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class Node { public: Node(int d, Node* p=O) : data(d), next(p) { } int data; Node* next; >; main0 -t int n; Node* p; Node* q=O; while (tin >> n) { p = new Node(n, q); q = p; > for ( ; pvnext; p = p->next ) tout CC p->data << ' --> "; tout CC "*\n"; )

1 4 f,

First note that the definition of the Node class includes two references to the class itself. This is allowed because each reference is actually a pointer to the class. Also note that the constructor initializes both data members. The whi 1 e loop continues reading in t s into n until the user enters the end-of-file character (Control-D on Mac and UNIX systems, and Control-Z on DOS and VAX systems). Within the loop, it gets a new node, inserts the int into its data member, and connects the new node to the previous node (pointed to by q). Finally, the for loop traverses the list, beginning with the node pointed to by p (which is the last node constructed) and continuing until p- >next is NUL (in which case, p will be pointing to the last node in the list). The list constructed in this example can be visualized like this:

CHAP. 9]

Sometimes a single value for a data member applies to all members of the class. In this case, it would be inefficient to store the same value in every object of the class. That can be avoided by declaring the data member to be static. This is done by including the static keyword at the beginning of the variable s declaration. It also requires that the variable be defined globally. So the syntax looks like this:

1 2 y.

Static variables are automatically initialized to 0, so the explicit initialization in the definition is unnecessary unless you want it to have a non-zero initial value.

d 25.

The Widget class maintains a s tat i c data member count which keeps track of the number of Widget objects in existence globally. Each time a widget is created (by the constructor) the counter is incremented, and each time a widget is destroyed (by the destructor) the counter is decremented.

1 4 p.

class Widget { public: Widget0 { ++count; } -Widget() { --count; } static int count; >; int Widget::count = 0;

(i) If Paul and Jack have equal amounts of money and Paul spends more than Jack, then Paul will have than Jack. (j) If Anne is now older than Helen, then 10 years ago, Anne was

main0 -C Widget w, x; tout << "Now there are -t Widget w, x, y, z; tout << "Now there > tout << "Now there are Widget y; tout << "Now there are

Applying triangle inequality theorems (Fig. 13-8) (a) Determine the integer values that the length of side a of the triangle can have if the other two sides have lengths 3 and 7.

' CC w.count CC ' widgets.\n";

Do While Not EOF(1) . . . . . Loop n = Len(Record) RecNo = Loc(1)

are ' C-C w.count -KC ' widgets.\n"; ' CC w.count << ' widgets.\n"; ' << w.count << ' widgets.\n";

(b) Determine which is the longest side of the triangle if two angles have measures 59 and 60 . (c) Determine which is the longest side of parallelogram ABCD if E is the midpoint of the diagonals and m/AEB m/AED.

Nuw there are .2 widget;s*

Fig. 13-8

.,..

(a) Since a must be less than 3 7 10 and greater than 7 3 4, a can have the integer values of 5, 6, 7, 8, 9. 60 ) 61 . Then the longest side is opposite AD or AB ( DC) is the longest

Notice how four widgets are created inside the inner block, and then they are destroyed when program control leaves that block, reducing the global number of widgets from 6 to 2.

(b) Since m/B 59 and m/C 60 , m/A 180 the largest angle, /A, so the longest side is BC. (c) In ABCD, AE CE and DE side (Principle 6).