This C++ Program consists of: operator overloading, as well as experience with managing dynamic memory allocation inside a class.

Task

One common limitation of programming languages is that the built-in types are limited to smaller finite ranges of storage. For instance, the built-in int type in C++ is 4 bytes in most systems today, allowing for about 4 billion different numbers. The regular int splits this range between positive and negative numbers, but even an unsigned int (assuming 4 bytes) is limited to the range 0 through 4,294,967,295. Heavy scientific computing applications often have the need for computing with larger numbers than the capacity of the normal integer types provided by the system. In C++, the largest integer type is long, but this still has an upper limit.

Your task will be to create a class, called MyInt, which will allow storage of any non-negative integer (theoretically without an upper limit -- although there naturally is an eventual limit to storage in a program). You will also provide some operator overloads, so that objects of type MyInt will act like regular integers, to some extent. There are many ways to go about creating such a class, but all will require dynamic allocation (since variables of this type should not have a limit on their capacity).

The class, along with the required operator overloads, should be written in the files "myint.h" and "myint.cpp". I have provided starter versions of these files, along with some of the declarations you will need. You will need to add in others, and define all of the necessary functions. Here are the files:

MYINT.H:

// starter file for MyInt class header class MyInt { // these overload starters are declared as friend functions friend MyInt operator+ (const MyInt& x, const MyInt& y); // add in multiplication, as well friend bool operator< (const MyInt& x, const MyInt& y); // add in the other comparison overloads, as well // declare overloads for input and output (MUST be non-member functions) // you may make them friends of the class public: MyInt(int n = 0); // first constructor // be sure to add in the second constructor, and the user-defined // versions of destructor, copy constructor, and assignment operator private: // member data (suggested: use a dynamic array to store the digits) }; 

MYINT.CPP:

#include  #include "myint.h" using namespace std; int C2I(char c) // converts character into integer (returns -1 for error) { if (c < '0' || c > '9') return -1; // error return (c - '0'); // success } char I2C(int x) // converts single digit integer into character (returns '\0' for error) { if (x < 0 || x > 9) return '\0'; // error return (static_cast(x) + '0'); // success } // Add in operator overload and member function definitions 

Details and Requirements

1) Your class must allow for storage of non-negative integers of any (theoretical) size. While not the most efficient in terms of storage, the easiest method is to use a dynamic array, in which each array slot is one decimal "digit" of the number. (This is the suggested technique). You may go ahead and assume that the number of digits will be bound by the values allowed in an unsigned int variable (it would be a good idea to keep track of the number of digits in such a variable). You should create appropriate member data in your class. All member data must be private.

2) There should be a constructor that expects a regular int parameter, with a default value of 0 (so that it also acts as a default constructor). If a negative parameter is provided, set the object to represent the value 0. Otherwise, set the object to represent the value provided in the parameter. There should be a second constructor that expects a C-style string (null-terminated array of type char) as a parameter. If the string provided is empty or contains any characters other than digits ('0' through '9'), set the object to represent the value 0. Otherwise, set the object to represent the value of the number given in the string (which might be longer than a normal int could hold).

Note that the two constructors described above will act as "conversion constructors" (as discussed previously in lecture class). Recall that such constructors will allow automatic type conversions to take place -- in this case, conversions from int to MyInt and from c-style strings to type MyInt -- when appropriate. This makes our operator overloads more versatile, as well. For example, the conversion constructor allows the following statements to work (assuming appropriate definitions of the assignment operator and + overloads described later):

MyInt x = 1234;

MyInt y = "12345";

MyInt z = x + 12;

3) Since dynamic allocation is necessary, you will need to write appropriate definitions of the special functions (the "automatics"): destructor, copy constructor, assignment operator. The destructor should clean up any dynamic memory when a MyInt object is deallocated. The copy constructor and assignment operator should both be defined to make a "deep copy" of the object (copying all dynamic data, in addition to regular member data), using appropriate techniques. Make sure that none of these functions will ever allow memory "leaks" in a program.

4) If you use a dynamic array, you will need to allow for resizing the array when needed. There should never be more than 5 unused array slots left in the array at the end of any public operation.

Operator overloads: (The primary functionality will be provided by the following operator overloads).

5) Create an overload of the insertion operator << for output of numbers. This should print the number in the regular decimal (base 10) format -- and no extra formatting (no newlines, spaces, etc -- just the number).

6) Create an overload of the extraction operator >> for reading integers from an input stream. This operator should ignore any leading white space before the number, then read consecutive digits until a non-digit is encountered (this is the same way that >> for a normal int works, so we want to make ours work the same way). This operator should only extract and store the digits in the object. The "first non-digit" encountered after the number may be part of the next input, so should not be extracted. You may assume that the first non-whitespace character in the input will be a digit. i.e. you do not have to error check for entry of an inappropriate type (like a letter) when you have asked for a number.

Example: Suppose the following code is executed, and the input typed is " 12345 7894H".

MyInt x, y;

char ch;

cin >> x >> y >> ch;

The value of x should now be 12345, the value of y should be 7894 and the value of ch should be 'H'.

7) Create overloads for all 6 of the comparison operators ( < , > , <= , >= , == , != ). Each of these operations should test two objects of type MyInt and return an indication of true or false. You are testing the MyInt numbers for order and/or equality based on the usual meaning of order and equality for integer numbers.

8) Create an overload version of the + operator to add two MyInt objects. The meaning of + is the usual meaning of addition on integers, and the function should return a single MyInt object representing the sum.

9) Create an overload version of the * operator to multiply two MyInt objects. The meaning of * is the usual meaning of multiplication on integers, and the function should return a single MyInt object representing the product. The function needs to work in a reasonable amount of time, even for large numbers. (So multiplying x * y by adding x to itself y times will take way too long for large numbers -- this will not be efficient enough).

10) Create the overloads of the pre-increment and post-increment operators (++). These operators should work just like they do with integers. Remember that the pre-increment operator returns a reference to the newly incremented object, but the post-increment operator returns a copy of the original value (before the increment).

11) Create an overloaded version of the - operator to subtract two MyInt objects. The meaning of - is the usual meaning of subtraction on integers, with one exception. Since MyInt does not store negative numbers, any attempt to subtract a larger number from a smaller one should result in the answer 0. The function should return a single MyInt object representing the difference.

12) Create an overloaded / operator and a % operator for division of two MyInt objects. The usual meaning of integer division should apply (i.e. / gives the quotient and % gives the remainder). Both functions should return their result in a MyInt object.

11) General Requirements:

As usual, no global variables, other than constants

All member data should be private

Use appropriate good programming practices as denoted on previous assignments

Since the only output involved with your class will be in the << overload, your output must match mine exactly when running test programs.

You may NOT use classes from the STL (Standard Template Library) -- this includes class -- as the whole point of this assignment is for you to learn how to manage dynamic memory issues inside of a class yourself, not rely on STL classes to do it for you. You may use standard I/O libraries like iostream and iomanip, as well as the common C libraries, like cstring and cctype

Hints and Tips:

1) Storage: The suggested storage mentioned above is a dynamic array, in which each array slot represents one digit of the number. There are multiple ways to manage this. One method would be to use an array of integers, another would be an array of characters. Both have advantages and disadvantages. An integer array would be easier for arithmetic calculations, but a character array would be easier for input purposes. You may choose your preference, but remember that an integer digit, like 6, and it's corresponding character representation, '6', are not stored as the same value! '6' is stored with its corresponding character code (usually ASCII).

In case you want to use them (not required), I have provided two functions for converting easily between single digit integers and their corresponding character codes. (These functions are only for use with single digits). You may use them if you like, as long as you cite the source in your comments. These are stand-alone functions (not members of any class). Copies of these functions are already placed in your starter "myint.cpp" file.

int C2I(char c)

// converts character into integer (returns -1 for error)

{

if (c < '0' || c > '9') return -1; // error

return (c - '0'); // success

}

char I2C(int x)

// converts single digit integer into character (returns '\0'

for error)

{

if (x < 0 || x > 9) return '\0'; // error

return (static_cast(x) + '0'); // success

}

6) Input: For the >> operator overload, there are some issues to be careful about. You will not be able to just use the normal version of >> for integers, because it attempts to read consecutive digits, until a non-digit is encountered. The problem here is that we will be entering numbers that go beyond the capacity of a normal int (like 6 trillion). So, you will probably find it necessary to read one digit at a time (which means one byte, or character, at a time). Because of this, you may find the above conversion functions useful.

You already know of some istream member functions (like getline), and you have used many versions of the >> operator on basic types. One thing worth keeping in mind is that when the >> operator is used for built-in types, any leading white space is automatically ignored. However, there are a couple of other istream functions you might find useful, including the ones with prototypes:

int get();

int peek();

The get() function extracts and returns the next single character on the stream, even if it is a white space character, like the newline. This function does not skip white space (or any other characters), like the built-in >> functions do. The peek() function just looks at and returns the next character on the stream, but it does not extract it. This function is good for seeing what the next character is (i.e. like whether it is a number or not), without actually picking it up.

7) Comparison Overloads: Writing all 6 comparison overloads may sound like a big task, but it really isn't. Always remember that once a function is written, other functions can call it. For example, once you have written the details of comparing two numbers with the less-than operator, it should be very easy to define the > operator (by calling upon the < operator to do the work). You can't write them ALL this way, because you don't want to get stuck in an infinite loop (i.e. don't make both < and > call each other -- you'll never get out!) But, you can certainly make the job easier by defining some of these in terms of others that are already written.

8) Addition operator: This is a more difficult function than it sounds like, in the MyInt class. However, the algorithm should be conceptually easy -- it can follow exactly the way you learned to add back in grade school! You'll probably find it most helpful to break this algorithm down to the grade school step-by-step level, performing an addition digit by digit. And don't forget that some additions can cause a carry!

9) Multiplication operator: This one will be a little more complex than the addition overload, but again, you should think back to the algorithm you learned for multiplying numbers in grade school. The same algorithm can be applied here. Again, don't forget the carries!

MAIN.CPP SAMPLE FILE:

// main.cpp // // Driver program to demonstrate the behavior of the MyInt class // // You can add more tests of your own, or write other drivers to test your // class #include  #include "myint.h" using namespace std; MyInt Fibonnaci(MyInt num); int main() { // demonstrate behavior of the two constructors and the << overload MyInt x(12345), y("9876543210123456789"), r1(-1000), r2 = "14H67", r3; char answer; cout << "Initial values: x = " << x << " y = " << y << " r1 = " << r1 << " r2 = " << r2 << " r3 = " << r3 << " "; // demonstrate >> overload cout << "Enter first number: "; cin >> x; cout << "Enter second number: "; cin >> y; cout << "You entered: "; cout << " x = " << x << ' '; cout << " y = " << y << ' '; // demonstrate assignment = cout << "Assigning r1 = y ... "; r1 = y; cout << " r1 = " << r1 << ' '; // demonstrate comparison overloads if (x < y) cout << "(x < y) is TRUE "; if (x > y) cout << "(x > y) is TRUE "; if (x <= y) cout << "(x <= y) is TRUE "; if (x >= y) cout << "(x >= y) is TRUE "; if (x == y) cout << "(x == y) is TRUE "; if (x != y) cout << "(x != y) is TRUE "; // demonstrating + and * overloads r1 = x + y; cout << "The sum (x + y) = " << r1 << ' '; r2 = x * y; cout << "The product (x * y) = " << r2 << " "; cout << "The sum (x + 12345) = " << x + 12345 << ' '; cout << "The product (y * 98765) = " << y * 98765 << ' '; // create Fibonacci numbers (stored as MyInts) using + cout << " Assuming that the Fibonnaci sequence begins 1,1,2,3,5,8,13..." << " The 10th Fibonnaci number = " << Fibonnaci(10) << " The 100th Fibonnaci number = " << Fibonnaci(100) << " The 1000th Fibonnaci number = " << Fibonnaci(1000) << " The 2000th Fibonnaci number = " << Fibonnaci(2000) << " "; } MyInt Fibonnaci(MyInt num) { MyInt n1 = 1, n2 = 1, n3; MyInt i = 2; while (i < num) { n3 = n1 + n2; n1 = n2; n2 = n3; i++; } return n2; } 

SAMPLE RUN EXAMPLES:

SAMPLE RUN 1:

Initial values: x = 12345 y = 9876543210123456789 r1 = 0 r2 = 0 r3 = 0 Enter first number: 987654987654 Enter second number: 987654987654 You entered: x = 987654987654 y = 987654987654 Assigning r1 = y ... r1 = 987654987654 (x <= y) is TRUE (x >= y) is TRUE (x == y) is TRUE The sum (x + y) = 1975309975308 The product (x * y) = 975462374637822892423716 The sum (x + 12345) = 987654999999 The product (y * 98765) = 97545744855647310 Assuming that the Fibonnaci sequence begins 1,1,2,3,5,8,13... The 10th Fibonnaci number = 55 The 100th Fibonnaci number = 354224848179261915075 The 1000th Fibonnaci number = 43466557686937456435688527675040625802564660517371 78040248172908953655541794905189040387984007925516929592259308032263477520968962 3239873322471161642996440906533187938298969649928516003704476137795166849228875 The 2000th Fibonnaci number = 42246963333923048787067256023414827825798528402506 81098010280137314308584370130707224123599639141511088446087538909603607640194711 64359602927198331259873732625355580260699158591522949245390499872225679531698287 44824729922639018337167780606070116154978867198798583114688708762645973690867228 84023654422295243347964480139515349562972087652656069529806499841977448720155612 802665404554171717881930324025204312082516817125 

SAMPLE RUN 2:

Initial values: x = 12345 y = 9876543210123456789 r1 = 0 r2 = 0 r3 = 0 Enter first number: 123456789123456789123456789123456789 Enter second number: 1010101010101010101010101010101010101 You entered: x = 123456789123456789123456789123456789 y = 1010101010101010101010101010101010101 Assigning r1 = y ... r1 = 1010101010101010101010101010101010101 (x < y) is TRUE (x <= y) is TRUE (x != y) is TRUE The sum (x + y) = 1133557799224466890133557799224466890 The product (x * y) = 1247038273974311001247038273974310999987529617260256889987 52961726025689 The sum (x + 12345) = 123456789123456789123456789123469134 The product (y * 98765) = 99762626262626262626262626262626262625265 Assuming that the Fibonnaci sequence begins 1,1,2,3,5,8,13... The 10th Fibonnaci number = 55 The 100th Fibonnaci number = 354224848179261915075 The 1000th Fibonnaci number = 43466557686937456435688527675040625802564660517371 78040248172908953655541794905189040387984007925516929592259308032263477520968962 3239873322471161642996440906533187938298969649928516003704476137795166849228875 The 2000th Fibonnaci number = 42246963333923048787067256023414827825798528402506 81098010280137314308584370130707224123599639141511088446087538909603607640194711 64359602927198331259873732625355580260699158591522949245390499872225679531698287 44824729922639018337167780606070116154978867198798583114688708762645973690867228 84023654422295243347964480139515349562972087652656069529806499841977448720155612 802665404554171717881930324025204312082516817125 

The following files need to be submitted:

myint.h

myint.cpp

You may take your time, and update me in the comments.

Thanks for your help!