Category Archives: C++

Flowchart to check whether the input no is Prime or Not

flowchartprimeornot

Flowchart’s Symbol

1. Terminal:-

terminal1terminal2

This Symbol represents the beginning (Start) and ending (Stop) in the Program Logic flow.

2. Input/Output:-

inputoutput

 

 

 

 

This Symbol denotes any function of Input/Output device in the program. Any Instruction which is input from keyboard,CD etc, is shown in this symbol. Any output instruction sending result to printer,mointor is shown in this box.

3. Process:-

process

This is used to represent airthmetic and data movement instructions. Operations like addition, subtraction etc is shown in this box. The logical process of moving data from one memory location to another is also shown here.

4. Decision:-

decision

This indicates a point where a decision has to be made.

5. Connector:-

connector

If flowchart becomes very long and the flow lines start crossing with each other,this symbol is used to avoid confusion.

6. Flow-Lines:-

flowlines

Arrow-headed lines indicate the flow of operation. It represents the sequence in which the instructions are to be executed.

Technical Aptitude Questions- Data Structure

  1. What is data structure?

A data structure is a way of organizing data that considers not only the items stored, but also their relationship to each other. Advance knowledge about the relationship between data items allows designing of efficient algorithms for the manipulation of data.

  1. List out the areas in which data structures are applied extensively?

  • Compiler Design,

  • Operating System,

  • Database Management System,

  • Statistical analysis package,

  • Numerical Analysis,

  • Graphics,

  • Artificial Intelligence,

  • Simulation

  1. What are the major data structures used in the following areas : RDBMS, Network data model & Hierarchical data model.

  • RDBMS – Array (i.e. Array of structures)

  • Network data model – Graph

  • Hierarchical data model – Trees

  1. If you are using C language to implement the heterogeneous linked list, what pointer type will you use?

The heterogeneous linked list contains different data types in its nodes and we need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So we go for void pointer. Void pointer is capable of storing pointer to any type as it is a generic pointer type.

  1. Minimum number of queues needed to implement the priority queue?

Two. One queue is used for actual storing of data and another for storing priorities.

  1. What is the data structures used to perform recursion?

Stack. Because of its LIFO (Last In First Out) property it remembers its ‘caller’ so knows whom to return when the function has to return. Recursion makes use of system stack for storing the return addresses of the function calls.

Every recursive function has its equivalent iterative (non-recursive) function. Even when such equivalent iterative procedures are written, explicit stack is to be used.

  1. What are the notations used in Evaluation of Arithmetic Expressions using prefix and postfix forms?

        Polish and Reverse Polish notations.

  1. Convert the expression ((A + B) * C – (D – E) ^ (F + G)) to equivalent Prefix and Postfix notations.

          Prefix Notation:

            ^ – * +ABC – DE + FG

         Postfix Notation:

          AB + C * DE – – FG + ^

  1. Sorting is not possible by using which of the following methods?

(a) Insertion

(b) Selection

(c) Exchange

(d) Deletion

(d) Deletion.

Using insertion we can perform insertion sort, using selection we can perform selection sort, using exchange we can perform the bubble sort (and other similar sorting methods). But no sorting method can be done just using deletion.

  1. A binary tree with 20 nodes has null branches?

10

21

Let us take a tree with 5 nodes (n=5)

It will have only 6 (ie,5+1) null branches. In general,

A binary tree with n nodes has exactly n+1 null nodes.

  1. What are the methods available in storing sequential files ?

  • Straight merging,

  • Natural merging,

  • Polyphase sort,

  • Distribution of Initial runs.

  1. How many different trees are possible with 10 nodes ?

1014

For example, consider a tree with 3 nodes(n=3), it will have the maximum combination of 5 different (ie, 23 – 3 = 5) trees.

12

In general:

If there are n nodes, there exist 2n-n different trees.

  1. List out few of the Application of tree data-structure?

  • The manipulation of Arithmetic expression,

  • Symbol Table construction,

  • Syntax analysis.

  1. List out few of the applications that make use of Multilinked Structures?

  • Sparse matrix, 

  • Index generation.

  1. In tree construction which is the suitable efficient data structure?

(a) Array (b) Linked list (c) Stack (d) Queue (e) none

(b) Linked list

  1. What is the type of the algorithm used in solving the 8 Queens problem?

Backtracking

  1. In an AVL tree, at what condition the balancing is to be done?

If the ‘pivotal value’ (or the ‘Height factor’) is greater than 1 or less than –1.

  1. What is the bucket size, when the overlapping and collision occur at same time?

One. If there is only one entry possible in the bucket, when the collision occurs, there is no way to accommodate the colliding value. This results in the overlapping of values.

  1. Traverse the given tree using Inorder, Preorder and Postorder traversals.

19

  • Inorder : D H B E A F C I G J

  • Preorder: A B D H E C F G I J

  • Postorder: H D E B F I J G C A

  1. There are 8, 15, 13, 14 nodes were there in 4 different trees. Which of them could have formed a full binary tree?

15.

In general:

There are 2n-1 nodes in a full binary tree.

By the method of elimination:

Full binary trees contain odd number of nodes. So there cannot be full binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete binary tree but not a full binary tree. So the correct answer is 15.

Note:

Full and Complete binary trees are different. All full binary trees are complete binary trees but not vice versa.

  1. In the given binary tree, using array you can store the node 4 at which location?

21

At location 6

1

2

3

4

5

Root

LC1

RC1

LC2

RC2

LC3

RC3

LC4

RC4

where LCn means Left Child of node n and RCn means Right Child of node n

  1. Sort the given values using Quick Sort?

65

70

75

80

85

60

55

50

45

Sorting takes place from the pivot value, which is the first value of the given elements, this is marked bold. The values at the left pointer and right pointer are indicated using L and R respectively.

65

70L

75

80

85

60

55

50

45R

Since pivot is not yet changed the same process is continued after interchanging the values at L and R positions

65

45

75 L

80

85

60

55

50 R

70

65

45

50

80 L

85

60

55 R

75

70

65

45

50

55

85 L

60 R

80

75

70

65

45

50

55

60 R

85 L

80

75

70

When the L and R pointers cross each other the pivot value is interchanged with the value at right pointer. If the pivot is changed it means that the pivot has occupied its original position in the sorted order (shown in bold italics) and hence two different arrays are formed, one from start of the original array to the pivot position-1 and the other from pivot position+1 to end.

60 L

45

50

55 R

65

85 L

80

75

70 R

55 L

45

50 R

60

65

70 R

80 L

75

85

50 L

45 R

55

60

65

70

80 L

75 R

85

In the next pass we get the sorted form of the array.

45

50

55

60

65

70

75

80

85

  1. For the given graph, draw the DFS and BFS?

23

  • BFS: A X G H P E M Y J

  • DFS: A X H P E Y M J G

  1. Classify the Hashing Functions based on the various methods by which the key value is found.

  • Direct method,

  • Subtraction method,

  • Modulo-Division method,

  • Digit-Extraction method,

  • Mid-Square method,

  • Folding method,

  • Pseudo-random method.

  1. What are the types of Collision Resolution Techniques and the methods used in each of the type?

  • Open addressing (closed hashing),

The methods used include:

Overflow block,

  • Closed addressing (open hashing)

The methods used include:

Linked list,

Binary tree…

  1. In RDBMS, what is the efficient data structure used in the internal storage representation?

B+ tree. Because in B+ tree, all the data is stored only in leaf nodes, that makes searching easier. This corresponds to the records that shall be stored in leaf nodes.

  1. Draw the B-tree of order 3 created by inserting the following data arriving in sequence – 92 24 6 7 11 8 22 4 5 16 19 20 78

27

  1. Of the following tree structure, which is, efficient considering space and time complexities?
  1. Incomplete Binary Tree

  2. Complete Binary Tree

  3. Full Binary Tree

(b) Complete Binary Tree.

By the method of elimination:

Full binary tree loses its nature when operations of insertions and deletions are done. For incomplete binary trees, extra storage is required and overhead of NULL node checking takes place. So complete binary tree is the better one since the property of complete binary tree is maintained even after operations like additions and deletions are done on it.

  1. What is a spanning Tree?

A spanning tree is a tree associated with a network. All the nodes of the graph appear on the tree once. A minimum spanning tree is a spanning tree organized so that the total edge weight between nodes is minimized.

  1. Does the minimum spanning tree of a graph give the shortest distance between any 2 specified nodes?

No.

Minimal spanning tree assures that the total weight of the tree is kept at its minimum. But it doesn’t mean that the distance between any two nodes involved in the minimum-spanning tree is minimum.

  1. Convert the given graph with weighted edges to minimal spanning tree.

31-a

the equivalent minimal spanning tree is:

31-b

  1. Which is the simplest file structure?

  1. Sequential

  2. Indexed

  3. Random

(a) Sequential

  1. Whether Linked List is linear or Non-linear data structure?

According to Access strategies Linked list is a linear one.

According to Storage Linked List is a Non-linear one.

  1. Draw a binary Tree for the expression :

A * B – (C + D) * (P / Q)

34

  1. For the following COBOL code, draw the Binary tree?

01 STUDENT_REC.

02 NAME.

03 FIRST_NAME PIC X(10).

03 LAST_NAME PIC X(10).

02 YEAR_OF_STUDY.

03 FIRST_SEM PIC XX.

03 SECOND_SEM PIC XX.

35

Technical Aptitude Question – C Language – Part-1

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:

5432 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 222223465

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:

0013 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<stdio.h>

main()

{
char s[]={‘a’,’b’,’c’,’\n’,’c’,”}; 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<stdio.h>

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<stdio.h>

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<stdio.h>

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.

21.

#define square(x) x*x

main()

{
int i;

i = 64/square(4);

printf(“%d”,i);

}

Answer:

64

Explanation:

the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()

{

char *p=”hai friends”,*p1; p1=p;
while(*p!=”) ++*p++;

printf(“%s %s”,p,p1);

}

Answer:

ibj!gsjfoet

Explanation:

++*p++ will be parse in the given order

  •   *p that is value at the location currently pointed by p will be taken
  •   ++*p the retrieved value will be incremented
  •   when ; is encountered the location will be incremented that is p++ will be executed

Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj! gsjfoet” and since p reaches ‘’ and p1 points to p thus p1doesnot print anything.

23.

#include <stdio.h>

#define a 10

main()

{

#define a 50

printf(“%d”,a);

}

Answer:

50

Explanation:

The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24.

#define clrscr() 100

main()

{
clrscr();

printf(“%d\n”,clrscr());

}

Answer:

100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :

main()

{

100;

printf(“%d\n”,100);

}

Note100; is an executable statement but with no action. So it doesn’t give any problem

25.

main()

{

       printf(“%p”,main);

}

Answer:

Some address will be printed.

Explanation:

Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

C Tutorial – structures, unions, typedef

In the C language structures are used to group together different types of variables under the same name. For example you could create a structure “Mobile”: which is made up of a string (that is used to hold the name of the person) and an integer (that is used to hold the Mobile number).

Take a look at the example:


struct mobile
{
char *name;
int number;
};

Note: Put the ; behind the last curly bracket.

With the declaration of the structure you have created a new type, called mobile. Before you can use the type mobile you have to create a variable of the typemobile. Take a look at the following example:

       #include<stdio.h>

	struct mobile
	{
		char *name;
		int number;
	};

	int main()
	{
	   struct mobile contact;
	   return 0;
        }

Note:

contact is now a variable of the type mobile .

To access the members of the structure mobile , you must use a dot between the structure name and the variable name(variables:name or number.) Take a look at the next example:

     
       #include<stdio.h>

	struct mobile 
	{
		char *name;
		int number;
	};

	int main()
	{
		struct mobile contact;

		contact.name = "Amar sg";
		contact.number = 997861;
		printf("Name: %s\n", contact.name);
		printf("Mobile number: %d\n", contact.number);

		return 0;
	}

TYPE DEFINITIONS AND STRUCTURES

Type definitions make it possible to create your own variable types. In the following example we will create a type definition called “intpointer” (a pointer to an integer):

      	

       #include<stdio.h>

	typedef int *int_ptr;	int main()
	{
		int_ptr myvar;
		return 0;
	}

     

It is also possible to use type definitions with structures. The name of
 the type definition of a structure is usually in uppercase letters. 
Take a look at the example:-
	#include<stdio.h>

	typedef struct mobile 
	{
		char *name;
		int number;
	}MOBILE;

	int main()
	{
		MOBILE contact;

		contact.name = "Amar sg";
		contact.number = 997861;
		printf("Name: %s\n", contact.name);
		printf("Telephone number: %d\n", contact.number);

		return 0;
	}

 

Note: The word struct is not needed before MOBILE contact;

POINTER TO STRUCTURES

If you want a pointer to a structure you have to use the -> (infix operator) instead of a dot.
Take a look at the following example:-

       

        #include<stdio.h>

	typedef struct mobile
	{
		char *name;
		int number;
	}MOBILE;

	int main()
	{
		MOBILE  contact;
		MOBILE *ptr_mycontact;

		ptr_mycontact = &contact;

		ptr_mycontact->name = "Amar sg";
		ptr_mycontact->number = 997861;

		printf("Name: %s\n", ptr_mycontact->name);
		printf("Mobile number: %d\n", ptr_mycontact->number);

		return 0;
	}

 

Note: The -> (infix operator) is also used in the printf  statement.

UNIONS

A union is like a structure in which all members are stored at the same address. Members of a union can only be accessed one at a time. The union data type was invented to prevent memory fragmentation. The union data type prevents fragmentation by creating a standard size for certain data. Just like with structures, the members of unions can be accessed with the . and -> operators. Take a look at the example:-

	

        #include<stdio.h>

	typedef union mobileunion
	{
		double PI;
		int B;
	}MOBILEUNION;

	int main()
	{
		MOBILEUNION numbers;
		numbers.PI = 3.14;
		numbers.B = 50;

	    return 0;
	}

That’s all for this tutorial. Happy Programming :-)

Technical Aptitude Question – C Language – Part-2

1) main()
{
    clrscr();
}

clrscr();

Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

2) enum colors {BLACK,BLUE,GREEN}
main()
{
printf(“%d..%d..%d”,BLACK,BLUE,GREEN);
return(1);

}

Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

3) void main()
{
     char far *farther,*farthest;
     printf(“%d..%d”,sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer

4.) main()
{

int i=400,j=300;

printf(“%d..%d”);

}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf’s may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

5) main()
{
char *p;
p=”Hello”;
printf(“%c\n”,*&*p);
}

Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string “Hello”. *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

6) main()
{
        int i=1;
        while (i<=5)

         {
                  printf(“%d”,i); 

                  if(i>2)

                        goto here;
                   i++;
         }
}
fun()
{
here:
printf(“PP”);
}
Answer:
Compiler error: Undefined label ‘here’ in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label ‘here’ is available in function fun() Hence it is not visible in function main.

7) main()
{
       static char names[5][20]={“pascal”,”ada”,”cobol”,”fortran”,”perl”};
       int i;
      char *t;
       t=names[3];
      names[3]=names[4]; names[4]=t;
             for (i=0;i<=4;i++)
                   printf(“%s”,names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

8) void main()
{
      int i=5;
      printf(“%d”,i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i

9) void main()
{
      int i=5;
     printf(“%d”,i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

10) #include<stdio.h>
main()
{
     int i=1,j=2;

      switch(i)
     {
            case 1: printf(“GOOD”);
           break;

           case j: printf(“BAD”);
           break;
      }
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that
we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

11) main()
{
      int i;
       printf(“%d”,scanf(“%d”,&i)); // value 10 is given as input here
}

Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

12) main ( )
{
     static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
     char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
      p = ptr;
     **++p;
      printf(“%s”,*–*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression
is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

13) main()
{
     int i=0;
     for(;i++;printf(“%d”,i)) ;
         printf(“%d”,i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is “evaluated”. Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

14) #include<stdio.h>
main()
{
     char s[]={‘a’,’b’,’c’,’\n’,’c’,”};
     char *p,*str,*str1;
     p=&s[3];
    str=p;
    str1=s;
    printf(“%d”,++*p + ++*str1-32);
}

Answer:
M
Explanation:
p is pointing to character ‘\n’.str1 is pointing to character ‘a’ ++*p meAnswer:”p is pointing to ‘\n’ and that is incremented by one.” the ASCII value of ‘\n’ is 10. then it is incremented to 11. the value of ++*p is 11. + +*str1 meAnswer:”str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII value of ‘b’ is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77(“M”);

15) #include main()
{
      struct xx
     {
              int x=3;
             char name[]=”hello”;
      };
          struct xx *s=malloc(sizeof(struct xx));
          printf(“%d”,s->x);
          printf(“%s”,s->name);
 }
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

16) #include<stdio.h>
main()
{
       struct xx
         {
               int x;
              struct yy
              {
                       char s;
                       struct xx *p;
               };
              struct yy *q;
          };
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

17)  main()
{
        extern int i;
        i=20;
       printf(“%d”,sizeof(i));
}
Answer:
Linker error: undefined symbol ‘_i’.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn’t find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

18) main()
{
      printf(“%d”, out);
}

int out=100;

Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

19) main()
{
      extern out;
      printf(“%d”, out);
}

int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.

20) main()
{
     show();
}
void show()
{
     printf(“I’m the greatest”);
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn’t know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

21) main( )
{
      int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
     printf(“%u %u %u %d \n”,a,*a,**a,***a);
     printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.

2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output. for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a
+1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

22) main( )
{
     int a[ ] = {10,20,30,40,50},j,*p;
      for(j=0; j<5; j++)
     {
             printf(“%d” ,*a);
            a++;
     }

     p = a;
     for(j=0; j<5; j++)
    {
         printf(“%d ” ,*p);
         p++;
    }
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

23) main( )
{
      static int a[ ] = {0,1,2,3,4};
      int *p[ ] = {a,a+1,a+2,a+3,a+4};
      int **ptr = p;
      ptr++;
      printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
     *ptr++;
     printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
     *++ptr;
     printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
      ++*ptr;
      printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111

222

333

344
Explanation:

Let us consider the array and the two pointers with some address

a

0 1 2 3 4
100 102 104 106 108

p

100 102 104 106 108
1000 1002 1004 1006 1008

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.

After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

24) main( )
{
     char *q;
     int j;
      for (j=0; j<3; j++)
         scanf(“%s” ,(q+j));
       for (j=0; j<3; j++)
             printf(“%c” ,*(q+j));
      for (j=0; j<3; j++)
            printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M O U S E ‘\O’

When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M T R A C K ‘\O’

The third input starts filling from the location 102

M T V I R T U A L ‘\O’

This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

25) main( )
{
       void *vp;
      char ch = ‘g’,
     *cp = “goofy”;
      int j = 20;
      vp = &ch;
       printf(“%c”, *(char *)vp);
       vp = &j;
       printf(“%d”,*(int *)vp);
       vp = cp; printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

C bits Manipulation

Bit manipulation is the act of algorithmically manipulating bits or other pieces of data shorter than a byte. C language is very efficient in manipulating bits.

Here are following operators to perform bits manipulation:

Bitwise Operators:

Bitwise operator works on bits and perform bit by bit operation.

Assume if B = 60; and B = 13; Now in binary format they will be as follows:

A = 0011 1100

B = 0000 1101

—————–

A&B = 0000 1000

A|B = 0011 1101

A^B = 0011 0001

~A  = 1100 0011

Try following example to understand all the Bitwise operators. Copy and paste following C program in test.c file and compile and run this program.
main()
{

   unsigned int a = 60;	/* 60 = 0011 1100 */  
   unsigned int b = 13;	/* 13 = 0000 1101 */
   int c = 0;           

   c = a & b;       /* 12 = 0000 1100 */ 
   printf("Line 1 - Value of c is %d\n", c );

   c = a | b;       /* 61 = 0011 1101 */
   printf("Line 2 - Value of c is %d\n", c );

   c = a ^ b;       /* 49 = 0011 0001 */
   printf("Line 3 - Value of c is %d\n", c );

   c = ~a;          /*-61 = 1100 0011 */
   printf("Line 4 - Value of c is %d\n", c );

   c = a << 2;     /* 240 = 1111 0000 */
   printf("Line 5 - Value of c is %d\n", c );

   c = a >> 2;     /* 15 = 0000 1111 */
   printf("Line 6 - Value of c is %d\n", c );
}

This will produce following result:-

Line 1 - Value of c is 12
Line 2 - Value of c is 61
Line 3 - Value of c is 49
Line 4 - Value of c is -61
Line 5 - Value of c is 240
Line 6 - Value of c is 15

There are following Bitwise operators supported by C language :-

 

Operator Description Example
& Binary AND Operator copies a bit to the result if it exists in both operands. (A & B) will give 12 which is 0000 1100
| Binary OR Operator copies a bit if it exists in eather operand. (A | B) will give 61 which is 0011 1101
^ Binary XOR Operator copies the bit if it is set in one operand but not both. (A ^ B) will give 49 which is 0011 0001
~ Binary Ones Complement Operator is unary and has the efect of ‘flipping’ bits. (~A ) will give -60 which is 1100 0011
<< Binary Left Shift Operator. The left operands value is moved left by the number of bits specified by the right operand. A << 2 will give 240 which is 1111 0000
>> Binary Right Shift Operator. The left operands value is moved right by the number of bits specified by the right operand. A >> 2 will give 15 which is 0000 1111

The shift operators perform appropriate shift by operator on the right to the operator on the left. The right operator must be positive. The vacated bits are filled with zero.

For example: x << 2 shifts the bits in x by 2 places to the left.

if x = 00000010 (binary) or 2 (decimal)

then: 
x >>= 2 => x = 00000000 or just 0 (decimal)

Also: if x = 00000010 (binary) or 2 (decimal) 
then
x <<= 2 => x = 00001000 or 8 (decimal)

 

Therefore a shift left is equivalent to a multiplication by 2. Similarly a shift right is equal to division by 2. Shifting is much faster than actual multiplication (*) or division (/) by 2. So if you want fast multiplications or division by 2 use shifts.

To illustrate many points of bitwise operators let us write a function, Bitcount, that counts bits set to 1 in an 8 bit number (unsigned char) passed as an argument to the function.

int bitcount(unsigned char x) 
{ 
   int count;

   for ( count=0; x != 0; x>>=1);
   {
      if ( x & 01)
         count++;
   }

   return count;
}

 

This function illustrates many C program points:

  • for loop not used for simple counting operation.
  • x >>= 1 => x = x>> 1;
  • for loop will repeatedly shift right x until x becomes 0
  • use expression evaluation of x & 01 to control if
  • x & 01 masks of 1st bit of x if this is 1 then count++

Bit Fields

Bit Fields allow the packing of data in a structure. This is especially useful when memory or data storage is at a premium. Typical examples:

  • Packing several objects into a machine word. e.g. 1 bit flags can be compacted.
  • Reading external file formats — non-standard file formats could be read in. E.g. 9 bit integers.

C allows us do this in a structure definition by putting :bit length after the variable.For example:

 

struct packed_struct {
  unsigned int f1:1;
  unsigned int f2:1;
  unsigned int f3:1;
  unsigned int f4:1;
  unsigned int type:4;
  unsigned int my_int:9;
} pack;

Here the packed_struct contains 6 members: Four 1 bit flags f1..f3, a 4 bit type and a 9 bit my_int.

C automatically packs the above bit fields as compactly as possible, provided that the maximum length of the field is less than or equal to the integer word length of the computer. If this is not the case then some compilers may allow memory overlap for the fields whilst other would store the next field in the next word.

 

 

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