Flexible array help

[Zasalamel]

What is it you are trying to achieve?
a box drawing program


What is the problem you are having?
it will not draw the boxes


Describe what you have tried to solve this problem
i have tried moving for loops around


Post any relevant code (You may choose to attach the file instead of posting the code if it is too long)

Turing:

var box : flexible array 1 .. 0 of     record         x1 : int         y1 : int         x2 : int         y2 : int     end record for i : lower (box) .. upper (box)     box (i).x1 := 0     box (i).y1 := 0     box (i).x1 := 0     box (i).y1 := 0 end for var x1, x2, y1, y2, x, y, b : int fcn b10 (num : int) : int     result round (num / 10) * 10 end b10 loop     Mouse.Where (x, y, b)     if b = 0 then         x1 := b10 (x) - 5         y1 := b10 (y) - 5         y2 := b10 (y) + 5         x2 := b10 (x) + 5     end if     for i : lower (box) .. upper (box)         if b = 1 then             box (i).x1 := x1             box (i).y1 := y1             y2 := b10 (y) + 5             x2 := b10 (x) + 5             box (i).x2 := x2             box (i).y2 := y2         end if         if box (i).x1 ~= 0 then             new box, 1         end if         drawbox (box (i).x1, box (i).y1, box (i).x2, box (i).y2, black)     end for     drawbox (x1, y1, x2, y2, brightred)     View.Update %I realize that there is no offscreenonly above     cls end loop

Please specify what version of Turing you are using
4.1.1

[Superskull85]

From what I understood you want a program that will draw a on the screen, and that each box would be store in memory. Is this right?

If so then the problem is that you are overwriting the information for every box in memory, and that you are constantly setting the upper bound of the array to one (hence only having one box.

What you want to do is something like the following:

Turing:

%Initialize the array with length one so that it can be written to (ie. var Array : flexible array 1 .. 1 of int). %Create a procedure to initialize the upper bound when called (passing upper (box) as the index of a array will return the last element in the array) %Declare your variables and procedures %Initialize your array by calling the procedure you setup to do this loop cls     Mouse.Where (x, y, b)     if b = 0 then         x1 := b10 (x) - 5         y1 := b10 (y) - 5         y2 := b10 (y) + 5         x2 := b10 (x) + 5     end if     %Draw your boxes by iterating from 1 to upper (box)     if b = 1 then         %Set the upper element of your box array ( box (upper (box)) ) to the current location     end if     if box (upper(box)).x1 ~= 0 then         new box, upper (box) + 1         %Initialize the last element of your array by calling the procedure you setup to do this     end if     drawbox (x1, y1, x2, y2, brightred)     View.Update end loop

Hope that helped.