7.3 Example
Here is an example that uses an RWGStack, a generic stack, to store a set of pointers to ints in a last-in, first-out (LIFO) stack. We will go through it line-by-line and explain what is happening:
#include <rw/gstack.h> //1
#include <iostream>
declare(RWGStack, int) //2
int main(){
RWGStack(int) gs; //3
gs.push(new int(1)); //4
gs.push(new int(2)); //5
gs.push(new int(3)); //6
gs.push(new int(4)); //7
cout << "Stack now has " << gs.entries()
<< " entries" << endl; //8
int* ip; //9
while( ip = gs.pop() ) //10
{
cout << *ip << "\n"; //11
delete ip;
}
return 0;
}
Program Output:
Stack now has 4 entries 4 3 2 1
Each line of the program is detailed below.
| //1 | This #include defines the preprocessor macro RWGStackdeclare(type). This macro is an elaborate and ugly-looking thing that continues for many lines and describes how a generic stack of objects of type type should behave. Mostly, the macro serves as a restricted interface to the underlying implementation. |
| //2 | This line invokes the macro declare, which is defined in the header file <generic.h> supplied with your compiler. If called with arguments declare(Class, type), it calls the macro Classdeclare with argument type. In this case, the macro RWGStackdeclare, defined in <rw/gstack.h>, will be called with argument int. In other words, the result of calling the declare(RWGStack, int) macro is to create a new class especially for your program. For all practical purposes, its name is RWGStack(int), a stack of pointers to int s. |
| //3 | At this line an instance gs of the new class RWGStack(int) is created. |
| //4-//7 | Four ints are created off the heap and inserted into the stack. After statement 8 executes, a pointer to the int 4 will be at the top of the stack, and a pointer to the int 1 at the bottom. |
| //8 | The member function entries() of class RWGStack(int) is called to verify how many items are on the stack. |
| //9 | A pointer to an int is declared and defined. |
| //10 | The stack is popped until empty. The member function pop() will return and remove a pointer to the item on the top of the stack. If there are no more items on the stack it will return zero, causing the while loop to terminate. |
| //11 | Each item is dereferenced and printed. |
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