Object-Oriented Design & Programming

Object-Oriented Design and Programming C++ Basic Examples Bounded Stack Example Linked List Stack Example UNIX File ADT Exam...

Object-Oriented Design and Programming

C++ Basic Examples

Bounded Stack Example

Linked List Stack Example

UNIX File ADT Example

Speci cation for a String ADT

String Class ADT Example

Circular Queue Class Example

Matrix Class Example, 

Bounded Stack Example

The following program implements a bounded

stack abstraction

{ This is the solution to the rst programming

assignment

e.g.,

/* The stack consists of ELEMENT TYPEs. */

typedef int ELEMENT TYPE;

class Stack f

private:

/* Bottom and maximum size of the stack. */

enum fBOTTOM = 0, SIZE = 100g;

/* Keeps track of the current top of stack. */

int stack top;

/* Holds the stack's contents. */

ELEMENT TYPE stack[SIZE];

2

Bounded Stack Example (cont'd)

/* The public section. */

public:

/* Initialize a new stack so that it is empty. */

Stack (void);

/* Copy constructor */

Stack (const Stack &s);

/* Assignment operator */

Stack &operator= (const Stack &s);

/* Perform actions when stack is destroyed. */

~Stack (void);

/* Place a new item on top of the stack

Does not check if the stack is full. */

void push (ELEMENT TYPE new item);

/* Remove and return the top stack item

Does not check if stack is empty. */

ELEMENT TYPE pop (void);

/* Return top stack item without removing it

Does not check if stack is empty. */

ELEMENT TYPE top (void);

/* Returns 1 if the stack is empty,

otherwise returns 0. */

int is empty (void);

/* Returns 1 if stack full, else returns 0. */

int is full (void);

g;

String Class ADT Example

/* Check whether String S is a substring

in this String. Return 􀀀1 if it is not, oth-

erwise return the index position where the

substring begins. */

int String:: nd (const String &s) const f

char rstc = s[0]; // s.operator[] (0);

int end index = this->len 􀀀 s.len + 1;

for (int i = 0;

i < end index

&& ((*this)[i] != rstc jj

memcmp (&this->str[i] + 1, s.str + 1, s.len 􀀀 1) !

i++)

;

return i < end index ? i : 􀀀1;

g

/* Extract a substring from this String of

size count starting at index pos. */

String String::substr (int index pos, int count) const f

if (index pos < 0 jj index pos + count > this->len)

return "";

else f

String tmp (count);

for (int i = 0; i < count; i++)

tmp[i] = (*this)[index pos++];

/* tmp.operator[] (i) =

this->operator[] (index pos++); */

return tmp;

g

g