, if it is in the other section then use the second one.
What you do is you calculate the variable on the left hand side by
putting the variable on the right hand side in a loop and solving. Below
is our finished line routine :
Pascal
Procedure Line x1,y1,x2,y2:integercol:byte
VAR x,y,xlength,ylength,dx,dy:integer
xslope,yslope:real
BEGIN
xlength:abs x1-x2
if x1-x20 then dx:-1
if x1-x20 then dx:0
if x1-x20 then dx:+1
ylength:abs y1-y2
if y1-y20 then dy:-1
if y1-y20 then dy:0
if y1-y20 then dy:+1
if dy0 then BEGIN
if dx0 then for x:x1 to x2 do
putpixel x,y1,col
if dx0 then for x:x2 to x1 do
putpixel x,y1,col
exit
END
if dx0 then BEGIN
if dy0 then for y:y1 to y2 do
putpixel x1,y,col
if dy0 then for y:y2 to y1 do
putpixel x1,y,col
exit
END
xslope:xlength/ylength
yslope:ylength/xlength
if yslope/xslope1 and yslope/xslope-1 then BEGIN
if dx0 then for x:x1 to x2 do BEGIN
y: round yslope*x
putpixel x,y,col
END
if dx0 then for x:x2 to x1 do BEGIN
y: round yslope*x
putpixel x,y,col
END
END
ELSE
BEGIN
if dy0 then for y:y1 to y2 do BEGIN
x: round xslope*y
putpixel x,y,col
END
if dy0 then for y:y2 to y1 do BEGIN
x: round xslope*y
putpixel x,y,col
END
END
END
C++
void Line2int x1, int y1, int x2, int y2, int col
int x, y, xlength, ylength, dx, dy
float xslope, yslope
xlength absx1-x2
if x1-x2 0 dx -1
if x1-x2 0 dx 0
if x1-x2 0 dx +1
ylength absy1-y2
if y1-y2 0 dy -1
if y1-y2 0 dy 0
if y1-y2 0 dy +1
if dy 0
if dx 0
for xx1 xx2+1 x++
Putpixel x,y1,col
if dx 0
for xx2 xx1+1 x++
Putpixel x,y1,col
if dx 0
if dy 0
for yy1 yy2+1 y++
Putpixel x1,y,col
if dy 0
for yy2 yy1+1 y++
Putpixel x1,y,col
if xlength ! 0 ylength ! 0
xslope floatxlength/floatylength
yslope floatylength/floatxlength
else
xslope 0.0
yslope 0.0
if xslope ! 0 yslope ! 0
yslope/xslope 1 yslope/xslope -1
if dx 0
for xx1 xx2+1 x++
y round yslope*x
Putpixel x,y,col
if dx 0
for xx2 xx1+1 x++
y round yslope*x
Putpixel x,y,col
else
if dy 0
for yx1 yx2+1 y++
x round xslope*y
Putpixel x,y,col
if dy 0
for yx2 yx1+1 y++
x round xslope*y
Putpixel x,y,col
Quite big, isnt it? Here is a much shorter way of doing much the same
thing :
Pascal
function sgna:real:integer
begin
if a0 then sgn:+1
if a0 then sgn:-1
if a0 then sgn:0
end
procedure linea,b,c,d,col:integer
var u,s,v,d1x,d1y,d2x,d2y,m,n:real
i:integer
begin
u: c - a
v: d - b
d1x: SGNu
d1y: SGNv
d2x: SGNu
d2y: 0
m: ABSu
n : ABSv
IF NOT MN then
BEGIN
d2x : 0
d2y : SGNv
m : ABSv
n : ABSu
END
s : INTm / 2
FOR i : 0 TO roundm DO
BEGIN
putpixela,b,col
s : s + n
IF not sm THEN
BEGIN
s : s - m
a: a +roundd1x
b : b + roundd1y
END
ELSE
BEGIN
a : a + roundd2x
b : b + roundd2y
END
end
END
C++
int sgn long a
if a 0 return +1
else if a 0 return -1
else return 0
void Lineint a, int b, int c, int d, int col
long u,s,v,d1x,d1y,d2x,d2y,m,n
int i
u c-a
v d-b
d1x sgnu
d1y sgnv
d2x sgnu
d2y 0
m absu
n absv
if mn
d2x 0
d2y sgnv
m absv
n absu
s intm / 2
for i0iroundmi++
Putpixela,b,col
s + n
if s m
s - m
a + d1x
b + d1y
else
a + d2x
b + d2y
This routine is very fast, and should meet almost all of your requirements
ASPHYXIA used it for quite a while before we made our new one.
In the end program, both the new line routine and the circle routine are
tested. A few of the procedures of the first parts are also used.
Line and circle routines may seem like fairly trivial things, but they are
a vital component of many programs, and you may like to look up other
methods of drawing them in books in the library I know that here at the
varsity they have books for doing this kind of stuff all over the place
A good line routine to look out for is the Bressenhams line routine ...
there is a Bressenhams circle routine too ... I have documentaiton for them
if anybody is interested, they are by far some of the fastest routines
you will use.
In closing
Varsity has started again, so I am shock going to bed before three in
the morning, so my quote this week wasnt written in the same wasted way
my last weeks one was For last weeks one, I had gotten 8 hours sleep in
3 days, and thought up and wrote the quote at 2:23 am before I fell asleep.
What does it do? she asks.
Its a computer, he replies.
Yes, dear, but what does it do?
It ..er.. computes! Its a computer.
What does it compute?
What? Er? Um. Numbers! Yes, numbers! He smiles
worriedly.
Why?
Why? Well ..um.. why? He starts to sweat.
I mean, is it just something to dust around, or does
it actually do something useful?
Um...you can call other computers with it! Hope lights
up his eyes. So you can get programs from other computers!
I see. Tell me, what do these programs do?
Do? I dont think I fol...
I see. They compute. Numbers. For no particular reason. He
withers under her gaze.
Yes, but...
She smiles, and he trails off, defeated. She takes another look
at the thing. Although, she says, with a strange look in
her eyes. He looks up, an insane look of hope on his
face. Does it come in pink? she asks.
- Grant Smith
Tue 27 July, 1993
9:35 pm.
See you next time,
- Denthor
circle and line procedures by xqtr
Procedure Circle1x1, y1, x2, y2 : integer color : byte Var Image: TScreenImage
Var
r,y,x,d : Integer
Begin
r : x2 - x1 Div 2
y : r
d : -r
Imagey1x1+r.Char:*
For x : 10 to truncr / Sqrt2*10 Do Begin
d : d + 2*x-1
if d 0 Then Begin
y : y - 1
d : d - 2* y
End
Imagey1+yx1+xChar:*
End
End
procedure Linex1, y1, x2, y2 : integer color : byte Var Image: TScreenImage C:Char
var i, deltax, deltay, numpixels,
d, dinc1, dinc2,
x, xinc1, xinc2,
y, yinc1, yinc2 : integer
begin
Calculate deltax and deltay for initialisation
deltax : absx2 - x1
deltay : absy2 - y1
Initialize all vars based on which is the independent variable
if deltax deltay then
begin
x is independent variable
numpixels : deltax + 1
d : 2 * deltay - deltax
dinc1 : deltay Shl 1
dinc2 : deltay - deltax shl 1
xinc1 : 1
xinc2 : 1
yinc1 : 0
yinc2 : 1
end
else
begin
y is independent variable
numpixels : deltay + 1
d : 2 * deltax - deltay
dinc1 : deltax Shl 1
dinc2 : deltax - deltay shl 1
xinc1 : 0
xinc2 : 1
yinc1 : 1
yinc2 : 1
end
Make sure x and y move in the right directions
if x1 x2 then
begin
xinc1 : - xinc1
xinc2 : - xinc2
end
if y1 y2 then
begin
yinc1 : - yinc1
yinc2 : - yinc2
end
Start drawing at
x : x1
y : y1
Draw the pixels
for i : 1 to numpixels do
begin
Imageyx.Char:C
Imageyx.Attr:Color
if d 0 then
begin
d : d + dinc1
x : x + xinc1
y : y + yinc1
end
else
begin
d : d + dinc2
x : x + xinc2
y : y + yinc2
end
end
end
by denthor of asphyxia. at the end you can find two procedures i use for
circles and lines in the blockart editor :
W E L C O M E
To the VGA Trainer Program
By
DENTHOR of ASPHYXIA
updated by Snowman
-- PART 3 --
Note: things in brackets have been added by Snowman. The original text
has remained mostly unaltered except for the inclusion of C++ material
Introduction
Greetings! This is the third part of the VGA Trainer series! Sorry it
took so long to get out, but I had a running battle with the traffic
department for three days to get my car registered, and then the MailBox
went down. Ahh, well, life stinks. Anyway, today will do some things
vital to most programs : Lines and circles.
Watch out for next weeks part : Virtual screens. The easy way to
eliminate flicker, doubled sprites, and subjecting the user to watch
you building your screen. Almost every ASPHYXIA demo has used a virtual
screen with the exception of the SilkyDemo, so this is one to watch out
for. I will also show you how to put all of these loose procedures into
units.
If you would like to contact me, or the team, there are many ways you
can do it : 1 Write a message to Grant Smith in private mail here on
the Mailbox BBS.
2 Write a message here in the Programming conference here
on the Mailbox Preferred if you have a general
programming query or problem others would benefit from
3 Write to ASPHYXIA on the ASPHYXIA BBS.
4 Write to Denthor, Eze or Livewire on Connectix.
5 Write to : Grant Smith
P.O.Box 270 Kloof
3640
6 Call me Grant Smith at 73 2129 leave a message if you
call during varsity
NB : If you are a representative of a company or BBS, and want ASPHYXIA
to do you a demo, leave mail to me we can discuss it.
NNB : If you have done/attempted a demo, SEND IT TO ME! We are feeling
quite lonely and want to meet/help out/exchange code with other demo
groups. What do you have to lose? Leave a message here and we can work
out how to transfer it. We really want to hear from you!
Circle Algorithim
You all know what a circle looks like. But how do you draw one on the
computer?
You probably know circles drawn with the degrees at these points :
0
270 ----+---- 90
180
Sorry about my ASCII - ... anyway, Pascal and C++ dont work that way ...
they work with radians instead of degrees. You can convert radians to
degrees, but Im not going to go into that now. Note though that in Pascal
and C++, the circle goes like this :
270
180 ----+---- 0
90
Even so, we can still use the famous equations to draw our circle ...
You derive the following by using the theorem of our good friend
Pythagoras
Sin deg Y/R
Cos deg X/R
This is standard 8? maths ... if you havent reached that level yet,
take this to your dad, or if you get stuck leave me a message and Ill
do a bit of basic Trig with you. I aim to please -
Where Y your Y-coord
X your X-coord
R your radius the size of your circle
deg the degree
To simplify matters, we rewrite the equation to get our X and Y values :
Y R*Sindeg
X R*Cosdeg
This obviousy is perfect for us, because it gives us our X and Y co-ords to
put into our Putpixel routine see Part 1. Because the Sin and Cos
functions return a Real long value, we use a round function to transform
it into an Integer.
Pascal
Procedure Circle oX,oY,rad:integerCol:Byte
VAR deg:real
X,Y:integer
BEGIN
deg:0
repeat
X:roundrad*COS deg
Y:roundrad*sin deg
putpixel x+ox,y+oy,Col
deg:deg+0.005
until deg6.4
END
C++
void Circleint X, int Y, int rad, int col
float deg 0
do
X roundrad * cosdeg
Y roundrad * sindeg
Putpixel X+160, Y+100, col
deg + 0.005
while deg 6.4
In the above example, the smaller the amount that deg is increased by,
the closer the pixels in the circle will be, but the slower the procedure.
0.005 seem to be best for the 320x200 screen. NOTE : ASPHYXIA does not use
this particular circle algorithm, ours is in assembly language, but this
one should be fast enough for most. If it isnt, give us the stuff you are
using it for and well give you ours.
Line algorithms
There are many ways to draw a line on the computer. I will describe one
and give you two. The second one you can figure out for yourselves it
is based on the first one but is faster
The first thing you need to do is pass what you want the line to look
like to your line procedure. What I have done is said that x1,y1 is the
first point on the screen, and x2,y2 is the second point. We also pass the
color to the procedure. Remember the screens top left hand corner is 0,0
see Part 1
Ie. o X1,Y1
ooooooooo
ooooooooo
oooooooo X2,Y2
Again, sorry about my drawings -
To find the length of the line, we say the following :
Pascal
XLength ABS x1-x2
YLength ABS y1-y2
C++
xlength absx1-x2
ylength absy1-y2
The ABS function means that whatever the result, it will give you an
absolute, or posotive, answer. At this stage I set a variable stating
wheter the difference between the two xs are negative, zero or posotive.
I do the same for the ys If the difference is zero, I just use a loop
keeping the two with the zero difference posotive, then exit.
If neither the xs or ys have a zero difference, I calculate the X and Y
slopes, using the following two equations :
Pascal
Xslope Xlength / Ylength
Yslope Ylength / Xlength
C++
Note: C++ is significantly different here. I had to add special divide
by zero checking. C++ doesnt like x/0. I also added in type-casting
which might not have been necessary. :
if xlength ! 0 ylength ! 0
xslope floatxlength/floatylength
yslope floatylength/floatxlength
else
xslope 0.0
yslope 0.0
As you can see, the slopes are real numbers.
NOTE : XSlope 1 / YSlope
Now, there are two ways of drawing the lines :
X XSlope * Y
Y YSlope * X
The question is, which one to use? if you use the wrong one, your line
will look like this :
o
o
o
Instead of this :
ooo
ooo
ooo
Well, the solution is as follows :
If the slope angle is in the area of the stars * then use the first
equation, if it is in the other section then use the second one.
What you do is you calculate the variable on the left hand side by
putting the variable on the right hand side in a loop and solving. Below
is our finished line routine :
Pascal
Procedure Line x1,y1,x2,y2:integercol:byte
VAR x,y,xlength,ylength,dx,dy:integer
xslope,yslope:real
BEGIN
xlength:abs x1-x2
if x1-x20 then dx:-1
if x1-x20 then dx:0
if x1-x20 then dx:+1
ylength:abs y1-y2
if y1-y20 then dy:-1
if y1-y20 then dy:0
if y1-y20 then dy:+1
if dy0 then BEGIN
if dx0 then for x:x1 to x2 do
putpixel x,y1,col
if dx0 then for x:x2 to x1 do
putpixel x,y1,col
exit
END
if dx0 then BEGIN
if dy0 then for y:y1 to y2 do
putpixel x1,y,col
if dy0 then for y:y2 to y1 do
putpixel x1,y,col
exit
END
xslope:xlength/ylength
yslope:ylength/xlength
if yslope/xslope1 and yslope/xslope-1 then BEGIN
if dx0 then for x:x1 to x2 do BEGIN
y: round yslope*x
putpixel x,y,col
END
if dx0 then for x:x2 to x1 do BEGIN
y: round yslope*x
putpixel x,y,col
END
END
ELSE
BEGIN
if dy0 then for y:y1 to y2 do BEGIN
x: round xslope*y
putpixel x,y,col
END
if dy0 then for y:y2 to y1 do BEGIN
x: round xslope*y
putpixel x,y,col
END
END
END
C++
void Line2int x1, int y1, int x2, int y2, int col
int x, y, xlength, ylength, dx, dy
float xslope, yslope
xlength absx1-x2
if x1-x2 0 dx -1
if x1-x2 0 dx 0
if x1-x2 0 dx +1
ylength absy1-y2
if y1-y2 0 dy -1
if y1-y2 0 dy 0
if y1-y2 0 dy +1
if dy 0
if dx 0
for xx1 xx2+1 x++
Putpixel x,y1,col
if dx 0
for xx2 xx1+1 x++
Putpixel x,y1,col
if dx 0
if dy 0
for yy1 yy2+1 y++
Putpixel x1,y,col
if dy 0
for yy2 yy1+1 y++
Putpixel x1,y,col
if xlength ! 0 ylength ! 0
xslope floatxlength/floatylength
yslope floatylength/floatxlength
else
xslope 0.0
yslope 0.0
if xslope ! 0 yslope ! 0
yslope/xslope 1 yslope/xslope -1
if dx 0
for xx1 xx2+1 x++
y round yslope*x
Putpixel x,y,col
if dx 0
for xx2 xx1+1 x++
y round yslope*x
Putpixel x,y,col
else
if dy 0
for yx1 yx2+1 y++
x round xslope*y
Putpixel x,y,col
if dy 0
for yx2 yx1+1 y++
x round xslope*y
Putpixel x,y,col
Quite big, isnt it? Here is a much shorter way of doing much the same
thing :
Pascal
function sgna:real:integer
begin
if a0 then sgn:+1
if a0 then sgn:-1
if a0 then sgn:0
end
procedure linea,b,c,d,col:integer
var u,s,v,d1x,d1y,d2x,d2y,m,n:real
i:integer
begin
u: c - a
v: d - b
d1x: SGNu
d1y: SGNv
d2x: SGNu
d2y: 0
m: ABSu
n : ABSv
IF NOT MN then
BEGIN
d2x : 0
d2y : SGNv
m : ABSv
n : ABSu
END
s : INTm / 2
FOR i : 0 TO roundm DO
BEGIN
putpixela,b,col
s : s + n
IF not sm THEN
BEGIN
s : s - m
a: a +roundd1x
b : b + roundd1y
END
ELSE
BEGIN
a : a + roundd2x
b : b + roundd2y
END
end
END
C++
int sgn long a
if a 0 return +1
else if a 0 return -1
else return 0
void Lineint a, int b, int c, int d, int col
long u,s,v,d1x,d1y,d2x,d2y,m,n
int i
u c-a
v d-b
d1x sgnu
d1y sgnv
d2x sgnu
d2y 0
m absu
n absv
if mn
d2x 0
d2y sgnv
m absv
n absu
s intm / 2
for i0iroundmi++
Putpixela,b,col
s + n
if s m
s - m
a + d1x
b + d1y
else
a + d2x
b + d2y
This routine is very fast, and should meet almost all of your requirements
ASPHYXIA used it for quite a while before we made our new one.
In the end program, both the new line routine and the circle routine are
tested. A few of the procedures of the first parts are also used.
Line and circle routines may seem like fairly trivial things, but they are
a vital component of many programs, and you may like to look up other
methods of drawing them in books in the library I know that here at the
varsity they have books for doing this kind of stuff all over the place
A good line routine to look out for is the Bressenhams line routine ...
there is a Bressenhams circle routine too ... I have documentaiton for them
if anybody is interested, they are by far some of the fastest routines
you will use.
In closing
Varsity has started again, so I am shock going to bed before three in
the morning, so my quote this week wasnt written in the same wasted way
my last weeks one was For last weeks one, I had gotten 8 hours sleep in
3 days, and thought up and wrote the quote at 2:23 am before I fell asleep.
What does it do? she asks.
Its a computer, he replies.
Yes, dear, but what does it do?
It ..er.. computes! Its a computer.
What does it compute?
What? Er? Um. Numbers! Yes, numbers! He smiles
worriedly.
Why?
Why? Well ..um.. why? He starts to sweat.
I mean, is it just something to dust around, or does
it actually do something useful?
Um...you can call other computers with it! Hope lights
up his eyes. So you can get programs from other computers!
I see. Tell me, what do these programs do?
Do? I dont think I fol...
I see. They compute. Numbers. For no particular reason. He
withers under her gaze.
Yes, but...
She smiles, and he trails off, defeated. She takes another look
at the thing. Although, she says, with a strange look in
her eyes. He looks up, an insane look of hope on his
face. Does it come in pink? she asks.
- Grant Smith
Tue 27 July, 1993
9:35 pm.
See you next time,
- Denthor
circle and line procedures by xqtr
Procedure Circle1x1, y1, x2, y2 : integer color : byte Var Image: TScreenImage
Var
r,y,x,d : Integer
Begin
r : x2 - x1 Div 2
y : r
d : -r
Imagey1x1+r.Char:*
For x : 10 to truncr / Sqrt2*10 Do Begin
d : d + 2*x-1
if d 0 Then Begin
y : y - 1
d : d - 2* y
End
Imagey1+yx1+xChar:*
End
End
procedure Linex1, y1, x2, y2 : integer color : byte Var Image: TScreenImage C:Char
var i, deltax, deltay, numpixels,
d, dinc1, dinc2,
x, xinc1, xinc2,
y, yinc1, yinc2 : integer
begin
Calculate deltax and deltay for initialisation
deltax : absx2 - x1
deltay : absy2 - y1
Initialize all vars based on which is the independent variable
if deltax deltay then
begin
x is independent variable
numpixels : deltax + 1
d : 2 * deltay - deltax
dinc1 : deltay Shl 1
dinc2 : deltay - deltax shl 1
xinc1 : 1
xinc2 : 1
yinc1 : 0
yinc2 : 1
end
else
begin
y is independent variable
numpixels : deltay + 1
d : 2 * deltax - deltay
dinc1 : deltax Shl 1
dinc2 : deltax - deltay shl 1
xinc1 : 0
xinc2 : 1
yinc1 : 1
yinc2 : 1
end
Make sure x and y move in the right directions
if x1 x2 then
begin
xinc1 : - xinc1
xinc2 : - xinc2
end
if y1 y2 then
begin
yinc1 : - yinc1
yinc2 : - yinc2
end
Start drawing at
x : x1
y : y1
Draw the pixels
for i : 1 to numpixels do
begin
Imageyx.Char:C
Imageyx.Attr:Color
if d 0 then
begin
d : d + dinc1
x : x + xinc1
y : y + yinc1
end
else
begin
d : d + dinc2
x : x + xinc2
y : y + yinc2
end
end
end
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