Use GraphicsPath. Unfortunately you can't serialize a GraphicsPath oblect, but you can save the PathPoints[^] array, PathTypes[^] array and the FillMode[^], and then restore the GraphicsPath by the GraphicsPath Constructor (Point[], Byte[], FillMode)[^]. Also keep extra info for the colors of pens or brushes etc, that you use to draw or fill a GraphicsPath.

As far as i know, vb uses the SafeArray (UnmanagedType enumeration[^])

You dont have problem with existing languages like c++.net or vb.net. But someone may develop a language that uses .net components, without support for unsigned integers. For such senarios stands the CLS.

Perhaps because uint type is not CLS[^]-compliant.

aksgh wrote:

Its a huffman tree.

Yes didnt see that, you are right. So, for a huffman tree you could the following

typedef struct _node_info {
int is_leaf;
char value;
}node_info;

typedef struct _node {
node_info inf;
node* l;
node* r;
} node;

// returns the root of the tree, from an array of node_info's structs, that
// represents the preorder traversal of the huffman tree. we assume a global stack s
// initially empty
node* built_tree(node_info* preorder_array, int preorder_array_size)
{
node *root, *n;
int i;
root = malloc(sizeof(node));
stack_push(s, root);
for (i = 0; i < preorder_array_size; i++) {
assert(!stack_empty(s));
n = stack_pop(s);
n->inf = preorder_array[i];
if (!n->inf.is_leaf) {
n->r = malloc(sizeof(node));
stack_push(s, n->r);
n->l = malloc(sizeof(node));
stack_push(s, n->l);
}
}
assert(stack_empty(s));
return root;
}

* i dont set r,l members to zero when i create a node, in order to save some space, but it is something you must do

modified on Friday, September 19, 2008 8:09 AM

You may get the same sequence from different binary trees. So its not possible.

You can use a pen instead. First adjust the width of your pen. Then set StartCap and EndCap of the pen to LineCap.Round. Finally draw lines, instead of 'points', from the previous point to the current point, in the mouse move event.

First four bytes of the header -> size of header

Member Function Pointers and the Fastest Possible C++ Delegates[^]

If you scroll down at the list[^] youll see the LOCALE_SISO639LANGNAME2 which is for the three letter iso, but its only for >=Vista. The xp compatible LOCALE_SISO639LANGNAME its the two letter version.

Set BufferDescription's GlobalFocus property to true

Use the following class

using System;
using System.Drawing;
using System.Drawing.Drawing2D;

namespace some_namespace
{
class Transformation
{
// world: the real world your bitmap represents
// bitmap: your (unscaled) bitmap
// screen: the portion of the screen that you display the image (a picture box maybe)

    Matrix matrix\_world;       // bitmap to world
    Matrix inv\_matrix\_world;   // world to bitmap
    Matrix matrix\_bitmap;      // screen to bitmap
    Matrix inv\_matrix\_bitmap;  // bitmap to screen

    Size bitmap\_size;
    Size screen\_size;


    public Transformation(PointF worldOrigin, SizeF worldSize, 
        Size bitmapSize, Size screenSize)
    {
        matrix\_world = new Matrix();
        matrix\_bitmap = new Matrix();
        bitmap\_size = bitmapSize;
        screen\_size = screenSize;

        // if you want the native y-points up coord system for the world, uncomment '-' and
        // '+ worldSize.Height' below
        matrix\_world.Scale(worldSize.Width / (float)bitmap\_size.Width, 
           /\* - \*/ worldSize.Height / (float)bitmap\_size.Height);                 
        matrix\_world.Translate(worldOrigin.X, 
            worldOrigin.Y /\* + worldSize.Height \*/, MatrixOrder.Append);
        inv\_matrix\_world = matrix\_world.Clone();
        inv\_matrix\_world.Invert();
        Reset();
    }

    // Reset transformation to the 'screen portion displays entire bitmap'
    public void Reset()
    {
        matrix\_bitmap.Reset();
        matrix\_bitmap.Scale((float)bitmap\_size.Width / (float)screen\_size.Width, 
            (float)bitmap\_size.Height / (float)screen\_size.Height);
        inv\_matrix\_bitmap = matrix\_bitmap.Clone();
        inv\_matrix\_bitmap.Invert();
    }

    // transform to 'screen portion displays the rectangular portion of the bitmap 
    // defined by the displayOrigin point and the displaySize'
    public void Transform(PointF displayOrigin, SizeF displaySize)
    {
        matrix\_bitmap.Reset();
        matrix\_bitmap.Scale(displaySize.Width / (float)screen\_size.Width, 
            displaySize.Height / (float)screen\_size.Height);
        matrix\_bitmap.Translate(displayOrigin.X, displayOrigin.Y, MatrixOrder.Append);
        inv\_matrix\_bitmap = matrix\_bitmap.Clone();
        inv\_matrix\_bitmap.Invert();
    }

You want to fit a 25000x25000 bitmap in a 1000x1000 bitmap without losing precision? :wtf:

How to: Draw Images with Transparency[^]

You must declare extern the g_Info in the MyInfoArrayClass.h header, under the declaration of the class (or add a forward declaration of your class above the const extern CMyInfoArrayClass g_Info;).

Googling 'plane equation three points' ... 1st result : Equation of a plane[^] Googling 'projection point plane' ... 3rd result : Projection of a point on a plane[^] You could do this, dont you?

raesa wrote:

If i take the min/max of all the three dimesnsions(x,y&z) i might get 6 different points (3points for minimum values of x/y/z and 3 for maximum values of x/y/z), right? Do you suggest me that i will have to compare and find out if my new point cordinates(x1, y1, z1) is either greater or less than the any of the (x,y,z)values of the 6 points? Which means that if any one value (either x1/y1/z1) is greater than the (x/y/z) of the 6 points, the new point lies outside the point cloud.

What I understood, is that Cedric suggested to test if the point p(x,y,z) is in the bounding box of this set of points. And you do this by just comparing if((x>=XMIN)&&(x<=XMAX)&&(y>=YMIN)&&(y<=YMAX)&&(z>=ZMIN)&&(z<=ZMAX)). So, if by the word "cloud" you mean bounding box, it's ok. If you mean "convex hull" it's a quite different problem. And although the 2d version of this problem is a common topic for an introductive to algorithms class, I don’t know much about the 3d case. Definitely this question belongs in an algorithms forum. Meanwhile search for “convex hull from set of 3d points”, and “point in convex hull of 3d points” algorithms. my intuition: linear programming problem

modified on Thursday, August 28, 2008 1:56 PM

1. With a System.Timers.Timer just use the Stop() method to stop the timer. 2. With a System.Threading.Timer, just don't use this timer (why? read documentation). Replace with a System.Timers.Timer 3. For good performance forget timers. Force rendering into app's main loop (as Greeeg mentioned)

modified on Wednesday, August 27, 2008 8:03 AM

What are you trying to do? (probably double buffering is the answer to your problem)

Your are welcome From the msdn (WM_COPYDATA) : "The receiving application should consider the data read-only. The lParam parameter is valid only during the processing of the message. The receiving application should not free the memory referenced by lParam. If the receiving application must access the data after SendMessage returns, it must copy the data into a local buffer. " From the msdn (SendMessage) : "The SendMessage function sends the specified message to a window or windows. It calls the window procedure for the specified window and does not return until the window procedure has processed the message. "

modified on Monday, August 25, 2008 1:08 PM