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MossyMossy

Tue Oct 20, 2009 7:45 pm


column problem.

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here is my code so far.



var column :int
locate (21, 1)
for count : 1..80

put "-"..
end for

loop
locate (23, 1)
put " "
locate (23, 1)

put "Choose a column for the line between 1 and 80, and with -1:"..

get column
exit when column = -1

for row : 1..20

locate (row, column)
put "+"..
end for
end loop


After doing this, I have to make it on a slant now,

I CANNOT figure out how to make a Decreasing angle, like a slant, so it go's down like " \ " instead of "|'

Please help.


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copthesaint

Wed Oct 21, 2009 12:16 am


RE:column problem.

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*deletia*

Mod Edit: Don't just give the answer. kthxbai


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Zren

Wed Oct 21, 2009 12:56 am


Re: column problem.

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BASIC:
This won't make an exact line, but it'll suffice for this.
Think about what column you start, then which you'll end in. Then find the difference (delta).

end - start = difference
col6 - col4 = 2 columns apart

Then consider the fact that a straight column ( | ) has a diffence of 0. We then think about the previous column (diff of 2) and realize we'll want to stretch the line over a total of 3 column values (4, 5, 6). So we'll be breaking it down into 3 different smaller lines (or breaking it up 2 times). We want to find out the y-value where we move to the next column.

http://fractalized.ca/gallery/breaking_up_a_line.png

Figured it out?


ADVANCED:
If I were you, I'd try this first with Draw.Dot() before implementing it into your program. However, here's how to essentially draw a Aliased line. If you need help with how the two are relevant or any of the math, just ask.
How do I draw a line?
You can use the Windows GDI to draw a line as DirectDraw doesn't provide a line drawing function. If you want to draw a line using DirectX then use Breshenham's Line Drawing Algorithm to draw lines by joining Pixels.

The pseudocode to draw a line between (x1, y1) and (x2, y2)

a) Assume x=x1 and y=y1. Plot (x, y) Pixel

b) Find the largest among (x2-x1) and (y2-y1) (To find the line is predominant on which axis)

c) If(x2-x1) is the biggest (Line is predominant on X Axis)



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MossyMossy

Wed Oct 21, 2009 8:16 am


Re: column problem.

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I've got it set for the user's input to make it for the vertical line, now i need to input it so that the user can select the row again but make it vertical.


var column : int                                            
    locate (21, 1)                                          
                                                            
for count : 1..80                                          
    put "-"..                                               
                                                            
end for   
   
    loop
    locate (23, 1)
    put " "
    locate (23, 1)
    put "Please enter a column for the line between 1 and 80, end it with -1:"..
get column
    exit when row > 25 or column > 80
       
            locate (row, column)
        put "+"..
    
    end for
    end loop
             


I've got the vertical line down correct, but im stuck on the mathematical problem that is here,


var row, column:int := 1

cls 
                                           
    locate (21, 1)                                         
                                                            
for count : 1..80                                           
    put "-"..                                               
                                                            
 end for
    
put "please enter a row"
get row
put "please enter column"
get column


loop 
exit when row > 25

    locate (column, row) 

    put "*" .. 

row := row +1
column := column +1
end loop



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Zren

Wed Oct 21, 2009 12:42 pm


Re: column problem.

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Right now you have the variables row and column. or x1,y1. Now you automatically assume that the line is going have a "width" of 25 or so (25-row actually) since that's how many times you repeat column+1. First we want to find out column and row of where we will stop (aka x2, y2).


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MossyMossy

Wed Oct 21, 2009 1:02 pm


Re: column problem.

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Right now you have the variables row and column. or x1,y1. Now you automatically assume that the line is going have a "width" of 25 or so (25-row actually) since that's how many times you repeat column+1. First we want to find out column and row of where we will stop (aka x2, y2).

now how can i find that,

make a variable called width?

then subtract the column with row?


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Zren

Wed Oct 21, 2009 2:45 pm


Re: column problem.

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

for x : x1 .. x2
    y := slope * x
end for

Slope is also usually in the form of a decimal. Also, this is for a line that is dependant on X (or Column). You need a formula that depends on Y (Rows).
hint: I've already posted the difference.