Question

3. (a) Outline any four features of Object-Oriented Programming OOP, giving examples in each case. [16 marks]

     (b) Consider the following code fragments: If   a = 10; Evaluate the new value of “b” in the following:

               (i)   b =   ++ a;

              (ii) b = a ++;

            What value would a and b store in (i) and (ii) after program execution?

                                                                                                           [4 marks]

4. Create a C++ program that makes use of three arrays; name, mark, grade. The program should accept the names and marks of a number of students to be entered by the user. Use the number of students as the size of the arrays.
Grades are computed and stored in the grade array using the criteria:
100 – 60 = A;
59 – 40 = B;
below 40 = C.
The program should continuously request a user who enters a mark below 0 or above 100 to enter a mark in the range 0 – 100. Finally, the program should display the list of students in this format: Name | Mark   |   Grade
[20 marks]

Q5. Create a structure called time. Its three members, all type int, should be called hours, minutes, and seconds. Write a program that prompts the user to enter a time value in hours, minutes, and
seconds. This can be in 12:59:59 format, or each number can be entered at a separate prompt
(“Enter hours:”, and so forth).

The program should then store the time in a variable of type struct time, and finally print out the total number of seconds represented by this time value:
long totalsecs = t1.hours*3600 + t1.minutes*60 + t1.seconds [20 marks]

Answer 1

ANSWER :

Answer for (3) :-

(a)

Four features of Object-Oriented Programming OOP are :-

1. Abstraction

This refers to the fact that units in Object-Oriented Programming (also refers to classes) do not indicate the details or implementation. That is to say, the classes only specify the functionality to the outside world using objects without mentioning the implementation details or how the functions are actually working.

Example :- Employee class will have a method getEmployeeDetails without mentioning from where this data is fetched or how to fetch this.

2. Polymorphism

This refers to the concept that the same entity can exist in many forms in OOP.

Example :- a function getArea() will return the area of the figure based on the number and type of arguments passed to it, Like area of circle, rectangle etc

getArea(radius);

getArea(length, breadth);

Same name many forms.

Similarly operators can be overloaded in OOP languages like Cpp.

3. Inheritence

This refers to the concept that classes can exist in parent child heirarchy. A child class can inherit attributes or properties/ functionality from a parent class.

Example :- Employee class can inherit properties like name, gender from class person.

4. Encapsulation

This refers to the concept that all the objects in OOP are wrapped up or encapsulated that is all the attributes and functionality are present at one place called a class.

Example :- Student class will have student attributes like name, class, roll number and also the methods to fetch information like getStudentName(), getGrades() etc.

(b)

If a = 10;

then for the

(i)

#include <stdio.h>

int main()
{
int a,b;
a = 10;
b = ++a;
printf("ans is %d",b);

return 0;
}

Output will be 11

It is incrementing the value

(ii)

#include <stdio.h>

int main()
{
int a,b;
a = 10;
b = a++;
printf("ans is %d",b);

return 0;
}

Output will be 10.The value of variable b will remain same as variable a.

answer for (4) :-

Code :-

#include <iostream>
using namespace std;
main()
{
int n;
cout<<"Enter no.of students:";
cin>>n; //stores no.of students
int marks[n]; //declaring integer array for storing marks
string names[n],grade[n];//declaring string arrays to store marks and grade
for(int i=0;i<n;i++)
{
cout<<"Enter name of student:";
cin>>names[i];//stores name of every student
cout<<"Enter marks of student:";
cin>>marks[i]; //stores marks of every student
while(marks[i]>100 || marks[i]<0) //if user enters negative values or above 100 then again asks the user for input
{
cout<<"Enter a marks in the range 0 - 100:";
cin>>marks[i];
}
if(marks[i]>=60) //marks>60 and below 100 are given to be A grade
{
grade[i]="A";
}
else if(marks[i]>=40 && marks[i]<=59) //marks above 40 and below 60 are B grade
{
grade[i]="B";
}
else //marks below 40 are C grade
{
grade[i]="C";
}
}
cout<<"name\tmarks\tgrade\n"; //\t is used for tab space and \n is used for new line
cout<<"----------------------\n";
for(int i=0;i<n;i++)
{
cout<<names[i]<<"\t"<<marks[i]<<"\t"<<grade[i]<<"\n";
}
}

Code Screenshot :-

1 2 #include <iostream> using namespace std; main() int n; cout<<"Enter no.of students:"; cin>>n; //stores no. of students int marks[n]; //declaring integer array for storing marks string names[n], grade[n];//declaring string arrays to store marks and grade for (int i=0;i<n;i++) { 4 { 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 cout<<"Enter name of student:"; cin>>names[i];//stores name of every student cout<<"Enter marks of student:"; cin>>marks[i]; //stores marks of every student while(marks[i]>100 | marks[i]<e) //if user enters negative values or above 190 then again asks the user for input { cout<<"Enter a marks in the range 0 - 100:"; cin>>marks[i]; if (marks[i]>=60) //marks>60 and below 100 are given to be A grade { grade[i]="A"; } else if(marks[i]>=40 && marks[i]<=59) //marks above 40 and below 60 are B grade 26 { grade[i]="B";/ } else //marks below 40 are C grade 28 29 30 31 32 grade[i]="C"; cout<<"name\tmarks\tgrade\n"; //\t is used for tab space and in is used for new Line cout<<" --\n"; for(int i=0;i<n;i++) { cout<<names[i]<<"\t"<<marks[i]<<"\t"<<grade[i]<<"\n"; 38 39 40 }

Output Screenshot :-

C:\Users\himanth\Documents\student.exe Enter no.of students:5 Enter name of student:himanth Enter marks of student: 88 Enter name of student:naidu Enter marks of student:900 Enter a marks in the range 0 - 100:67 Enter name of student: junnu Enter marks of student:76 Enter name of student:vijay Enter marks of student:43 Enter name of student:ravi Enter marks of student:12 name marks grade hinanth 88 naidu 67 junnu vijay ravi 76 43 12 A A A B с Process exited after 34.62 seconds with return value o Press any key to continue

answer for (5) :-

Code :-

#include <iostream>
using namespace std;
//structure defination is here
struct time
{
int hour;
int minute;
int seconds;
};

int main()
{ struct time d;
//taking values fro user
cout<<"Enter hours:\n";
cin>>d.hour;
cout<<"Enter minutes:\n";
cin>>d.minute;
cout<<"Enter Seconds:\n";
cin>>d.seconds;
//printing the results
cout<<"Total numbers of seconds is "<<d.hour*3600+d.minute*60+d.seconds;

return 0;
}

Code Screenshot :-

6 8 1 #include <iostream> 2 using namespace std; 3 //structure defination is here 4 struct time 5- { int hour; 7 int minute; int seconds; 9 }; 10 11 int main() 12. { struct time d; 13 //taking values fro user 14 cout<<"Enter hours: \n"; 15 cin>>d.hour; 16 cout<<"Enter minutes:\n"; 17 cin>>d.minute; 18 cout<<"Enter Seconds:\n"; 19 cin>>d.seconds; 20 //printing the results cout<<"Total numbers of seconds is "«<d. hour 3600+d.minute*60+d. seconds; 22 23 return 0; 24 } 25 21

Output Screenshot :-

Enter hours: 1 Enter minutes: 35 Enter Seconds: 25 Total numbers of seconds is 5725 .Program finished with exit code 0 Press ENTER to exit console.

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