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The System Interface (user s view) shows what is done abstract base class pure virtual functions The System Implementation (implementor s view) shows how it is done concrete derived classes functions
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11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8
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What is the difference between composition and inheritance What is the difference between protected and private members How do the default constructors and destructors behave in an inheritance hierarchy What is a virtual member function What is a pure virtual member function What is a memory leak How can virtual destructors plug a memory leak What is an abstract base class What is a concrete derived class
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11.10 What is the difference between static binding and dynamic binding 11.11 What is polymorphism
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11.12 How does polymorphism promote extensibility
11.13 What is wrong with the following definitions: class X { protected: int a; > ; class Y : public X { public: void set(X x, int c) { x.a = > ;
Solved Programming Problems
11.14 Implement a
Card class, a composite Hand class, and a composite Deck class for play-
ing poker.
COMPOSITION AND INHERITANCE
[CHAP. 11
First we implement a Card class:
enum Rank {two, three, four, five, six, seven, eight, nine, ten, jack, queen, king, ace}; enum Suit {clubs, diamonds, hearts, spades}; class Card { friend class Hand; friend class Deck; friend ostream& operatorcc(ostream&, const Card&); public: char rank0 { return rank-; } char suit0 { return suit-; } private: Card0 1 1; Card(Rank rank, Suit suit) : rank (rank), suit (suit) { }; Card(const Card& c) : rank-(c.rank- >, suit - (c.suit - ) { }; -Card0 1 1; Rank rank-; Suit suit-;
This class uses enumeration types for a card s 13 possible ranks and 4 possible suits. Anticipating the implementation of Hand and Deck classes, we declare them here to be friend classes to the Card class. This will allow them to access the private members of the Card class. Notice that all three constructors and the destructor are declared to be private. This will prevent any cards to be created or destroyed except by the Card's two friend classes. Here is the implementation of the overloaded insertion operator CC for cards:
ostream& operator<c(ostream& { switch (card.rank-) { case two : ostr cc case three : ostr CC case four : ostr cc case five : ostr cc case six : ostr c< case seven : ostr CC case eight : ostr cc case nine : ostr cc case ten : ostr cc case jack : ostr cc case queen : ostr CC case king : ostr cc case ace : ostr cc
ostr, const Card& card)
"two of II; "three of "; "four of "; "five of "; "six of "; "seven of "; "eight of "; "nine of '; "ten of "; "jack of "; "queen of "; "king of "; "ace of ";
break; break; break; break; break; break; break; break; break; break; break; break; break;
switch (card.suit-) { case clubs : ostr CC "clubs"; case diamonds : ostr C< "diamonds"; case hearts : ostr C< "hearts"; case spades : ostr cc "spades"; return ostr;
break; break; break; break;
CHAP. 1 l]
COMPOSITION AND INHERITANCE
Here is the implementation of the Hand class:
#include "Card.h" class Hand { friend class Deck; public: Hand(unsigned n=5) : size(n) { cards = new CardhI; ) &Hand0 { delete [] cards; } void display(); int isPair(); int isTwoPair(); int isThreeOfKind.0; int isStraight(); int isFlush(); int isFullHouse(); int isFourOfKind(); int isStraightFlush(); private: unsigned size; Card* cards; void sort(); 1;
It uses an array to store the cards in the hand. The sort ( ) function is a private utility that is called by the Deck class after dealing the hand. It can be implemented by any simple sort algorithm such as the Bubble Sort. The di splay ( ) function is also straightforward, using the insertion operator CC that is overloaded in the Card class. The eight boolean functions that identify special poker hands are not so straightforward. Here is the implementation of the i sThree0 f Kind ( ) function:
int Hand: :isThreeOfKind() if (cards[O].rank- == &SC cards[l].rankSC& cards[2].rank&SC cards[3].rankif (cards[O].rank- != &SC cards[l].rank&& cards[2].rank&SC cards[3].rankif (cards[O].rank- != && cards[l].rank&& cards[2].rank&& cards[3].rankreturn 0; cards[l].rank== cards[2].rank!= cards[3].rank!= cards[4].rank-) cards[l].rank== cards[2].rank== cards[3].rank!= cards[4].rank-) cards[l].rank!= cards[2].rank== cards[3].rank== cards[4].rank-)
return 1;
return 1;
return 1;
Since the hand is sorted by rank-, the only way there could be three cards of the same rank with the other two cards of different rank would be one of the three forms: xxxyz, xyyyz, or xyzzz. If any of these three forms is identified, then the function returns 1. If not it returns 0.
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