Programming For Problem Solving Lecture Notes

Syllabus, Introduction
PROGRAMMING FOR
PROBLEM SOLVING (PPS)
B.Tech I Sem CST

Course Outcomes
 CO1: Understand fundamentals of problem solving concepts with
various data types and operators
 CO2: Apply conditional and iterative statements for solving a
given problem
 CO3: Illustrate the applications of functions and storage classes.
 CO4: Apply the concepts of pointers and dynamic memory
management in problem solving.
 CO5: Understand the purpose of structures, unions and files.

Unit 1
 General Problem Solving Concepts
 Algorithm, Flowchart for problem solving with Sequential Logic Structure,
Decisions and Loops.
 Imperative Languages
 Introduction ; syntax and constructs of a specific language (ANSI C)–
Types Operator and Expressions with discussion of variable naming and
Hungarian Notation: Variable Names, Data Type and Sizes (Little
Endian Big Endian), Constants, Declarations, Arithmetic Operators,
Relational Operators, Logical Operators, Type Conversion, Increment
Decrement Operators, Bitwise Operators, Assignment Operators and
Expressions, Precedence and Order of Evaluation, Formatted input/output

Unit 2
 Control Flow with discussion on structured and
unstructured programming
 Statements and Blocks, If-Else-If, Switch,
 Loops–while, do, for, break and continue, goto labels,
 structured and un-structured programming

Unit 3
 Functions and Program Structure with discussion on
standard library
 Basics of functions, parameter passing and returning
type, C main return as integer,
 External, Auto, Local, Static, Register Variables,
 Scope Rules, Block structure, Initialization,
 Recursion, Pre-processor, Standard Library Functions and
return types.

Unit 4
 Pointers and Arrays:
 Pointers and address, dynamic memory management,
Pointers and Function Arguments, Pointers and Arrays,
Address Arithmetic, character Pointers and Functions,
Pointer Arrays,
 Pointer to Pointer, Multi-dimensional array and Row/column
major formats, Initialization of Pointer Arrays, Command
line arguments, Pointer to functions, complicated declarations
and how they are evaluated

Unit 5
 Structures and Unions:
 Basic Structure, Structures and Functions,
 Array of structures, Pointer of structures, Self-referral
structures, Table look up, typedef, Unions, Bit-fields.
 Files:
 Introduction to Files, Opening and Closing files, Reading
and Writing files, File I/O functions, Error Handling in files

Textbooks & References
 1.The C Programming Language, B. W. Kernighan
and D. M. Ritchie, Second Edition, PHI.
 2.Programming in C, B. Gottfried, Second Edition,
Schaum Outline Series.
 1. C: The Complete Reference, Herbert Schildt,
Fourth Edition, McGraw Hill.
 2. Let Us C, Yashavant Kanetkar, BPB Publications

PROGRAMING FOR PROBLEM
SOLVING LAB [PPS(P)]
 CO1:Implementprograms using conditional and loop statements in C.
 CO2:Develop programs using 1-Dimensional and 2-Dimensional arrays.
 CO3:Perform Call by value, Call by reference and Recursion through
functions.
 CO4:Implement programs using pointers.
 CO5:Develop programs using structures and file concepts

List of Experiments
1. Conditional Statements:
Quadratic equations, usage of switch statement.
2. Loop Statements :
Adam Number, Cosine series
3. Arrays:
Max Min problem, standard deviation and variance.
4. Character Arrays:
Palindrome, implementation of string handling functions.

List of Experiments
5. Functions and Recursion :
Matrix operations, Towers of Hanoi, GCD
6. Pointers:
Interchanging problem,
implementation of dynamic memory allocation.
7. Structures:
Usage of structures in various applications.
8. Files:
Reading contents from files and writing contents to files

C is a programming
language developed
at Bell Laboratories of
USA in 1972 by Dennis
Ritchie

 C is a programming language developed at Bell
Laboratories of USA in 1972 by Dennis Ritchie

Programming For Problem Solving Lecture Notes

Decimal to Binary
156 (10)
10011100 (2)

Binary representation (Decimal to Binary
Decimal no.= 123
Binary no. = 1111011
Decimal no. = 255
Binary no. = 1111 1111
Decimal no. = 486
Binary no. = 111100110

Binary to Decimal
Binary no. = 110011
Decimal no. = 51
Binary no. = 1111 0001
Decimal no. = 241

Unit 1
 General Problem Solving Concepts
 Algorithm, Flowchart for problem solving with Sequential Logic Structure, Decisions
and Loops.
 Imperative Languages
 Introduction ; syntax and constructs of a specific language (ANSI C)–Types
Operator and Expressions with discussion of variable naming and Hungarian
Notation: Variable Names, Data Type and Sizes (Little Endian Big Endian),
Constants, Declarations, Arithmetic Operators, Relational Operators, Logical
Operators, Type Conversion, Increment Decrement Operators, Bitwise Operators,
Assignment Operators and Expressions, Precedence and Order of Evaluation,
Formatted input/output

Algorithm
 Algorithm is a finite set of instructions that if followed
accomplishes a particular task.
Algorithm
Input Output
Problem
Computer

Problem: Shortest distance
Input
n nodes
connecting nodes
distance between nodes
Output
Shortest distance: 18

Characteristics of Algorithm
 Input
 Output
 Definiteness
 Effectiveness
 Finiteness
Input − An algorithm should have 0 or
more well-defined inputs.
Output − An algorithm should have 1 or
more well-defined outputs, and should
match the desired output.
Definiteness −Should be clear and
unambiguous. Each of its steps and their
inputs/outputs should be clear and must
lead to only one meaning.
Effectiveness − An algorithm should have
step-by-step directions, which should be
independent of any programming code
Finiteness − An algorithm must terminate
after a finite number of steps.

Characteristics of Algorithm
 Input − Should have 0 or more well-defined inputs.
 Output Should have 1 or more well-defined outputs, and
should match the desired output.
 Definiteness −Should be clear and unambiguous.
 Effectiveness − Should have step-by-step directions, which
should be independent of any programming code.
 Finiteness − Must terminate after a finite number of steps.

Program to display “Welcome to C”
#include<stdio.h>
int main()
{
printf(“Welcome to C”);
return 0;
}

Problem 1: Addition of two numbers
 Input :
 two numbers (num1 = 4,
num2 = 6)
 Output:
 Sum = 10
Algorithm
 Step 1: Start
 Step 2: Read the first number(num1)
 Step 3: Read second number(num2)
 Step 4: Sum  num1+num2
 Step 5: Print Sum
 Step 6: Stop
Sum = num1 + num2

Program to accept two numbers from the user and display the
sum using C language
/*
Program to accept two numbers and display the sum
Programmer : --Your name--
Roll no: --your roll number--
Date of compilation: 27Jan2021
Class : B.Tech I Sem „CST'
*/

#include<stdio.h>
// Main function begins
int main()
{
int num1,num2,sum;
printf("Enter first value:");
scanf("%d",&num1);
printf("Enter Second value:");
scanf("%d",&num2);
sum=num1+num2;
printf(“Addition of %d and %d is: %d",num1,num2,sum);
return 0;
}

Problem 2: Algorithm to find the area of rectangle
 Input :
 two numbers. (length = 8
breadth = 4)
 Output:
 Area of rect = 32
Algorithm
 Step 1: Start
 Step 2: Read first number(length)
 Step 3: Read second number (breadth)
 Step 4: Area  length * breadth
 Step 5: Print “Area of rect”, Area
 Step 6: Stop
Area = length * breadth

Ex3: Problem: Find greater of two no‟s
 Algorithm:
 Step 1: Start
 Step 2: Declare variables A, B
 Step 3: Accept two values from user and store in A and B
respectively.
 Step 4: Check whether A > B; True: 4.1; False: 4.2
4.1 Yes  Print A is greater
4.2 No  Print B is greater
 Step5: Stop

Ex4: Find Largest of three numbers
Step 1: Start
Step 2: Declare variables a,b and c.
Step 3: Read variables a,b and c.
Step 4: If a>b ; True: 4.1; False: 4.2
4.1 If a>c True: 4.1.1; False: 4.1.2
4.1.1 Display a is the largest number.
else
4.1.2 Display c is the largest number.
4.2 Else
If b>c True: 4.2.1; False: 4.2.2
4.2.1 Display b is the largest number.
else
4.2.2 Display c is the greatest number.
Step 5: Stop

Ex5: Calculate Gross, Net Salary
 Problem:
 Accept the employee details such as empid and
basic salary. Consider DA with 120% of basic, HRA
with 10% of basic. Calculate the gross salary and
net salary considering tax deduction of Rs. 2% of
gross.

Example: Gross, Net
 Empid: 1001
 Basic: 10000
 DA: 120% of Basic=1.20*10000= 12000
 HRA: 10% of Basic=0.1*10000 = 1000
 Gross=Basic+DA+HRA = 10000+12000+1000=23000
 Tax: 2% of Gross = 0.02*23000=460
 Net = Gross-Tax = 23000-460 = 22540

Algorithm: Gross, Net
Step 1: Start
Step 2: Declare variables empid,basic,da,hra,tax,gross,net
Step 3: Print "Enter Employee ID"
Step 4: Accept empid
Step 5: Print "Enter basic salary"
Step 6: Accept basic
Step 7: Compute da=(120/100)*basic
Step 8: Compute hra=(10/100)*basic
Step 9: Compute gross=(basic+da+hra)
Step 10: Print "Gross Salary"
Step 11: Compute tax=(2/100)*gross
Step 12: net=gross-tax;
Step 13: Print "Net Salary"
Step 14: Stop

#include<stdio.h>
int main()
{
int empid;
float basic,da,hra,tax,gross,net;
printf("Enter Employee id:");
scanf("%d",&empid);
printf("Enter Basic Salary:");
scanf("%f",&basic);
da=1.2*basic;
hra=0.1*basic;
gross=(basic+da+hra);
printf("Gross Salary=%f",gross);
tax=0.02*gross;
net=gross-tax;
printf("Net Salary=%f",net);
return 0;
}

Ex 6: Calculate Simple Interest
 Si = (P * T * R) / 100

Flowchart
 A Flowchart is a type of diagram (graphical or symbolic) that
represents an algorithm or process.
 Each step in the process is represented by a different symbol
and contains a short description of the process step.
 The flow chart symbols are linked together with arrows showing
the process flow direction. A flowchart typically shows the flow
of data in a process, detailing the operations/steps in a
pictorial format which is easier to understand than reading it in
a textual format.

Flowchart
 Flowchart is the pictorial representation of an algorithm

49
Identify
 What do each of the following symbols represent?
Terminal
Input/Output
Operation
Process
Decision
Connector
Module

Example 1: Algorithm & Flowchart
 Problem: Display the Sum, Average, Product of
three given numbers
Algorithm
 Step1: Start
 Step 2: Read X, Y, Z
 Step 3: Compute Sum (S)  X + Y + Z
 Step 4: Compute Average (A)  S / 3
 Step 5: Compute Product (P) as X x Y x Z
 Step 6: Print S, A, P
 Step 7: Stop

Example 2: Algorithm & Flowchart
 Problem: Display the largest of two given numbers
Algorithm
 Step1: Start
 Step 2: Read A, B
 Step 3: If A is less than B
 True: Assign A to BIG and B to SMALL
 False: Assign B to BIG and A to SMALL
 Step 6: Print S, A, P
 Step 7: Stop

Control Structures
 Sequence: A series of steps or statements that are executed in
the order they are written in an algorithm.
 Selection: Defines two courses of action depending on the
outcome of a condition. A condition is an expression that is,
when computed, evaluated to either true or false. (ex: if, if
else, nested if)
 Repetition/Loop: Specifies a block of one or more statements
that are repeatedly executed until a condition is satisfied. (Ex:
for, while, do while)

Ex: Sequence Control Structure
 A series of steps or statements that are executed in the
order they are written in an algorithm.
 The beginning and end of a block of statements can be
optionally marked with the keywords begin and end.
begin
statement 1
statement 2
statement n
.....
end

Sequence Ex: Calculate Person‟s age

Sequence Example
int main()
{
int first, second, temp;
printf("Enter first number: ");
scanf("%d", &first);
printf("Enter second number: ");
scanf("%d", &second);
temp = first;
first = second;
second = temp;
printf("After swapping, firstno. = %dn", first);
printf("After swapping, secondno. = %d", second);
}
Swap two
numbers
using temp
var

Sequence Example
int main()
{
int a, b;
printf("Enter a and b: ");
scanf("%d %d", &a, &b);
a = a - b;
b = a + b;
a = b - a;
printf("After swapping, a = %dn", a);
printf("After swapping, b = %d", b);
return 0;
}
Swap two
numbers
without
using temp
var

Selection (Decision) Control structure
is
condition
Y N
Print
stmt1
Print
stmt2

Syntax: if
if (condition)
stmt1
else
stmt2
C programming
if (a>b)
printf(“a is greater than b”);
else
printf(“b is greater than a”);
Selection (Decision) Control structure

Selection Ex: Program to print the given
number is negative or positive using if else
int main()
{
int num;
printf("Enter a number to check.n");
scanf("%d",&num);
if (num<0)
printf(“Given Number = %d is negativen",num);
else
printf(“Given Number = %d is positiven",num);
return 0;
}

Selection Ex:
 Problem: Write an algorithm to determine a student‟s final grade
and display pass or fail. The final grade is calculated as the
average of four subject marks.
Algorithm
 Step1: Start
 Step 2: Input M1,M2,M3,M4
 Step 3: GRADE  (M1+M2+M3+M4)/4
 Step 4: if (GRADE > 60) then Print “Pass” else “Fail”
 Step 5: Stop

START
Input
M1,M2,M3,M4
GRADE(M1+M2+M3+M4)/4
IS
GRADE>60
STOP
Y
N
Print
“Fail”
Print
“Pass”
Algorithm
Step1: Start
Step 2: Input M1,M2,M3,M4
Step 3: GRADE  (M1+M2+M3+M4)/4
Step 4: if (GRADE > 60) then Print “Pass” else “Fail”
Step 5: endif
Step 6: Stop
Selection Ex:

Selection Ex:
Program to check for odd or even
int main()
{
int num;
printf("Enter an integer: ");
scanf("%d", &num);
if (num % 2 == 0)
printf("%d is even.", num);
else
printf("%d is odd.", num);
return 0;
}

Selection Ex: Program to check odd or
even using ternary operator ? :
int main()
{
int num;
printf("Enter an integer: ");
scanf("%d", &num);
(num % 2 == 0) ? printf("%d is even.", num) : printf("%d is odd.", num);
return 0;
}

Selection Ex: Problem: Find greater of two no‟s
 Algorithm:
 Step 1: Start
 Step 2: Declare variables A, B
 Step 3: Accept two values from user and store in A and B
respectively.
 Step 4: Check whether A > B; True: 4.1; False: 4.2
4.1 Yes  Print A is greater
4.2 No  Print B is greater
 Step5: Stop

Selection Ex: Find Largest of three numbers
Step 1: Start
Step 2: Declare variables a,b and c.
Step 3: Read variables a,b and c.
Step 4: If a>b ; True: 4.1; False: 4.2
4.1 If a>c True: 4.1.1; False: 4.1.2
4.1.1 Display a is the largest number.
else
4.1.2 Display c is the largest number.
4.2 Else
If b>c True: 4.2.1; False: 4.2.2
4.2.1 Display b is the largest number.
else
4.2.2 Display c is the greatest number.
Step 5: Stop

int main()
{
double n1, n2, n3;
printf("Enter three different numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
if (n1 >= n2 && n1 >= n3)
printf("%.2f is the largest number.", n1);
if (n2 >= n1 && n2 >= n3)
if (n3 >= n1 && n3 >= n2)
return 0;
}
Selection: if

int main()
{
double n1, n2, n3;
printf("Enter three numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
if (n1 >= n2 && n1 >= n3)
printf("%.2lf is the largest number.", n1);
else if (n2 >= n1 && n2 >= n3)
else
return 0;
}
Selection: if else if

int main()
{
double n1, n2, n3;
printf("Enter three numbers: ");
scanf("%lf %lf %lf", &n1, &n2, &n3);
if (n1 >= n2)
{
if (n1 >= n3) printf("%.2lf is the largest number.", n1);
else printf("%.2lf is the largest number.", n3);
}
else
{
if (n2 >= n3) printf("%.2lf is the largest number.", n2);
else printf("%.2lf is the largest number.", n3);
}
return 0;
}
Selection: nested if

Loop: for
for(initialization, condition, incrementation)
{
code statements;
}
int main()
{
int i;
for (i=0; i<10; i++)
{
printf("i=%dn",i);
}
return 0;
}

Loop Example: Multiplication table
int main()
{
int n, i;
printf("Enter no. to print multiplication table: ");
scanf("%d",&n);
for(i=1;i<=10;++i)
{
printf("%d * %d = %dn", n, i, n*i);
}
}

for(i = 0; i < 5; i++)
{
printf("tttt");
for(j = 0; j < 5; j++)
printf("* ");
printf("n");
}

Practice Programs in Lab
 1. Program to swap two numbers.
 2. Program to accept student id (integer), marks of four subjects
and calculate the sum and average. If the average is greater than
60 Print “Pass” else “Fail”.
 3. Program to display whether the given integer is odd or even.
 4. Program to find simple interest.
 5. Program to find the largest of three given numbers.
 6. Program to accept two numbers and check whether both are
same.

C basic elements
 Valid character set
 Identifiers
 Keywords
 Basic data types
 Constants
 Variables

C Valid character set
 uppercase English alphabets A to Z,
 lowercase letters a to z,
 digits 0 to 9,
 certain special characters as building blocks to form
basic program elements viz. constants, variables,
operators, expressions and statements.

C Identifiers
 Identifiers are names given to various items in the program, such as
variables, functions and arrays.
 An identifier consists of letters and digits, in any order, except that the
first character must be a letter.
 Both upper and lowercase letters are permitted.
 C is a case sensitive, the upper case and lower case considered
different, for example code, Code, CODE etc. are different
identifiers.
 The underscore character _ can also be included.
 Keywords like if, else, int, float, etc., have special meaning and they
cannot be used as identifier names.

Valid and invalid C identifiers
 ANSI standard recognizes 31 characters
 valid identifiers: A, ab123, velocity, stud_name,
circumference, Average, TOTAL.
 Invalid identifiers:
 1st
 "Jamshedpur"
 stud-name
 stud name

C Data types and sizes
Data type Description Size Range
char single character 1 byte 0 - 255
int integer number 4 bytes
-2147483648 to
2147483647
float
single precision floating point
number (number containing
fraction & or an exponent)
4 bytes 3.4E-38 to 3.4E+38
double
double precision floating point
number
8 bytes 1.7E-308 to 1.7E+308

C Data types and sizes
Data type Size Range
short int 2 bytes -32768 to 32767
long int 4 bytes
-2147483648 to
2147483647
unsigned short int 2 bytes 0 to 65535
unsigned int 4 bytes 0 to 4294967295
unsigned long int 4 bytes 0 to 4294967295
long double (extended precision) 8 bytes 1.7E-308 to1.7E+308

C Constants
 C can be classified into four categories namely integer
constants, floating point constants, character constants
and string constants.
 A character constant is written as for example - 'A'
 A normal integer constant is written as 1234.
 A long int uses L (uppercase or lowercase) at the end of
the constant, e.g. 2748723L
 C also supports octal and hexadecimal data. Ex: 0xC

Escape sequence characters
Character Escape Sequence ASCII Value
Bell a 007
Backspace b 008
Null 0 000
Newline n 010
Carriage return r 013
Vertical tab v 011
Horizontal tab t 009
Form feed f 012

Symbolic constants
 #define PI 3.141593
 #define TRUE 1
 #define PROMPT "Enter Your Name :"

Accept name from the user
Method 1
char name[20];
printf("Enter your name:");
scanf("%s",name);
printf("Your name is: %s",name);
Method 2
char name[20]
printf(“Enter your name:”)
gets(name);
printf(“Your name is:”);
puts(name);

Accept name from the user
Method 3
#define MAX_LIMIT 20
int main()
{
char name[MAX_LIMIT];
fgets(name,MAX_LIMIT,stdin);
Method 4
char name[20];
scanf("%[^n]%*c",name);

Operators in C
 Arithmetic operators
 Relational operators
 Logical operators
 Bitwise operators
 Assignment operators
 Misc operators

int a = 10, b = 20, c = 25, d = 25;
printf(“ %d" (a + b) );
printf(“ %d" (a - b) );
printf(“%d “ (a * b) );
printf(“ %d” (b / a) );
printf(“ %d” (b % a) );
printf(“ %d” (c % a) );
printf (“%d“ (a++) );
printf(“%d “ (a--) );
printf(“%d “ (d++) );
printf(“%d “ (++d) );
OUTPUT
30
-10
200
2
0
5
10
11
25
27

Example: Bitwise Operators
int a = 60;
int b = 13;
int c = 0;
c = a & b; printf(“%d" + c );
c = a | b; printf(“%d" + c );
c = a ^ b; printf(“%d" + c );
c = ~a; printf(“%d" + c );
c = a << 2; printf(“%d" + c );
c = a >> 2; printf(“%d" + c );
OUTPUT
a= 60 = 0011 1100
b= 13 = 0000 1101
c = a & b; 0000 1100
= 12
c = a | b; 0011 1101
= 61
c = a ^ b; 0011 0001
= 49
c = ~a; 1100 0011
= -61

Misc Operators
sizeof() : Returns the size of the variable
& : Returns the address of a variable
* : Pointer variable
?: : Conditional / Ternary operator

Operator Precedence
e = (a + b) * c / d;
// Print value of e
e = ((a + b) * c) / d;
// Print value of e
e = (a + b) * (c / d);
// Print value of e
e = a + (b * c) / d;
// Print value of e
int a = 20, b = 10, c = 15, d = 5;
int e;
Value of (a + b) * c / d is : 90
Value of ((a + b) * c) / d is : 90
Value of (a + b) * (c / d) is : 90
Value of a + (b * c) / d is : 50
( 30 * 15 ) / 5
(30 * 15 ) / 5
(30) * (15/5)
20 + (150/5)
associativity of operators determines the direction in which an
expression is evaluated. Example, b = a;
associativity of the = operator is from right to left (RL).

Operator Description Associativity
()
[ ]
.

Parentheses (grouping)
Brackets (array subscript)
Member selection
Member selection via pointer
LR
++ --
+ -
! ~
(type)
*
&
sizeof
Unary preincrement/predecrement
Unary plus/minus
Unary logical negation/bitwise
complement
Unary cast (change type)
Dereference
Address
Determine size in bytes
RL
= Assignment operator

Arithmetic Operators
Multiplication operator, Divide
by, Modulus
*, /, % LR
Add, Subtract +, – LR
Relational Operators
Less Than <
LR
Greater than >
Less than equal to <=
Greater than equal to >=
Equal to ==
Not equal !=
Logical Operators
AND && LR
OR || LR
NOT ! RL
1 == 2 != 3
operators == and
!= have the same
precedence, LR
Hence, 1 == 2 is
executed first

Hungarian Notation
 Hungarian is a naming convention for identifiers. Each identifier
would have two parts to it, a type and a qualifier.
 Each address stores one element of the memory array. Each
element is typically one byte.
 For example, suppose you have a 32-bit quantity written as
12345678, which is hexadecimal.
 Since each hex digit is four bits, eight hex digits are needed to
represent the 32-bit value. The four bytes are: 12, 34, 56, and 78.
There are two ways to store in memory: Bigendian and little endian

Endianness
 The endianness of a particular computer system is
generally described by whatever convention or set of
conventions is followed by a particular processor or
combination of processor/architecture and possibly
operating system or transmission medium for the
addressing of constants and the representations of
memory addresses.
 Often referred to as byte order

Big Endian storage
 Big-endian: Stores most significant byte in smallest address.
 The following shows how 12345678 is stored in big endian
Big Endian Storage
Address Value
1000 12
1001 34
1002 56
1003 78

Little Endian storage
 Little-endian: Stores least significant byte in smallest
address.
 The following shows how 12345678 is stored in big
endian Little Endian Storage
Address Value
1000 78
1001 56
1002 34
1003 12

 For example 4A3B2C1D at address 100, they store
the bytes within the address range 100 through 103
in the following order:m

Type Casting
 Type casting is a way to convert a variable from
one data type to another data type.
 Implicit Conversions
 Explicit Conversions

Implicit Conversions
int main()
{
int i = 17;
char c = 'c'; /* ascii value is 99 */
int sum;
sum = i + c;
printf("Value of sum : %dn", sum );
}

Explicit Conversions
General format / Syntax:
(type_name) expression
Example:
main()
{
int sum = 17, count = 5;
double mean;
mean = (double) sum / count;
printf("Value of mean : %lfn", mean );
}

Examples: Type Conversions
float a = 5.25;
int b = (int)a;
char c = ‟A‟;
int x = (int)c;
int x=7, y=5 ;
float z;
z=x/y;
int x=7, y=5;
float z;
z = (float)x/(float)y;

break Statement
The break statement in C programming language has the
following two usages:
When the break statement is encountered inside a loop, the loop
is immediately terminated and program control resumes at the
next statement following the loop.
It can be used to terminate a case in the switch statement
If you are using nested loops, the break statement will stop the
execution of the innermost loop and start executing the next line
of code after the block.
Syntax:
break;

int a = 10;
while( a < 20 )
{
printf("value of a: %dn", a);
a++;
if( a > 15)
{
break;
}
}
OUTPUT
value of a: 10
value of a: 11
value of a: 12
value of a: 13
value of a: 14
value of a: 15

int num =0;
while(num<=100)
{
printf("value of variable num is: %dn", num);
if (num==2)
{
break;
}
num++;
}
printf("Out of while-loop");
Output:
value of variable num is: 0
Out of while-loop

int var;
for (var =100; var>=10; var --)
{
printf("var: %dn", var);
if (var==99)
{
break;
}
}
printf("Out of for-loop");
Output:
var: 100
var: 99
Out of for-loop

int a = 4;
while( a < 10 )
{
++a;
if( a > 8)
{
break;
printf("Breakn");
}
printf("Hellon");
}
printf("Out of While loopn");
}

continue
 The continue statement in C programming
language works somewhat like the break
statement.
 Instead of forcing termination, continue
statement forces the next iteration of the loop
to take place, skipping any code in between.

int a = 10;
do {
if( a == 15)
{
a = a + 1;
continue;
}
a++;
} while( a < 20 );
value of a: 10
value of a: 11
value of a: 12
value of a: 13
value of a: 14
value of a: 16
value of a: 17
value of a: 18
value of a: 19

for (int j=0; j<=8; j++)
{
if (j==4)
{
continue;
}
printf("%d ", j);
}
Output:
0 1 2 3 5 6 7 8

int counter=10;
while (counter >=0)
{
if (counter==7)
{
counter--;
continue;
}
printf("%d ", counter);
counter--;
}
Output:
10 9 8 6 5 4 3 2 1 0

int a = 10;
do
{
if( a == 15)
{
a = a + 1;
continue;
}
a++;
} while( a < 20 );

goto statement
 The goto statement is used to alter the normal sequence of
program execution by transferring control to some other part of
the program unconditionally.
 Syntax: goto label;
 Control may be transferred to anywhere within the current
function.
 The target statement must be labeled, and a colon must follow
the label.
 label : statement;

int w;
printf("Enter the input 1 or 0:n");
scanf("%d", &w);
if (w == 1)
goto CST;
if (w == 0) printf("Value entered is 0n");
return 0;
CST : printf("You belong to CST Sectionn");
goto end;
end : printf("Have a nice Dayn");
return 0;
}

Example: Nested Loop
int i, j;
for(i=2; i<10; i++)
{
for(j=2; j <= (i/j); j++)
if(!(i%j)) break;
if(j > (i/j)) printf("%d is primen", i);
}

Program to check alphabet, digit or special character
if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))
printf("'%c' is alphabet.", ch);
else if(ch >= '0' && ch <= '9')
printf("'%c' is digit.", ch);
else
printf("'%c' is special character.", ch);
Alphabet: a to z (or) A to Z
Digit : 0 to 9
else it is special character

Program to check vowel of consonant
if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u' ||
ch=='A' || ch=='E' || ch=='I' || ch=='O' || ch=='U')
{
printf("'%c' is Vowel.", ch);
}
else if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))
printf("'%c' is Consonant.", ch);
else printf("'%c' is not an alphabet.", ch);
Vowel : a (or) e (or) i (or) o (or) u
Consonant : a to z or A to Z
else: not an alphabet

Sum of digits of given number using while
sum = 0;
// Accept number and store in n
num = n;
while( n > 0 )
{
rem = n % 10;
sum += rem;
n /= 10;
}
printf("Sum of digits of %d is %d", num, sum);
OUTPUT
Enter a number: 456
Sum of digits of 456 is 15

Print all ODD numbers from 1 to N using while loop.
number=1;
while(number<=n)
{
if(number%2 != 0)
printf("%d ",number);
number++;
}

Print its multiplication table
i=1;
while(i<=10)
{
printf("%dn",(num*i));
i++;
}

count total digits in a given integer using loop
do
{
count++;
num /= 10;
} while(num != 0);
printf("Total digits: %d", count);

Program to print numbers between 1 and 100 which
are multiple of 3 using the do while loop
int i = 1;
do
{
if(i % 3 == 0)
{
printf("%d ", i);
}
i++;
}while(i < 100);

find sum of odd numbers from 1 to n
for(i=1; i<=n; i+=2)
{
sum += i;
}
printf("Sum of odd numbers = %d", sum);

Exponential series
int i, n;
float x, sum=1, t=1;
// Accept x
// Accept n
for(i=1;i<=n;i++)
{
t=t*x/i;
sum=sum+t;
}
printf(“Exponential Value of %f = %.4f", x, sum);

Program to print its multiplication table
i=1;
while(i<=10)
{
i++;
}
i=1;
do
{
i++;
}while(i<=10);
for(i=1;i<=10;i++)
{
}

k = 1;
for(i=1; i<=rows; i++)
{
for( j=1; j<=cols; j++, k++)
{
printf("%-3d", k);
}
printf("n");
}
Print number pattern as shown

int main()
{
int i;
for (i=0; i<10; i++)
{
printf("i=%dn",i);
}
return 0;
}

Factorial using for loop
fact=1;
for(i=num; i>=1; i--)
fact=fact*i;
printf("Factorial of %d is = %ld",num,fact);

Sum of n natural no‟s using for
sum = 0;
for(i = 1; i <= n; i++)
sum += i;
printf("Sum is: %dn", sum);

Ex: 2: Nested for loop
for (int i=0; i<2; i++)
{
for (int j=0; j<4; j++)
{
printf("%d, %dn",i ,j);
}
}

Ex: 3: Nested for loop
k=1;
for (i=0; i<2; i++)
{
for (j=0; j<4; j++,k++)
{
printf("%d, %d, %dn",i ,j, k);
}
}

Print number pattern as shown
Output:
1
12
123
1234
for(i=1; i<=rows; i++)
{
for( j=1; j<=i; j++)
printf("%d", j);
printf("n");
}

Structured vs Unstructured Programming
Structured Programming is a
programming paradigm which divides
the code into modules or function.
Unstructured Programming is the
paradigm in which the code is
considered as one single block.
Readability
Structured Programming based
programs are easy to read.
Unstructured Programming based
programs are hard to read.
Purpose
Structured Programming is to make the
code more efficient and easier to
understand.
Unstructured programming is just to
program to solve the problem. It does
not create a logical structure.
Complexity
Structured Programming is easier
because of modules.
Unstructured programming is harder
when comparing with the structured
programming

Application
Structured programming can be used for
small and medium scale projects.
Unstructured programming is not applicable
for medium and complex projects.
Modification
It is easy to do changes in Structured
Programming.
It is hard to do modifications in Unstructured
Programming.
Data Types
Structured programming uses many data
types.
Unstructured programming has a limited
number of data types.
Code Duplication
Structured programming avoids code
duplication.
Unstructured programming can have code
duplication.
Testing and Debug
It is easy to do testing and debugging in
Structured Programming.
It is hard to do testing and debugging in
Unstructured programming.

Palindrome number
rev=0
origin=n;
while (n != 0)
{
rem = n % 10;
rev = rev * 10 + rem;
n /= 10;
}
if (orig == rev) printf("%d is a palindrome.“,orig);
else printf("%d is not a palindrome.", orig);

int i, j;
for(i = 2; i<100; i++)
{
for(j = 2; j <= (i/j); j++)
if(!(i%j)) break; // if factor found, not prime
if(j > (i/j)) printf("%d is primen", i);
}
Print Prime numbers upto 100

Fibonacci series
for(i=3;i<=n;i++)
{
trm=prv+pre;
printf("% 5d",trm);
prv=pre;
pre=trm;
}
printf("n");

Palindrome number
rev=0
origin=n;
while (n != 0)
{
rem = n % 10;
rev = rev * 10 + rem;
n /= 10;
}
if (origin == rev) printf("%d is a palindrome.“,orig);
else printf("%d is not a palindrome.", orig);

goto statement: Unstructured programming
 The goto statement is used to alter the normal sequence
of program execution by transferring control to some
other part of the program unconditionally.
 Syntax: goto label;
Control may be transferred to anywhere within the
current function.
 The target statement must be labeled, and a colon must
follow the label.
 label : statement;

Programming For Problem Solving Lecture Notes

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