Programming for Problem Solving

PROGRAMMING FOR
PROBLEM SOLVING (PPS)
06Dec2022: nested if, switch, for; Lab Programs
PROBLEM SOLVING (PPS)
B.Tech I Sem CST
Dr. C. Sreedhar

Today’s Topics
 Operators in C
 Control Structures:
if
 Loops
for
do while
if
if else
if else if ladder
nested if
switch
do while

Operators in C
 Arithmetic operators
+ - * / %
Relational operators
 Bitwise operators
& | ^ ~ << >>
Assignment operators
 Relational operators
== > < >= <= !=
 Logical operators
&& "" !
 Assignment operators
= += -= *= /= %=
<<= >>= &= ^= |=
 Misc operators
sizeof() & * , ?:

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) );
OUTPUT
30
-10
200
2
0
printf(“ %d” (b % a) );
printf(“ %d” (c % a) );
printf (“%d“ (a++) );
printf(“%d “ (a--) );
printf(“%d “ (d++) );
printf(“%d “ (++d) );
0
5
10
11
25
27

int a = 10, b = 100;
float c = 10.5, d = 100.5;
printf("++a = %d n", ++a);
printf("--b = %d n", --b);
++a = 11
--b = 99
printf("++c = %f n", ++c);
printf("--d = %f n", --d);
--b = 99
++c = 11.500000
--d = 99.500000

int a = 10, b = 4, res;
res = a++;
printf("a is %d and res is %dn", a, res);
res = a--;
printf("a is %d and res is %dn", a,res);
a is 11 and res is 10
printf("a is %d and res is %dn", a,res);
res = ++a;
res = --a;

Bitwise operators
Operator Description
& Bitwise AND
| Bitwise OR
^ Bitwise exclusive OR
<< left shift
>> right shift
~ Bitwise Not

Bitwise operators
a b a & b a | b a ^ b ~a
0 0 0 0 0 1
0 0 0 0 0 1
0 1 0 1 1 1
1 0 0 1 1 0
1 1 1 1 0 0

Example: Bitwise Operators
int a = 60;
int b = 13;
int c = 0;
c = a & b; printf(“%d“, c );
c = a | b; printf(“%d" , c );
a= 60 = 0011 1100
b= 13 = 0000 1101
OUTPUT
c = a & b; 0000 1100 = 12
c = a | b; 0011 1101 = 61
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 );
c = a | b; 0011 1101 = 61
c = a ^ b; 0011 0001 = 49
c = ~a; 1100 0011 = -61
c = a << 2; 1111 0000 =
240
c = a >> 2; 1111 =15

Misc Operators
sizeof() : Returns the size of the variable
& : Returns the address of a variable
* : Pointer variable
* : Pointer variable
?: : Conditional / Ternary operator
Ex:
(a>b) ? printf("a is greater") : printf("b is greater");

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);
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
( 30 * 15 ) / 5
(30 * 15 ) / 5
(30) * (15/5)
e = (a + b) * (c / d);
// Print value of e
e = a + (b * c) / d;
// Print value of e
Value of (a + b) * (c / d) is : 90
Value of a + (b * c) / d is : 50
(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
++ --
+ -
Unary preincrement/predecrement
Unary plus/minus
+ -
! ~
(type)
*
&
sizeof
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 <
Greater than >
Less than equal to <=
1 == 2 != 3
operators == and
!= have the same
precedence, LR
Hence, 1 == 2 is
executed first
LR
Less than equal to <=
Greater than equal to >=
Equal to ==
Not equal !=
Logical Operators
AND && LR
OR || LR
NOT ! RL
executed first

If if else nested if
 Program to find the largest of three given numbers
using if else if ladder
using nested if

Flow chart : Largest
of three no’s
if (a >= b)
{
if (a >= c)
printf("a");
else
printf("c");
}
}
else
{
if (b >= c)
printf("b");
else
printf("c");
}

#include <stdio.h>
int main()
{
double a, b, c;
printf("Enter three numbers: ");
scanf("%lf %lf %lf", &a, &b, &c);
if (a >= b)
{
if (a >= c)
printf("%.2lf is the largest number.", a);
else
printf("%.2lf is the largest number.", c);
}
else
{
if (b >= c)
printf("%.2lf is the largest number.", b);
else
}
return 0;
}

Largest of three no’s
if (a >= b && a >= c)
printf("%d is largest", a);
else if (b >= a && b >= c)
else if (b >= a && b >= c)
printf("%d is largest", b);
else
printf("%d is largest ", c);

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')
else if(ch >= '0' && ch <= '9')
printf("'%c' is digit.", ch);
else
printf("'%c' is special character.", ch);

Program to check vowel or 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);

switch
switch (expression)
{
case constant1:
stmt1;
stmt2;
break;
case constant2:
case constant2:
stmt1;
stmt2;
break;
default:
// default statements
}

Ex:1: Print choice
based on input using
switch
Input
Choices available:
1. CST
// Print choices available
//Accept number
switch (num)
{
case 1:
printf("You selected CST "); break;
case 2:
printf("You selected CSE "); break;
2. CSE
3. ECE
4. CSBS
Enter your choice: 1
Output
You selected CST
printf("You selected CSE "); break;
case 3:
printf("You selected ECE "); break;
case 4:
printf("You selected CSBS "); break;
default:
printf("Wrong choice ");
}

 Program to perform arithmetic operations on two
given numbers using swtich case

switch(op)
{
case '+':
printf("Additionn");
c=a+b;
printf("Sum=%dn",c);
break;
case '-':
printf("Subtractionn");
case '/':
printf("Divisionn");
c=a/b;
printf("Quotient=%dn",c);
break;
case '%':
printf("Remaindern");
printf("Subtractionn");
c=a-b;
printf("Difference=%dn",c);
break;
case '*':
printf("Multiplicationn");
c=a*b;
printf("Product=%dn",c);
break;
printf("Remaindern");
c=a%b;
printf("Remainder=%dn",c);
break;
default:
printf("Invalid Optionn");
break;
}

Lab Program 2:
 Program to read angles or sides of a triangle
(based on options: 1) angles, 2) sides) and print if
it is equilateral or isosceles or isosceles
it is equilateral or isosceles or isosceles
perpendicular or just perpendicular or scalene
triangle.

Test Case 1
Find the type of triangle based on angles or sides.
1. Angles
2. Sides
2. Sides
Enter the first angle: 60
Enter the second angle: 60
Enter the third angle: 60
The triangle is: Equilateral triangle

Test Case 2
1. Angles
2. Sides
2. Sides
Enter the first side: 60
Enter the second side: 50
Enter the third side: 60
The triangle is: Isosceles triangle

Test Case 3
1. Angles
2. Sides
2. Sides
Invalid Choice! ! !

Test Case 4
1. Angles
2. Sides
2. Sides
Enter the first angle: 45
Enter the second angle: 90
Enter the third angle: 45
 The triangle is: Isosceles Perpendicular triangle

Test Case 5
1. Angles
2. Sides
2. Sides
Enter the first side: 70
Enter the second side: 80
Enter the third side: 90
The triangle is: Scalene triangle

If choice = 1 ie., Angles
 Equilateral triangle
 Isosceles Perpendicular triangle
Isosceles triangle
 Isosceles triangle
 Perpendicular triangle
 Scalene triangle
 It is not a triangle

a1==a2 && a2 == a3
a1==a2 && a2 == a3
a1
a2
a3

(a1==a2||a2==a3||a1==a3)
(a1==a2||a2==a3||a1==a3)
&&
(a1==90||a2==90||a3==90)

a1==a2||a2==a3||a1==a3

Isosceles triangle
(a1==90||a2==90||a3==90)
 Perpendicular triangle
 Scalene triangle
 It is not a triangle

Choice =2 ie., sides
if(s1==s2 && s2==s3) Equilateral triangle
else if((s1==s2||s2==s3||s1==s3)&&
(s1*s1==s2*s2+s3*s3 || s2*s2==s1*s1+s3*s3||
s3*s3==s1*s1+s2*s2))
Isoceles Perpendicular triangle

Choice =2 ie., sides
 else if(s1==s2||s2==s3||s1==s3)
Isoceles triangle
else if(s1*s1==s2*s2+s3*s3 || s2*s2==s1*s1+s3*s3 ||
s3*s3==s1*s1+s2*s2)
Perpendicular triangle

Start
Print 1. Angles 2. Sides
Accept choice
ch==1
T
A
B
F
Ch=2
T
Invalid choice

C1
C2
P1
C3
P2
P6
T
T
T
F
F
F
F
C1
if(a1+a2+a3==180)
C2
if(a1==a2 && a2==a3)
P1
Equilateral Triangle
C3
else if((a1==a2||a2==a3||a1==a3) &&
(a1==90||a2==90||a3==90))
P2
Isosceles Perpendicular
C4
else if(a1==a2||a2==a3||a1==a3)
P3
C5
else if(a1==90||a2==90||a3==90)
P4
P5
P6
A
P2
C4
C5
P3
P4 P5
T
T
F
F
Triangle
Isosceles Triangle
Scalene triangle
Not a triangle
B

C1
C2
P1
C3
P2
P6
T
T
T
F
F
F
F
C1
if((s1+s2>s3)&&(s2+s3>s1)&&(s1+s3>s2))
C2
if(s1==s2&&s2==s3)
P1
C3
else if((s1==s2||s2==s3||s1==s3)&&(s1*s1==s2*s2+s3*s3||
s2*s2==s1*s1+s3*s3||s3*s3==s1*s1+s2*s2))
P2
Triangle
C4
else if(s1==s2||s2==s3||s1==s3)
P3
Isosceles Triangle
C5
else if(s1*s1==s2*s2+s3*s3||s2*s2==s1*s1+s3*s3||
s3*s3==s1*s1+s2*s2)
P4
P5
Scalene triangle
P6
Not a triangle
B
P2
C4
C5
P3
P4 P5
T
T
F
F
Triangle
Isosceles Triangle
Scalene triangle
Not a triangle
c

if(ch==1)
{
//accept three angles and store in a1,a2,a3
if(a1+a2+a3==180)
{
if(a1==a2 && a2==a3)
{
printf("The triangle is: Equilateral trianglen");
}
else if((a1==a2||a2==a3||a1==a3) && (a1==90||a2==90||a3==90))
printf("The triangle is: Isosceles Perpendicular trianglen");
printf("The triangle is: Isosceles Perpendicular trianglen");
else if(a1==a2||a2==a3||a1==a3)
printf("The triangle is: Isosceles trianglen");
else if(a1==90||a2==90||a3==90)
printf("The triangle is: Perpendicular trianglen");
else
printf("the triangle is: Scalene trianglen");
}
else printf("It is not a triangle.n");
}

else if(ch==2)
{
//accept three sides and store in s1,s2,s3
if((s1+s2>s3)&&(s2+s3>s1)&&(s1+s3>s2))
{
if(s1==s2&&s2==s3)
{
printf("The triangle is: Equilateral trianglen");
}
else if((s1==s2||s2==s3||s1==s3)&&(s1*s1==s2*s2+s3*s3||s2*s2==s1*s1+s3*s3||s3*s3==s1*s1+s2*s2))
printf("The triangle is: Isosceles perpendicular trianglen");
else if(s1==s2||s2==s3||s1==s3)
else if(s1==s2||s2==s3||s1==s3)
printf("The triangle is: Isosceles trianglen");
else if(s1*s1==s2*s2+s3*s3||s2*s2==s1*s1+s3*s3||s3*s3==s1*s1+s2*s2)
printf("The triangle is: Perpendicular trianglen");
else
printf("The triangle is: Scalene trianglen");
}
else printf("It is not a triangle.n");
else
{
printf("Invalid choice!!!n");
}

Lab Program 3
 Write a C program to calculate the area and
perimeter of different shapes using switch statement.
 Triangle
 Triangle
 Rectangle
 Square
 Circle

while(op<5)
{
switch(op)
{
case 1:// Accept b & h
a=0.5*b*h;
// Display Area
// Accept s1,s2,s3
p=s1+s2+s3;
// Display Perimeter
case 1:
case 2:
case 3:
case 2:
//Accept l
a=l*l;
p=4*l;
// Display Area , Perimeter
case 3:
// Accept r
a=3.14*r*r;
p=2*3.14*r;
case 4:
// Accept l,b
a=l*b;
p=2*(l+b);
}
}
printf("Enter your optionn");
printf("1. trianglen2. squaren3. circlen4. rectanglen5. exitn");
scanf("%d",&op);
// Display Perimeter
break;
case 3:
case 4:
break;
break;

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

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

Print the number format shown using for
for(i = 1; i < 5; i++)
{
printf("n");
printf("n");
for(j = i; j > 0; j--)
{
printf("%d", j);
}
}

Print the number format shown using for
int r=0, c=0;
for(r=0; r<10;r++)
{
for(c=0; c<r; c++)
{
printf(" * ");
}
printf("n");
}

Program to print its multiplication table
i=1;
while(i<=10)
{
printf("%dn",(num*i));
i++;
i=1;
do
{
i++;
i++;
}
i++;
}while(i<=10);
for(i=1;i<=10;i++)
{
}

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

for(i=1; i<=n; i++)
{
if(i%2 == 0)
Print even numbers upto n
for(i=2; i<=n; i+=2)
{
printf("%dn",i);
if(i%2 == 0)
printf("%dn", i);
}
printf("%dn",i);
}

Print even for a given range
if(start%2 != 0)
start++;
start++;
for(i=start; i<=end; i+=2)
printf("%dn",i);

Factorial of a given no.
fact=1;
for(i=num; i>=1; i--)
fact=fact*i;

Sum of n natural no’s
sum=0;
for (i = 1; i <= n; ++i)
sum += i;

Draw flowchart and Find the output
for(i = 2; i <= 6; i = i + 2)
printf("%dt", i + 1);
Output
3 5 7

Draw flowchart and Find the output
for(i = 2; i != 11; i = i + 3)
Output
3 6 9

Print even numbers upto n
for(i=1; i<=n; i++)
{
if(i%2 == 0)
for(i=2; i<=n; i+=2)
{
if(i%2 == 0)
printf("%dn", i);
}
{
printf("%dn",i);
}

Loop: while
Syntax:
while (test_expression)
while (test_expression)
{
statement/s to be executed.
}

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

Lab Program
 Program to generate a series of ‘N’ numbers based
on the pattern of the numbers as : 9 13 22 36 55
79
79

Test Case - 1
User Output
Enter the number of terms you want: 6
The series is: 9 13 22 36 55 79
Test Case - 2
User Output
The series is: 9 13 22 36 55 79 108 142 181 225
Test Case - 3
User Output
The series is: 9 13 22 36 55 79 108 142 181 225 274 328 387 451 520 594 673 757 846 940

int n,term=9,gap=4,i;
scanf("%d",&n);
for(i=1;i<=n;i++)
for(i=1;i<=n;i++)
{
printf(" %d",term);
term=term+gap;
gap=gap+5;
}
printf("n");

rev=0;
while (n != 0)
{
Reverse of given no.
{
remainder = n % 10;
rev = rev * 10 + remainder;
n /= 10;
}

Count no. Of digits
count=0;
do {
n /= 10;
++count;
} while (n != 0);

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

Lab Program 4
Program to read monthly salary of an employee and
calculate Income tax to be paid based on the following
criteria:
criteria:
 < 100000 per year- no tax
 100001 to 200000 per year- 5%
 200001 to 300000 per year- 10%
 300001 to 500000 per year- 20%
 > 500000 per year- 30%

Expected Output
Enter your monthly salary: 35000
You have to pay 39000.000000/- as income tax

ysal=12*msal;
if(ysal<0)
{
printf("Invalid Salary!!!n");
exit(0);
}
else if(ysal<=100000)
{
printf("You are exempted from Income Tax.n");
{
it=5000+(ysal-200000)*0.1;
}
{
it=15000+(ysal-300000)*0.2;
}
printf("You are exempted from Income Tax.n");
}
{
it=(ysal-100000)*0.05;
}
}
else
{
it=55000+(ysal-500000)*0.3;
}
printf("You have to pay %lf/- as income
taxn",it);

Accept name from the user
Method 1
char name[20];
printf("Enter your name:");
Method 2
char name[20]
printf(“Enter your name:”)
scanf("%s",name);
printf("Your name is: %s",name);
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()
Method 4
char name[20];
int main()
{
char name[MAX_LIMIT];
fgets(name,MAX_LIMIT,stdin);
scanf("%[^n]%*c",name);

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.
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
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
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

Programming for Problem Solving

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Programming for Problem Solving