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#include <iostream> using namespace std; class figure { protected: double x, y; public: void set_dim(double i, double j) { x = i; y = j; } virtual void show_area() { cout << "No area computation defined "; cout << "for this class\n"; } } ; class triangle : public figure { public: void show_area() { cout << "Triangle with height "; cout << x << " and base " << y; cout << " has an area of "; cout << x * 05 * y << "\n"; } }; class square : public figure { public: void show_area() { cout << "Square with dimensions "; cout << x << "x" << y; cout << " has an area of "; cout << x * y << "\n";

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menuEditLabel.Enabled = (listViewMain.SelectedItems.Count == 1); menuProperties.Enabled = (listViewMain.SelectedItems.Count == 1); if (this._albumsShown) menuEditLabel.Text = "&Name"; else menuEditLabel.Text = "&Caption"; }

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23:

REDISPLAYING THE ALBUMS As a final change, we need to give our user the opportunity to redisplay the album view. We may as well provide a menu to display the photo view as well, as an alternative to double-clicking on the album.

} }; int main() { figure *p; /* create a pointer to base type */ triangle t; /* create objects of derived types */ square s; p = &t; p->set_dim(100, 50); p->show_area(); p = &s; p->set_dim(100, 50); p->show_area(); return 0; }

ALLOW USER SELECTION OF THE KIND OF OBJECT TO DISPLAY Action 1 In the MainForm.cs [Design] window, add three menu items to the bottom of the View menu.

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As you can see by examining this program, the interface to both square and triangle is the same even though both provide their own methods for computing the area of each of their objects Given the declaration for figure, it is possible to derive a class called circle that computes the area of a circle given its radius To do so, you must create a new derived type that computes the area of a circle The power of virtual functions is based in the fact that you can easily derive a new type that shares the same common interface as other related objects For example, here is one way to do it:

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class circle : public figure { public: void show_area() { cout << "Circle with radius "; cout << x; cout << " has an area of "; cout << 314 * x * x; } } ;

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Settings Menu separator Albums Property Text (Name) Text Photos (Name) Text Value menuAlbums &Albums menuPhotos &Photos private void menuAlbums_Click (object sender, System.EventArgs e) { // Display albums in the list if (!_albumsShown) { LoadAlbumData(PhotoAlbum.DefaultDir); } } private void menuPhotos_Click (object sender, System.EventArgs e) { // Activate the selected album listViewMain_ItemActivate(sender, e); } private void menuView_Popup (object sender, System.EventArgs e) { View v = listViewMain.View; . . . if (_albumsShown && listViewMain. SelectedItems.Count > 0) menuPhotos.Enabled = true; else menuPhotos.Enabled = false; }

Before trying to use circle, look closely at the definition of show_area( ) Notice that it uses only the value of x, which is assumed to hold the radius (Remember that the area of a circle is computed using the formula R2) However, the function set_dim( )

This completes our discussion of the ListView class. In this chapter we discussed list views in detail, and created a new MyAlbumExplorer interface to display the collection of albums available in our default album directory. We supported all four possible views available in a ListView control, and provided support for column

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as defined in figure assumes that it will be passed not just one, but two values Since circle does not require this second value, what is the best course of action There are two ways to resolve this problem First, you can call set_dim( ) using a dummy value as the second parameter when using a circle object This has the disadvantage of being sloppy as well as requiring you to remember a special exception, which violates the one interface, many methods approach A better way to resolve this problem is to give the y parameter inside set_dim( ) a default value In this way, when calling set_dim( ) for a circle, you need specify only the radius When calling set_dim( ) for a triangle or a square, you would specify both values The expanded program is shown here:

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