New Draw Module

[copthesaint]

I have been working on something to replace the Turing Draw Module with something that will hopefully be an improvement. The whole idea is to use a picture as the buffer for the screen, and since you can edit the values of the pixels in Turing in 24bit format, I thought it would be awesome if people could ditch the crappy 8 bit colors that are given in the Draw Module and replace it with this! The reason why I posted this in the help, is because I have never attempted making drawing methods before and I would like some help. Currently I added a few fun things into the class like Color Filter, Color Inverting. If anyone can help, it would be appreciated.

Turing:

View.Set ("Graphics:640;440,offscreenonly,nobuttonbar") type ColorValues :     record         CLR : array 1 .. 3 of int     end record type PictureVals :     record         RGB : ColorValues         RGBBuffer : ColorValues     end record class Layer     import ColorValues, PictureVals     var colorBack : ColorValues     colorBack.CLR (1) := 255 % Blue     colorBack.CLR (2) := 255 % Green     colorBack.CLR (3) := 255 % Red end Layer module Render     import ColorValues, PictureVals     export Update, ColorBack, DrawDot, Initialize, DrawRect, ColorInvert, ColorInvertArea, Cls, ColorFilter, ColorFilterArea, DrawLine     var imageX : int := 0     var imageY : int := 0     var pictureBound : int := 0     var picture : flexible array 1 .. 0 of PictureVals     var colorBack : ColorValues     colorBack.CLR (1) := 0 % Blue     colorBack.CLR (2) := 0 % Green     colorBack.CLR (3) := 0 % Red     procedure NewLayer     end NewLayer     procedure Initialize (sizeX, sizeY : int)         pictureBound := sizeX * sizeY         if pictureBound > upper (picture) then             new picture, pictureBound         end if         for i : 1 .. pictureBound             picture (i).RGB := colorBack         end for         imageX := sizeX         imageY := sizeY         put picture (upper (picture)).RGB.CLR (1)     end Initialize     procedure ColorBack (R, B, G : int)         if R < 256 and B < 256 and G < 256 and R > -1 and B > -1 and G > -1 then             colorBack.CLR (1) := B             colorBack.CLR (2) := G             colorBack.CLR (3) := R         end if         Error.Halt ("Color Must stay within the range of 0-255.")     end ColorBack     procedure ColorInvert         for i : 1 .. pictureBound             picture (i).RGB.CLR (1) := 255 - picture (i).RGB.CLR (1)             picture (i).RGB.CLR (2) := 255 - picture (i).RGB.CLR (2)             picture (i).RGB.CLR (3) := 255 - picture (i).RGB.CLR (3)         end for     end ColorInvert     procedure ColorInvertArea (x1, y1, x2, y2 : int)         for i : x1 .. x2             for j : y1 .. y2                 if i < imageX and i > -1 and j < imageY and j > -1 then                     picture ((j - 1) * imageX + i).RGB.CLR (1) := 255 - picture ((j - 1) * imageX + i).RGB.CLR (1)                     picture ((j - 1) * imageX + i).RGB.CLR (2) := 255 - picture ((j - 1) * imageX + i).RGB.CLR (2)                     picture ((j - 1) * imageX + i).RGB.CLR (3) := 255 - picture ((j - 1) * imageX + i).RGB.CLR (3)                 end if             end for         end for     end ColorInvertArea     procedure ColorFilter (R, G, B : int, p : real)         for i : 1 .. pictureBound             picture (i).RGB.CLR (1) := (picture (i).RGB.CLR (1) * (1.0 - p) + B * p) div 2             picture (i).RGB.CLR (2) := (picture (i).RGB.CLR (2) * (1.0 - p) + G * p) div 2             picture (i).RGB.CLR (3) := (picture (i).RGB.CLR (3) * (1.0 - p) + R * p) div 2         end for     end ColorFilter     procedure ColorFilterArea (x1, y1, x2, y2, R, G, B : int, p : real)         for i : x1 .. x2             for j : y1 .. y2                 if i < imageX + 1 and i > -1 and j < imageY + 1 and j > -1 then                     picture ((j - 1) * imageX + i).RGB.CLR (1) := (picture ((j - 1) * imageX + i).RGB.CLR (1) * (1.0 - p) + B * p) div 2                     picture ((j - 1) * imageX + i).RGB.CLR (2) := (picture ((j - 1) * imageX + i).RGB.CLR (2) * (1.0 - p) + G * p) div 2                     picture ((j - 1) * imageX + i).RGB.CLR (3) := (picture ((j - 1) * imageX + i).RGB.CLR (3) * (1.0 - p) + R * p) div 2                 end if             end for         end for     end ColorFilterArea         procedure DrawDot (x, y, R, G, B : int)         if x < imageX + 1 and x > -1 and y < imageY + 1 and y > -1 then             picture ((y - 1) * imageX + x).RGB.CLR (1) := B             picture ((y - 1) * imageX + x).RGB.CLR (2) := G             picture ((y - 1) * imageX + x).RGB.CLR (3) := R         end if     end DrawDot     procedure DrawLine (x1, y1, x2, y2, R, G, B : int)         var distX : real := x2 - x1 % 5         var distY : real := y2 - y1 % 10         var totDist : real := sqrt (distX * distX + distY * distY)         distX := distX / distY % 0.5         distY := 1 % 1         var inc : real := 0         for y : 0 .. round (totDist) %25 + 100 = 11.18                DrawDot (x1 + round (inc), y1 + y, round (R * (inc rem 1)), round (G* (inc rem 1)), round (B* (inc rem 1)))             DrawDot (x1 + round (inc + (1-(inc rem 1))), y1 + y, round (R * (inc rem 1)), round (G* (inc rem 1)), round (B* (inc rem 1)))             inc := inc + distX         end for     end DrawLine     procedure DrawRect (x1, y1, x2, y2, R, G, B : int)         for i : x1 .. x2             for j : y1 .. y2                 DrawDot (i, j, R, G, B)             end for         end for     end DrawRect     procedure Update         var b : array 1 .. sizepic (1, 1, imageX, imageY) of int         cls         takepic (1, 1, imageX, imageY, b)         for i : 19 .. upper (b)             b (i) := 0         end for         b (7) := 48         var id := 19         var imageID : int := 0         var n : int := 0         for j : 1 .. imageY             for i : 1 .. imageX                 imageID := imageID + 1                 for p : 1 .. 3                     b (id)| = picture (imageID).RGB.CLR (p) shl n     %b                     if n < 24 then                         n += 8                     else                         n := 0                         id += 1                     end if                 end for             end for             if n > 0 then                 id += 1                 n := 0             end if         end for         drawpic (1, 1, b, 1)         var pic1 : int := Pic.New (1, 1, imageX, imageY)         var pic2 : int := Pic.Scale (pic1, maxx, maxy)         Pic.Draw (pic2, 0, 0, picCopy)         View.Update         Pic.Free (pic1)         Pic.Free (pic2)     end Update     procedure Cls         for i : 1 .. pictureBound             picture (i).RGB := colorBack         end for     end Cls end Render Render.Initialize (maxx div 2, maxy div 2) Render.DrawDot (maxx div 2, maxy div 2, 255, 255, 255) Render.DrawDot (1, 1, 255, 255, 255) Render.DrawLine (1, 1, maxx div 2, 50, 255,255, 255) Render.DrawRect (32, 30, 50, 50, 255, 55, 55) Render.ColorInvertArea (20, 20, 40, 40) Render.ColorFilterArea (50, 50, 256, 256, 55, 250, 105, 0.2) Render.Update

[evildaddy911]

i wouldnt mind some rounded rectangles, such as:

Turing:

proc drawfillroundbox (x1, y1, x2, y2, xr, yr, clr : int)     var minx : int := min (x1, x2)     var miny : int := min (y1, y2)     var Maxx : int := max (x1, x2)     var Maxy : int := max (y1, y2)     var XR : int := min (Maxx - minx, xr)   %   so it doesnt end up drawing a box     var YR : int := min (Maxy - miny, yr)     drawfillarc (minx + XR, miny + YR, XR, YR, 180, 270, clr)     drawfillarc (minx + XR, Maxy - YR, XR, YR, 90, 180, clr)     drawfillarc (Maxx - XR, miny + YR, XR, YR, 270, 360, clr)     drawfillarc (Maxx - XR, Maxy - YR, XR, YR, 0, 90, clr)     drawfillbox (minx, miny + YR, Maxx, Maxy - YR, clr)     drawfillbox (minx + XR, miny, Maxx - XR, Maxy, clr) end drawfillroundbox proc drawroundbox (x1, y1, x2, y2, xr, yr, clr : int)     var minx : int := min (x1, x2)     var miny : int := min (y1, y2)     var Maxx : int := max (x1, x2)     var Maxy : int := max (y1, y2)     var XR : int := min (Maxx - minx, xr)   %   so it doesnt end up drawing a box     var YR : int := min (Maxy - miny, yr)     drawarc (minx + XR, miny + YR, XR, YR, 180, 270, clr)     drawarc (minx + XR, Maxy - YR, XR, YR, 90, 180, clr)     drawarc (Maxx - XR, miny + YR, XR, YR, 270, 360, clr)     drawarc (Maxx - XR, Maxy - YR, XR, YR, 0, 90, clr)     drawline (minx + XR, miny, Maxx - XR, miny, clr)     drawline (minx + XR, Maxy, Maxx - XR, Maxy, clr)     drawline (minx, miny + YR, minx, Maxy - YR, clr)     drawline (Maxx, miny + YR, Maxx, Maxy - YR, clr) end drawroundbox

edit for your module's style of course

also, the program either crashes ("segment violation" on line 163), says Halt is not on the export list of Error, or crashes Turing because of a bug in the environment
(depending on the version)

ps, what will newLayer do?

[copthesaint]

I have updated the module,
Currently you can draw:
- Ovals
- FillRectangles
- lines (Problem with rounding I think, if anyone wants to look at that.)
- polygons (Problem with lines causes problem with polygons)
Currently you can also:
-Invert Colors
-Fixed Color Filter

I wish I could have done this sooner but I have been busy.

@evildaddy that can definatly be done but right now Im just figuring the math out behind the draw modules. Its something ive never spent the time to learn or figure out.
@compsci, if anyone wants anything like evildaddy Its more then fine to post it here, but right now I just need reasorces for the mathematics of drawing objects by each individual pixel.

[Amarylis]

As I mentioned in the other post, I'd love 2D array buffers as opposed to 1D array buffers

[copthesaint]

Although I could do that, it would be useless because turing doesnt allow flexible arrays of fleixble array. plus manipulating a one dimensional array that holds only values specific to the pixel at that index is a hell of alot easier then dealing with a 1 dimensional buffer that holds more then one pixels values at a certain index point.

[Raknarg]

@copthesaint

var arr : flexible array 1 .. 3, 1 .. 3 of int

new arr, 2, 3

Works as long as you only change the first element. Maybe that helps.

[Amarylis]

Works with both of them

new ar, 1, 0 chances the array bounds to 1, 0

[Raknarg]

You're sure it works? Whenever I try to change the second element, it gives me an error, saying that it hasnt been implemented yet

[Amarylis]

What version of Turing are you using?

[Dreadnought]

http://compsci.ca/holtsoft/doc/flexible.html wrote:

In the current implementation (1999), with a multi-dimensional array with a non-zero number of total elements, it is a run-time error to change any but the first dimension (unless one of the new upper bounds is one less than the corresponding lower bound, giving 0 elements in the array) as the algorithm to rearrange the element memory locations has not yet been implemented.

Turing:

% Basically var test : flexible array 1..3, 1..3 of int new test, 4, 3  % Is allowed since only the first dimension in modified new test, 0, 0  % Are all allowed since the upper bound of at least one  new test, 3, 0  % of the dimensions is lower than its lower bounds new test, 5, 0  % (this creates an array of size 0) new test, 3, 6  % Are not allowed since a dimension other than the first new test, 2, 9  % (the second in this case) is modified without producing an array of size 0

Hope this clear things up.

[Raknarg]

So basically you can either make it smaller or change the first one.

[Dreadnought]

There is only one kind of "smaller" that is allowed and that is size 0.

Note that a size 0 flexible array can be changed into an array of any size you desire.

[copthesaint]

You cannot do flexible array of flexible arrays.

Its not happening. The buffer for drawpic is an array 1 .. * of int, It would be completly pointless to convert the values then set them to a 2D array so then it can be set as a 1D array for the buffer again. Plus I already have done the math so you Can draw as if it were a 2D array...

for the amount of work for this to work would be ridiculous since you can already, easily used a single dimension array. :

Turing:

var x : flexible array 1 .. 0, 1 .. 0 of int new x , 1,1 x(1,1) := 0 put x(1,1) new x , 1,2 x(1,2) := 0 put x(1,2)

[Amarylis]

Dreadnought @ Thu May 10, 2012 1:49 pm wrote:

http://compsci.ca/holtsoft/doc/flexible.html wrote:

In the current implementation (1999), with a multi-dimensional array with a non-zero number of total elements, it is a run-time error to change any but the first dimension (unless one of the new upper bounds is one less than the corresponding lower bound, giving 0 elements in the array) as the algorithm to rearrange the element memory locations has not yet been implemented.

Turing:

% Basically var test : flexible array 1..3, 1..3 of int new test, 4, 3  % Is allowed since only the first dimension in modified new test, 0, 0  % Are all allowed since the upper bound of at least one  new test, 3, 0  % of the dimensions is lower than its lower bounds new test, 5, 0  % (this creates an array of size 0) new test, 3, 6  % Are not allowed since a dimension other than the first new test, 2, 9  % (the second in this case) is modified without producing an array of size 0

Hope this clear things up.

I tried using this:

[syntax="turing]var ar : flexible array 1 .. 0, 1 .. 0 of int

new ar, 1, 1
new ar, 1, 5

put upper (ar, 2)[/syntax]

Didn't give me any errors

[Dreadnought]

I feel bad for steering this post off-topic, but your code (copy-pasted as is) produces the same error in both Turing 4.0.5, 4.1.1 and 4.1.2. What version might you be using?