Programming Questions - Research Paper Example

 A constructor is constructed by the class in case it is not written by the programmer. In this case time() is the default constructor which will be used in case time( int hh, int mm, int ss) is not used.  

1. d) Instance variables are the variables hh, mm, ss that are used to pass the values given by the user to the constructor.
2. e) A setter method sets a value to a variable of an object. In this example void setTime (int hh,int mm, int ss) is a setter method.

   Question 27                                                 

1. Time ( ) and Time (int hh, int mm, int ss) are both constructors. Time() is the default that initializes random values in case Time (int hh, int mm, int ss) is not defined by the user in the program.
2.  

void Time::setTime (int hh, int mm, int ss);

{

            hour = (hh>=0 && hh<24) ? h :0;

            mnute = (mm>=0 && mm<=60) ? m : 0;

            second = (ss>=0 && ss <60 ) ? s : 0;

}

Time::Time()

{

      hh=mm=ss=0;

}

1. Int Time::timesAreEqual (Time &tl, Time &t2);

{

            if ( t1.hour= t2.hour && t1.minute == t2.minute && t1.second == t2.second );

            return 1;

            else

            return 0;

}

Question 28

1. a) A pointer is a variable that holds some memory address.
2. b) An array in C++ is a variable that holds subsequent memory locations up to the number declared by the user. The first element in the array has the index 0.
3. d) when the statements q1 = &x; and q2 = &y; are executed pointer q1 get the value of x and q2 gets the value of y.

x

78

y

96

    x=78         y=96

Question 29

1. Dynamic data structures can grow or shrink according to the wish of the programmer. Static values do not hold this property; once declared they remain fixed to the defined length.
2. After the assignment chPtr = new char[10]; chPtr content will be the memory location where char[10] is stored.

Question 30

1. Encapsulation in object-oriented programming is a technique that integrates data and operations to make an object. The benefits of encapsulation are abstraction and security. Abstraction in the sense that the operations and properties of an object are hidden from the outside world who is not bogged down with the inner workings of the object and security because the outside world can access the object only through its interface and does not have access to the inner workings of the object the outside world, therefore, cannot mess with the object.
2. They include data and operations into a single entity.
3. Inheritance and Polymorphism, with inheritance classes, get properties from other classes, and with polymorphism, object operations are dynamically defined during program execution.
4. Inheritance allows reusing previously defined classes. There are three benefits of inheritance use: (i) reuse of existing code from the base class, (ii) all children of the base class have the same methods, and (iii) the programmer can redefine in the child class methods from the parent class.
5. “Signature” defines parameters overloading of the parents' class. Functions’ signature has to be unique to avoid confusion of overloaded parent class parameters.

Question 31

struct Zebra

{

            char * name;

            int age;

            char * zoo;

            int weight;

            char * vet;

            void setName(char * nameIn);

            void setAge(int ageIn);

            void setZoo(char * zooIn);

            void setWeight(int weightIn);

            void setVet(char * vetIn);

            void getName();

            void getAge();

            void getZoo();

            void getWeight();

            void getVet();

}

#include Zebra.h

void Zebra::setName(char * nameIn)

{

            int size;

            size=strlen(nameIn);

            name=new char[size+1];

            strcpy(name, nameIn);

}

void Zebra::setAge(int ageIn)

{

            age=ageIn;

}

void Zebra::setZoo(char * zooIn)

{

            int size;

            size=strlen(zooIn);

            zoo=new char[size+1];

            strcpy(zoo, zooIn);

}

void Zebra::setWeight(int weightIn)

{

            weight=weightIn;

}

void Zebra::setVet(char * vetIn)

{

            int size;

            size=strlen(vetIn);

            vet=new char[size+1];

            strcpy(vet, vetIn);

}

void Zebra::getName()

{

            cout<<name;

}

void Zebra::getAge()

{

            cout<<age;

}

void Zebra::getZoo()

{

            cout<<zoo;

}

void zebrea::getWeight()

{

cout<<weight;

}

void getVet()

{

cout<<vet;

}

int main()

{

            Zebra array[3];

            for (int i=0; i<3; i++)

{

            char * nameIn, *zooIn, *vetIn;

            int ageIn, weightIn;

            cout<<” Please enter zebra name, age, zoo it resides, weight and name of vet.”<<endl;

            cin>>nameIn>>ageIn>>zooIn>>weightIn>>vetIn;

            array[i].name=nameIn;

            array[i].age=ageIn;

            array[i].zoo=zooIn;

            array[i].weight=weightIn;

            array[i].ver=vetIn;

}

for (int j=0; j<3; j++)

{

            cout<<array[i].getName();

            cout<<array[i].getAge();

            cout<<array[i].getZoo();

            cout<<array[i].getWeight();

            cout<<array[i].getVet();

}

}