Nagarro Placement Papers..

Ques.1 :- Seat Reservation prog for the theatre. Write a function for seat allocation for the movie tickets. Total no of seats available are 200. 20 in each row. Each row is referred by the Character, "A" for the first row and 'J' for the last. And each seat in a row is represented by the no. 1-20. So seat in diffrent rows would be represented as A1,A2....;B1,B2.....;........J1,J2... Each cell in the table represent either 0 or 1. 0 rep would seat is available , 1 would represent seat is reserved. Booking should start from the last row (J) to the first row(A). At the max 20 seats can be booked at a time. if seats are available, then print all the seat nos like "B2" i.e (2 row, 3 col) otherwise Print "Seats are not available." and we must book consecutive seats only.

Ques.2 :- A string of charaters were given. Find the highest occurance of a character and display that character.
eg.: INPUT: AEGBCNAVNEETGUPTAEDAGPE
OUTPUT: E
Ques.3 :- Int Matrix of certain size was given, We had few values in it like this.
1 4 5 45
3 3 5 4
34 3 3 12
3 3 4 3
3 3
4 4 3
We were supposed to move back all the spaces in it at the end.
Note: If implemented this prog using recursion, would get higher preference.


Ques.4 :- write a function to give demostrate the functionality of 3D matrix in 1D matirx.
function prototype :-
void set (int value,int indexX,int indexY,int indexZ, int [] 1dArray);
void get (int value,int indexX,int indexY,int indexZ, int [] 1dArray);

Ques.5 :- A chessboard was given to us. Where in there was a Knight and King was placed on certain positions. Our aim is to reach the king from the knight in minimum no of counts.As we know, knight can either move 2 steps vertical/horizontal and 1 step horizontal/vertical. same goes here as well.
Proper image of the chess board was given in the question paper, and all the positions(max 8) were given that knight can take in the first step.

Sol : I've implemented using recursive func.I was appreciated by the interviewer.They will ask for the dry run becz there was 8 recursions.So Be prepare.

Ques.6 :-
Struct person
{
char * name;
person[] friends;
};
We were given the networklist of friends. Each has set of friends which was unidirectional i.e, if you are my frnd, then i may or may not be in ur frnds list. okie. Network was like this:
Amit - ->Rahul -> Aman -> kumar
Rahul- ->Vipin->Ankit->Reena->kumar
Kumar- ->Rahul->Reena->Tanmay

We need to identify whether 1st person being passed is a frnd of another person or not. Frnds can be frnd's friend also and so on. And we need to identify the distance.
For example :-
Input: Amit, Kumar
Output Distance 1
Input Amit, Tanmay
Output: Distance 2
Input: Rahul, Aman
Not frnds.

Ques.7 :- There was a 2D matrix given, we were supposed to sort the all diagnols elements. diagnols of Top left corner and Top right corner were to be sorted in the same matrix in an efficient way.

Ques.8 :- We need to write the function to check the password entered is correct or not based on the
following conditions..
a) It must have atleast one lower case character and one digit.
b)It must not have any Upper case characters and any special characters
c) length should be b/w 5-12.
d) It should not have any same immediate patterns like

abcanan1 : not acceptable coz of an an pattern
abc11se: not acceptable, coz of pattern 11
123sd123 : acceptable, as not immediate pattern
adfasdsdf : not acceptable, as no digits
Aasdfasd12: not acceptable, as have uppercase character

Ques.9 :- Wrie a function which returns the most frequent number in a list of integers. Handle the case of more than one number which meets this criterion.

public static int[] GetFrequency(int[] list)

Ques:10 :- Counting in Lojban, an artificial language developed over the last fourty years, is easier than in most
languages
The numbers from zero to nine are:
0 no
1 pa
2 re
3 ci
4 vo
5 mk
6 xa
7 ze
8 bi
9 so
Larger numbers are created by gluing the digit togather.
For Examle 123 is pareci

Write a program that reads in a lojban string(representing a no less than or equal to 1,000,000) and output it in numbers.

Ques.11 :- Where now stands that small knot of villages known as the Endians, a mighty forest once stood.Indeed, legand has it that you could have stoodon the edge of the wood and seen it stretch out for miles,were it not for the trees getting in the way.
In one section of the forest, the trees stood in a row and were of hight from 1 to n, each hight occurring once and once only.

Like:--
A tree was only visible if there were no higher trees before it in the row.
For example, if the heights were 324165, the only visible trees would have been those of height 3,4 &6.

Ques:12 :- Given an array containing k nos in the range 1..n and another scratch array of size n. Write an program to remove the duplicates from the array.

Written Test Pattern:--
The written exam consisted of 25 general maths questions on trigonometry, circles, area, perimeter,mensuration, time, work, distance etc ques.
and aptitude consisted of 10 ques with several parts. they wer twisty ques including the puzzle test ques
given in R.S.Aggarwal verbal-nonverbal book....
The second written exam included 4 coding ques which wer to be written in any programming languages .It was something like this:
Instructions:
1.The code should emphasize on min time and then min memory requirements.
2.The time limit was 1hr 30min.

Ques1: Given an array containing k nos in the range 1..n and another scratch array of size n. Write an program to remove the duplicates from the array.

Ques2: Given a table of the form:
Product Sold on
A 1/1/1980
B 1/1/1980
C 1/1/1980
A 1/1/1980
B 1/1/1980
C 2/1/1980
A 2/1/1980
There are 30 products and 10,000 records of such type. Also the month period during which sales happened is given to u.
Write the program to display the result as:
Product Month No. of copies
A January 12
A February 15
A March 27
B January 54
B February 15
B March 10
C January 37


Ques3: Definition of priority queue was given. We have to implement the priority queue using array of pointers with the priorities given in the range 1..n.
The array could be accessed using the variable top. The list corresponding to the array elements contains the items having the priority as the array index.
Adding an item would require changing the value of top if it has higher priority than top. Extracting an item would require deleting the first element from the corresponding queue.
The following class was given:
class PriorityQueue
{
int *Data[100];
int top;
public:
void put(int item, int priority); // inserts the item with the given priority.
int get(int priority); // extract the element with the given priority.
int count(); // returns the total elements in the priority queue.
int isEmpty(); // check whether the priority queue is empty or not.
};
We had to implement all these class functions.


Ques4: An array of size 5X5 is given to us. The elements from 1 to 25 are to be inserted in the array, such that starting from a particular position for an element i, the next element i+1can be inserted only at the mentioned positions (u,v), and if these all positions are occupied then it returns giving a count of how many positions have been occupied in the array:

(u,v) = (x+/-3 , y)
(u,v) = (x , y+/-3)
(u,v) = (x+/-2 , y+/-2).

Example: if the starting element is 1 with the given positions (1,2), then next element 2 can be placed at
any one of the positions marked with *.
_ _ _ _ _
1 _ _ _ *
_ _ _ _ _
_ _ * _ _
* _ _ _ _
Function to be implemented is fun(int start, int end) where start and end will give the start and end coordinate of first element i.e.


Ques.5 :- suppose u r given a 4*3 rectangle like (take these values from user)
Now u have to calculate the no. of squares in this rectangle like:
No. of squares of dimension 1 is 12
No. of squares of dimension 2 is 6
No. of squares of dimension 3 is 2
No. of squares of dimension 4 is 0
Total no. of squares are 20.

b) Suppose you are given a string. you have to find the occurance of the characters A-Z in that string.Each character must appear in the string and must appear only once. If It that occurs in string more than one time return 1 showing it is a perfect string otherwise return 0 showing it is not a perfect string.

c) Suppose you arr given 10000 marks. you have to pick up top 20 top marks from them and display it on the screen.(Use the optimal sorting algorithm)

d) Suppose you have a chess board. you have to insert 8 queens on the chessboard in the style that the queens don’t intersect in the diagonals, columns and rows. If they intersect return 1 else return 0.(that is no more than one queen should be present either in row or column or diagonals.)
If the queen is inserted at a position in the chessboard, its count is 1.

Then the next round was technical interview. The interviewer was very cool and friendly. He just asked some general questions and then u have to justify the code that u have written in the test line by line. If he is satisfied from ur code then HR interview is only a formality.

Ques.1 :- A Class Structure was given with function names only.
Using one dimensional array make the fuctionality of two dimensional array?
We have to write the function body and the main program which calls them.
the function attributes and return type was given.
some already defined variables were also there.

Ques.2 :- If employee B is the boss of A and C is the boss of B and D is the boss of C and E is the boss of D. Then write a program using the Database such that if an employee name is Asked to Display it also display his bosses with his name.
For eg. If C is displayed it should also display D and E with C?

Ques.3 :- Arrange Doubly linked list in the ascending order of its integral value and replace integer 5 with 7?

Ques.4 :- Three tables were given and we had to link them.


Ques.5 :- You have given a string containing a binary number and you have to calculate the decimal number from it.

Ques.6 :- It was the password validation question mentioned above.

Ques.7 :- You have given two list of names containing name first list is very large and second one is very small, Now you have to conclude whether names appers in the second list are in same order as in the frist or not.
Example :
First: Ram, Mohan, Babu, Shyam, Komal, Rani, Sita
Second: Mohan, Komal, Sita.
Ans : yes.

Questions Asked on 31 May 2008:
Ques.1 :- Subset sum problem.

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C Aptitude Questions...

C Questions Check Your Ability In C

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
 Programs run under DOS environment,
 The underlying machine is an x86 system,
 Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer: I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer: 5 4 3 2 1

Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++)
{
printf(" %d ",*c);
++q;
}
for(j=0;j<5;j++)
{
printf(" %d ",*p);
++p;
}
}

Answer: 2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.


6. main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer: Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i; specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer: 0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer: 1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer: Compiler Error : Type mismatch in redeclaration of function display

Explanation :

In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12. main()
{
int c=- -2;
printf("c=%d",c);
}
Answer: c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer: sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char

14.
main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer: i=0
Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer: 77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16.
#include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer: SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer: Compiler Error
Explanation:
You should not initialize variables in declaration

18.
#include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer: Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19.
main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer: hai
Explanation:
\n - newline
\b - backspace
\r - linefeed

20.
main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

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C code to Check the string has valid identifier or not in.

#include
#include
#include

char keyword[][10]={"auto","break","case","char","const","continue","default","printf", "double","else","enum","extern","float","for","goto","if","int","do", "long","register","return","short","signed","sizeof","static","struct","switch","typedef","union","unsigned","void","volatile","while"};

#define SIZE 25

int main()
{
char str[SIZE];
int len,i,flag=0;
printf("Enter The C Identifier :> ");
gets(str);

for( i = 0 ; i <= 32 ; i++ )
{
if(strcmp(str,keyword[i])==0)
{
printf(" Given string is invalid identifier\n");
exit(0);
}
}
if( str[0]=='_' || isalpha(str[0]) )
{
flag=1;
len=strlen(str);
for(i = 1 ; i < len ; i++ )
{
if( str[i]=='_' || isalpha(str[i]) || isdigit(str[i]) )
continue;
else
{
flag=0;
//printf(" H!!!!!!!! ");
break;
}
}
}

if( flag == 1 )
printf("Given string is a valid C Identifier\n");
else
printf("Given string is invalid C identifier\n");
return 0;
}

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Code for matrix problem in C++

#include
using namespace std;
#define row 10
#define col 10
int row1,row2,col1,col2;
class matrix
{
int a[row][col];
int b[row][col];
int c[row][col];
public :
void input();
void addtion();
void subtraction();
void multiply();
void show();
};
void matrix::input()
{
int i,j;
cout<<"enter the number of row of first matrix\n";
cin>>row1;
cout<<"enter the number of coloumn of first matrix\n";

cout<<"enter the elements of first matrix";
for loop from i=0 upto row1;
{
for loop from j=0 upto col1;
{
cin>>a[i][j];
}
}
cout<<"enter the number of row & coloumn of second matrix";
cin>>row2>>col2;
cout<<"enter the elements of second matrix";
for loop from i=0 upto row2;
{
for loop from j=0 upto col2;
{
cin>>b[i][j];
}
}
}
void matrix::addtion()
{
int i,j;
for loop from i=0 upto row1;
{
for loop from j=0 upto col2;
{
c[i][j]=a[i][j]+b[i][j];
}
}
}
void matrix::subtraction()
{
int i,j;
for loop from i=0 upto row1;
{
for loop from j=0 upto col1;
{
c[i][j]=a[i][j]-b[i][j];
}
}
}
void matrix::multiply()
{
int i,j,k,sum=0;
for loop from k=0 upto col2;
{
for loop from i=0 upto row1;
{
sum=0;
for loop from j=0 upto col1;
{
sum=sum+a[i][j]*b[j][k];
}
c[i][k]=sum;
}
}
}
void matrix::show()
{
int i,j;
for loop from i=0 upto row1;
{
for loop from j=0 upto col1;
{
print two dimension array c according to i and j<<' ';
}
cout<<"\n";
}
}
int main()
{
matrix m;
int ch;
m.input();
while(1)
{
cout<<"options:-\npress 1 for addition\npress 2 for subtraction\npress 3 for multiplication\npress any other key for exit\n";
cin>>ch;
switch(ch)
{
case 1:
if(row1!=row2 ||col1!=col2)
cout<<"addition is not possible\n";
else
{
m.addtion();
m.show();
}
break;
case 2:
if(row1!=row2 ||col1!=col2)
cout<<"subtraction is not possible\n";
else
{
m.subtraction();
m.show();
}
break;
case 3:
if(col1!=row2)
cout<<"multiplicaton is not possible\n";
else
{
m.multiply();
m.show();
}
break;
default:
exit(1);
}
}

}

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Implementation of a queue in C++

#include
using namespace std;
class queue
{
int front,rear;
int item[3];
public:
void insert(int);
int remove();
void display();
void setvalues()
{
front=-1;
rear=0;
}
};
void queue::insert(int m)
{

item[rear++]=m;
}
int queue::remove()
{
int x;

if(front==-1)
cout<<"empty";
x=item[++front];
return(x);
}
void queue::display()
{
int i=front+1;
for loop upto rear then print the items according to item[i]

}
int main()
{
queue s;
int item,ch;
s.setvalues();
while(1)
{
cout<<"optoins:-\npress 1 for inserting the elements\npress 2 for removing the element\npress 3 for display the result\npress any key for exit\n";
cin>>ch;
switch(ch)
{
case 1:
cout<<"enter your item";
cin>>item;
s.insert(item);
break;
case 2:
item=s.remove();
print the item.
break;
case 3:
s.display();
break;
default:
exit(1);
}
}
}

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A linked list program in C++

#include
using namespace std;
struct node
{
int info;
struct node *next;

struct node *getnode();
struct node *insert_last(struct node *);
struct node *insert_begin(struct node *);
struct node *insert_after(struct node *);
struct node *delete_after(struct node *);
void display(struct node *);

};
struct node* node:: getnode()
{
int x;
struct node *p;
p=new node;
cout<<"enter the information ";
cin>>x;
p->info=x;
p->next=NULL;
return(p);
}
struct node* node::insert_last(struct node *ptr)
{
struct node *temp;
temp=ptr;
while(temp->next!=NULL)
temp=temp->next;
temp->next=getnode();
return(ptr);
}
struct node* node::insert_begin(struct node *ptr)
{
struct node *temp;
temp=ptr;
temp=getnode();
temp->next=ptr;
ptr=temp;
return(ptr);
}
struct node* node::insert_after(struct node *ptr)
{
struct node *temp1,*temp2;
int x,i;
temp2=temp1=ptr;
cout<<"enter the position";
cin>>x;
for(i=1;i<(x-1);i++)
{
temp1=temp1->next;
temp2=temp2->next;
}
temp2=temp2->next;
temp1->next=getnode();
temp1=temp1->next;
temp1->next=temp2;
return(ptr);
}
struct node* node::delete_after(struct node *ptr)
{
int x,i;
struct node *temp1,*temp2;
temp1=temp2=ptr;
cout<<"enter your position";
cin>>x;
for(i=1;i<(x-1);i++)
{
temp1=temp1->next;
temp2=temp2->next;
}
temp2=temp2->next;
temp1->next=temp2->next;
return(ptr);
}
void node:: display(struct node *ptr)
{
struct node *temp;
temp=ptr;
while(ptr!=NULL)
{
cout<info<<" ";
ptr=ptr->next;
}
}
int main()
{
struct node *head,a;
int ch;
head=a.getnode();
while(1)
{
cout<< "options:-\npress 1 for insert at last\npress 2 for insert at begin\n press 3 for insert in between\n press 4 for delete\n press 5 for display\n press any key for exit"<<"\n\n";
scanf("%d",&ch);
switch(ch)
{
case 1://head=n.getnode();
head=a.insert_last(head);
break;
case 2:
head=a.insert_begin(head);
break;
case 3:
head=a.insert_after(head);
break;
case 4:
head=a.delete_after(head);
break;
case 5:
a.display(head);
break;
default:
exit(1);
}
}
}

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OOPS Concepts..

Q:-What is Function Overloading ?

A:-Function overloading:
C++ enables several functions of the same name to be defined, as long as
these functions have different sets of parameters (at least as far as their types are concerned). This
capability is called function overloading. When an overloaded function is called, the C++ compiler
selects the proper function by examining the number, types and order of the arguments in the call.
Function overloading is commonly used to create several functions of the same name that perform
similar tasks but on different data types.

Q:-What do you mean by Operator Overloading?

A:-Operator Overloading:
Operator overloading allows existing C++ operators to be redefined so that they work on objects
of user-defined classes. Overloaded operators are syntactic sugar for
equivalent function calls. They form a pleasant facade that doesn't add anything fundamental to the
language (but they can improve understandability and reduce
maintenance costs).


Q:-Difference between realloc() and free()?

A:-The free subroutine frees a block of memory previously allocated by the malloc subroutine.
Undefined results occur if the Pointer parameter is not a valid pointer. If the Pointer parameter is a
null value, no action will occur. The realloc subroutine changes the size of the block of memory
pointed to by the Pointer parameter to the number of bytes specified by the Size parameter and
returns a new pointer to the block. The pointer specified by the Pointer parameter must have been
created with the malloc, calloc, or realloc subroutines and not been deallocated with the free or
realloc subroutines. Undefined results occur if the Pointer parameter is not a valid pointer


Q:- What do you mean by binding of data and functions?

A:-Encapsulation.

Q:- What is abstraction?

A:-Abstraction is of the process of hiding unwanted details from the user.

Q:- What is encapsulation?

A:-Packaging an object’s variables within its methods is called encapsulation.

Q:- What is the difference between an object and a class?

A:- Classes and objects are separate but related concepts. Every object belongs to a class and every class contains one or more related objects.
1: A Class is static. All of the attributes of a class are fixed before, during, and after the execution of a program. The attributes of a class don't change.
2: The class to which an object belongs is also (usually) static. If a particular object belongs to a
certain class at the time that it is created then it almost certainly will still belong to that class right up until the time that it is destroyed.
3: An Object on the other hand has a limited lifespan. Objects are created and eventually destroyed. Also during that lifetime, the attributes of the object may undergo significant change.

Q:- What is polymorphism? Explain with an example?

A:-"Poly" means "many" and "morph" means "form". Polymorphism is the ability of an object (or
reference) to assume (be replaced by) or become many different forms of
object.
Example: function overloading, function overriding, virtual functions. Another example can be a plus ‘+’ sign, used for adding two integers or for using it to concatenate two strings.

Q:- What do you mean by inheritance?

A:-Inheritance is the process of creating new classes, called derived classes, from existing classes or base classes. The derived class inherits all the capabilities of the base class, but can add embellishments and refinements of its own.

Q:- What is a scope resolution operator?

A:-A scope resolution operator (::), can be used to define the member functions of a class outside the class.

Q:- What are virtual functions?

A:-A virtual function allows derived classes to replace the implementation provided by the base class. The compiler makes sure the replacement is always called whenever
the object in question is actually of the derived class, even if the object is accessed by a base pointer rather than a derived pointer. This allows algorithms in the base class to be replaced in the derived class, even if users don't know about the derived class.

Q:- What is friend function?

A:-As the name suggests, the function acts as a friend to a class. As a friend of a class, it can access its private and protected members. A friend function is not a member of the class. But it must be listed in the class definition.

Q:- What is the difference between class and structure?

A:- Structure:
Initially (in C) a structure was used to bundle different type of data types together to perform a particular functionality. But C++ extended the structure to contain functions also. The major difference is that all declarations inside a structure are by default public.
Class:
Class is a successor of Structure. By default all the members inside the class are private

Q:- What is public, protected, private?

A:-
1:- Public, protected and private are three access specifiers in C++.
2:-Public data members and member functions are accessible outside the class.
3:-Protected data members and member functions are only available to derived classes.
4:-Private data members and member functions can’t be accessed outside the class.
However there is an exception can be using friend classes.

Q:- What is an object?

A:- Object is a software bundle of variables and related methods. Objects have state and behavior.

Q:- What is a class?

A:-Class is a user-defined data type in C++. It can be created to solve a particular kind of problem.
After creation the user need not know the specifics of the working of a class.

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OOPS Concepts...

Q:-What is namespace?

A:-Namespaces allow us to group a set of global classes, objects and/or functions under a name. To say
it somehow, they serve to split the global scope in sub-scopes known
as namespaces.
The form to use namespaces is:
namespace identifier { namespace-body }
Where identifier is any valid identifier and namespace-body is the set of classes, objects and
functions that are included within the namespace.

For example:

namespace general { int a, b; } In this case, a and b are normal variables integrated within the
general namespace. In order to access to these variables from outside the namespace we have to
use the scope operator ::. For example, to access the previous variables we would have to put:
general::a general::b
The functionality of namespaces is specially useful in case that there is a possibility that a global
object or function can have the same name than another one, causing a
redefinition error.

Q:- What is RTTI?

A:- Runtime type identification (RTTI) lets you find the dynamic type of an object when you have only a
pointer or a reference to the base type. RTTI is the official way in
standard C++ to discover the type of an object and to convert the type of a pointer or reference
(that is, dynamic typing). The need came from practical experience with
C++. RTTI replaces many homegrown versions with a solid, consistent approach.


Q:- What is a template?

A:-Templates allow to create generic functions that admit any data type as parameters and return value
without having to overload the function with all the possible data types. Until certain point they fulfill
the functionality of a macro. Its prototype is any of the two following ones:
template function_declaration; template
function_declaration;
The only difference between both prototypes is the use of keyword class or typename, its use is
indistinct since both expressions have exactly the same meaning and behave exactly the same way.

Q:- What do you mean by inline function?

A:-The idea behind inline functions is to insert the code of a called function at the point where the
function is called. If done carefully, this can improve the application's
performance in exchange for increased compile time and possibly (but not always) an increase in the
size of the generated binary executables.

Q:-What is virtual class and friend class?

A:-Friend classes are used when two or more classes are designed to work together and need access
to each other's implementation in ways that the rest of the world
shouldn't be allowed to have. In other words, they help keep private things private. For instance, it may be desirable for class DatabaseCursor to have more privilege to the internals of class Database than main() has.

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1. What is virtual constructors/destructors?

Virtual destructors:

If an object (with a non-virtual destructor) is destroyed explicitly by applying
the delete operator to a base-class pointer to the object, the base-class destructor function
(matching the pointer type) is called on the object.
There is a simple solution to this problem – declare a virtual base-class destructor. This makes all
derived-class destructors virtual even though they don’t have the same name as the base-class
destructor. Now, if the object in the hierarchy is destroyed explicitly by applying the delete operator
to a base-class pointer to a derived-class object, the destructor for the appropriate class is called.

Virtual constructor:

Constructors cannot be virtual. Declaring a constructor as a virtual function is
a syntax error. Does c++ support multilevel and multiple inheritance?
Yes.
What are the advantages of inheritance?
• It permits code reusability.
• Reusability saves time in program development.
• It encourages the reuse of proven and debugged high-quality software, thus reducing problem
after a system becomes functional.
What is the difference between declaration and definition?
The declaration tells the compiler that at some later point we plan to present the definition of this
declaration.
E.g.: void stars () //function declaration
The definition contains the actual implementation.
E.g.: void stars () // declarator
{
for(int j=10; j>=0; j--) //function body
cout<<”*”;
Print New Line
}

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