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Note: Visual Studio uses the naming convention for all event handlers consisting of the variable name, followed by an underscore, followed by the event name. 3 Add our code to handle the

De-packetizer Blu-ray Disc Demystified Matrix 0 2-ch FIFO Buffer MLP De-packetizer Data Matrix 1 6-ch De-packetizer FIFO Buffer FIFO Buffer De-packetizer Metadata Matrix 2 8-ch

As this examples shows, the keyword virtual precedes the rest of the inherited class s specification Now that both d1 and d2 have inherited base as virtual, any multiple inheritance involving them will cause only one copy of base to be present Therefore, in d3, there is only one copy of base, so di = 10 is perfectly valid and unambiguous One further point to keep in mind: Even though both d1 and d2 specify base as virtual, base is still present in any objects of either type For example, the following sequence is perfectly valid:

protected void btnLoad_Click(object sender, System.EventArgs e) { OpenFileDialog dlg = new OpenFileDialog(); dlg.Title = "Open Photo"; dlg.Filter = "jpg files (*.jpg)|*.jpg" + "|All files (*.*)|*.*"; if (dlg.ShowDialog() == DialogResult.OK) { pbxPhoto.Image = new Bitmap(dlg.OpenFile()); } dlg.Dispose(); }

// define a class of type d1 d1 myclass; myclassi = 100;

Virtual base classes and normal ones differ only when an object inherits the base more than once If virtual base classes are used, only one base class is present in the object Otherwise, multiple copies will be found

Note: Some of these lines do not fit this table. The dlg.Filter line, in particular, should be a single string. Here and throughout the book, we will reformat the code to fit the table in a way that is equivalent to the code in the online examples.

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Class member functions may be declared as const, which causes this to be treated as a const pointer Thus, a const function cannot modify the object that invokes it Also, a const object may not invoke a nonconst member function However, a const member function can be called by either const or nonconst objects To specify a member function as const, use the form shown in the following example

IMPLEMENT A CLICK HANDLER FOR THE BTNLOAD BUTTON (continued) Action 4 Set the SizeMode property for the PictureBox control to StretchImage.

class X { int some_var; public: int f1() const; // const member function };

BindingUnitData 1 1 new_trailer Fall2006mpls Root BDMV indexbdmv MovieObjectbdmv PLAYLIST CLIPINF 00000mpls 01000clpi 02000clpi STREAM 01000m2ts 02000m2ts AUXDATA soundbdmv 12345otf BDMV indexbdmv MovieObjectbdmv PLAYLIST CLIPINF STREAM AUXDATA 00000mpls 01000clpi 01000m2ts soundbdmv Root

As you can see, the const follows the function s parameter declaration The purpose of declaring a member function as const is to prevent it from modifying the object that invokes it For example, consider the following program

Result When an image is displayed, the entire image will now be stretched and distorted to fit within the box. Note: In the Properties window, notice how nondefault properties for a control are displayed in bold type.

/* Demonstrate const member functions This program won't compile */ #include <iostream> using namespace std; class Demo { int i; public: int geti() const { return i; // ok }

a. Display the designer window. b. Right-click the PictureBox control. c. Select Properties. d. Locate the SizeMode property. e. Set its value to StretchImage.

void seti(int x) const { i = x; // error! } }; int main() { Demo ob; obseti(1900);

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