A little help with this function problem.

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Hey fellas and/or ladies. Quick couple. I have to write a function that draws an octagon. It is 40 by 40 pixels. This function should accept arguments for x and y coordinates for the center point of the octagon. The function ultimately should draw the octagon with the center point located athose coordinates.

Next I have to call this function 6 times in order to draw a pattern.

Can anyone show me how to do this exactly? I was told by my professor to draw dotted lines horizontally and vertically in fourths. Since the radius is 40, each dotted line represents 20 of the 40. I would have to add or subract from the center point to figure out the x,y coordinates for each point. However I have no idea how to go about this. I have Visual 2010 with the DarkGDK header file which allows me to draw these shapes. I can draw up the octagon but don't know how to get those points.

This is my 2nd time posting b/c my earlier question wasn't very clear. Please and thank you :)

Posted 23-Sep-11 6:04am

Member 8210099

Updated 23-Sep-11 6:13am

André Kraak

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André Kraak 23-Sep-11 12:13pm

It would be best if you did your own homework, it is given to you so that you will learn something. Read your text books and give it a try.

Simon Bang Terkildsen 23-Sep-11 12:37pm

please do not repost, click "improve question" to add more details.
Can't seem to figure this function question

Richard MacCutchan 23-Sep-11 13:13pm

Try drawing the shapes on a piece of paper so you can visualise what needs to be done to get the final result in the way you want. You can even use a ruler to get the different measurements. Converting that to code should then be much easier.

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Sergey Chepurin · Answer 1 · 2011-09-23T08:26:00

I am not sure if i understood the question correctly, but to calculate the vertices of octagon given the center(x,y) and radius R you can use this:

void Octagon(HDC hDC, int x, int y, int R)
{
int R2 = int(R/sqrt(2));

POINT pt[8];
pt[0].x = x; pt[0].y = y - R;
pt[1].x = x + R2 pt[1].y = y - R2;
pt[2].x = x + R; pt[2].y = y;
pt[3].x = x + R2; pt[3].y = y + R2;
pt[4].x = x; pt[4].y = y + R;
pt[5].x = x - R2; pt[5].y = y + R2;
pt[6].x = x - R; pt[6].y = y;
pt[7].x = x - R2; pt[7].y = y - R2;

Polygon(hDC, pt, 8);
}

(from codeguru.com/forum)

Now, you can draw any polygon (including octagon) on GDI+ with:

PointF myPointFArray[] = {PointF(1, 1), 
   PointF(20, 10), PointF(5, 4), PointF(100, 2),
   PointF(200, 50), PointF(39, 45)};
g->DrawPolygon(myPen, myPointFArray);

(Drawing Lines and Shapes with GDI+)

using sample Drawing a Line as a template.

Stefan_Lang · Answer 2 · 2011-09-23T06:20:00

Solution 1

http://en.wikipedia.org/wiki/Coordinate_system[^]

Posted 23-Sep-11 6:20am

Stefan_Lang