1. C++ Handbook for
Class IX
Computer Science
Programming is like playing a game. The coach can assist
the players by explaining the rules and regulations of the
game. But it is when the players practice on the field,
they get the expertise and nuances of the game. Same
way, the teacher explains the syntax and semantics of
the programming language in detail. It is when the
students explore and practice programs on the
computer; they get the grip of the language.
Programming is purely logical and application oriented.
Learning the basics of C++ in IX standard would be
helpful if the student wishes to take up Computer
Science in the Senior Secondary Level.
Priya Kumaraswami
10/27/2013
2. Acknowledgement
I should thank my family members for adjusting with my time schedules and giving inputs and
comments wherever required.
Next, my thanks are due to my colleagues and friends who provided constant support.
My sincere thanks to all my students, their queries and curiosity have helped me compile the
book in a student friendly manner.
Special thanks to Gautham, Nishchay and Mohith for reviewing the book and providing valuable
comments.
Thanks to the Almighty for everything.
Priya Kumaraswami
2
3. Table Of Contents
Chapter – 1...........................................................................................................................4
Introduction to Programming...............................................................................................4
Chapter – 2...........................................................................................................................7
First Program in C++...........................................................................................................7
Chapter – 3.........................................................................................................................11
C++ Variables and Constants.............................................................................................11
Chapter – 4.........................................................................................................................18
C++ Datatypes...................................................................................................................18
Chapter – 5.........................................................................................................................25
Operators and Expressions.................................................................................................25
Chapter – 6.........................................................................................................................31
Input and Output................................................................................................................31
Chapter – 7........................................................................................................................36
Conditions (if and switch)..................................................................................................36
Chapter – 8.........................................................................................................................55
Loops..................................................................................................................................55
Chapter – 9.........................................................................................................................74
Number Arrays...................................................................................................................74
Chapter – 10.......................................................................................................................85
Char Arrays........................................................................................................................85
Chapter – 11.......................................................................................................................94
Functions............................................................................................................................94
Worksheet - 1..................................................................................................................108
Worksheet – 2 (if..else)....................................................................................................111
Worksheet – 3 (switch).....................................................................................................113
Worksheet – 4 (Loops).....................................................................................................120
Worksheet – 5 ..................................................................................................................123
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4. Chapter – 1
Introduction to Programming
•
Program – Set of instructions / command given to the
computer to complete a task
•
Programming – The process of writing the
instructions / program using a computer language to
complete the given task.
Stages in Program Development
1. Analysis – Deciding the inputs required, features
offered and presentation of the outputs
2. Design – Algorithm or Flowchart can be used to design
the program.
1. The given task / problem is broken down into
simple steps. These steps together make an
algorithm.
2. If the steps are represented diagrammatically
using special symbols, it is called Flowchart.
3. The steps to arrive at the specified output from
the given inputs are identified.
3. Coding – The simple steps are translated into high
level language instructions.
1. For this, the rules of the language should be
learnt (syntax and semantics).
4. Compilation – The high level language instructions are
converted to machine language instructions.
– At this point, the syntax & semantic errors in
the program can be removed.
– Syntax Error: error due to missing colon,
semicolon, parenthesis, etc.
– Semantic Error: it is a logical error. It is due
to wrong logical statements. (statements that do
not give proper meaning.)
4
5. 5. Running / Execution – The instructions written are
carried out in the CPU.
6. Debugging / Testing – The process of removing all the
logical errors in the program by checking all the
conditions that the program needs to satisfy.
Types / Categories of
Programming techniques
1. Procedural Programming
2. Object Oriented Programming (OOP)
Comparison
• In Procedural Programming, importance is given to the
sequence of things that needs to be done and in OOP,
importance is given to the data.
• In Procedural Programming, larger programs are divided
into functions / steps and in OOP, larger programs are
divided into objects.
• In Procedural Programming, data has no security,
anyone can modify it. In OOP, mostly the data is
private and only functions / steps inside the object
can modify the data.
Why should we learn a
programming language?
•
To write instructions / commands to the computer to
perform a task
All the programs and processes running inside the computer
are written in some programming language.
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6. Components / Elements of a
programming language
1. Constants – Entities that do not change their values
during program execution. Example - 4 remains 4
throughout the program. “Hello” remains “Hello”
throughout the program
2. Variables – Entities that change their values during
program execution. These are actually memory locations
which can hold a value.
3. Operators
1. Arithmetic – Operations like +, -, *, /, ^
2. Relational – Compares two values (<, >, <=, >=, !
=, ==)
3. Logical – AND, OR, NOT, used in conditions
4. Expressions – Built with constants, variables and
operators.
5. Conditions – The statements that alter the flow of the
program based on their truth values (true / false).
6. Loops – A piece of code that repeats itself until a
condition is satisfied.
7. Arrays – Continuous memory locations which store same
type of data. Used to store bulk data. Single data is
stored in variables.
8. Functions – Portion of code within a
large program that performs a specific task and which
can be called as required
6
7. Chapter – 2
First Program in C++
// my first program in C++
#include <iostream.h>
void main()
{
cout << "Hello World!";
}
Explanation
// my first program in C++
•
This is a comment statement which is used for giving
remarks in between the program.
• Comments are not compiled / executed.
• This is a single line comment.
• For multi line comments, /* ……………… */ should be used.
• Example
/* This is a multi line comment
spanning three lines to demonstrate
the syntax of it */
#include <iostream.h>
•
•
•
#include <iostream.h> is written in a program because
it is required to do input and output operations.
The functions for input and output operations are
available in iostream.h.
iostream.h is a standard library.
void main()
•
•
•
Every program should have a main() function.
A function will start with { and end with }.
All the statements inside the function should end with
a semicolon.
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8. •
•
•
The syntax of main function is given below.
void main()
{
…
…
}
Inside the main function, other statements can be
added.
We’ll learn how to declare, define and use other
functions in Chapter 11.
cout << "Hello World!";
•
•
•
•
•
•
•
•
•
•
•
cout statement is used for printing on the screen.
The syntax of cout statement is cout << “Message”;
We can combine many messages in one cout statement
using cascading technique as shown below (using
multiple <<).
cout << “Message 1” << “Message 2”;
If the Message 2 has to be printed in next line, endl
keyword should be added in between as follows.
cout << “Message 1” << endl << “Message 2”;
endl denotes end of line.
One more way to print the Message 2 in the next line
is using ‘n’ character.
‘n’ is called a newline character.
cout << “Message 1” << ‘n’ << “Message 2”;
cout << “Message 1” << “n” << “Message 2”;
Compilation
•
•
After writing the code, the code has to be compiled.
Compilation is the process of converting high level
language instructions to machine language instructions.
Execution
•
8
Execution is the process of running the instructions
one by one in the CPU and getting the results.
9. Integrated Development
Environment
•
•
An integrated development environment (IDE) is
a software application that provides
facilities for program development.
An IDE normally consists of:
o a source code editor where code can be written
o a compiler
o a provision to run the program
o a provision to debug the program
Exercise
1. Fill in the blanks
#include ____________
void main _____
{
_______ “ Hello” ;
________
“ How are you?”;
/* This program prints Hello in the first line
and How are you in the second line */
}
2. Write a program to print 3 lines of 10 stars.
3. Write a program to print your name, class, section and
age.
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11. Chapter – 3
C++ Variables and Constants
Before we learn C++ variables and constants, we should know
these terms – integer, floating point, character and string.
Integer represents any whole number, positive or negative.
The number should not contain decimal point or commas. They
can be used in arithmetic calculations.
Examples – 14, -19, 34, -504
Floating point represent any number (positive or negative)
containing decimal point. No commas. They can be used in
arithmetic calculations.
Examples – 14.025, -13.65, 506.505, -990.0, 0.65
Character represents any character that can be typed
through the keyboard. They cannot be used in arithmetic
calculations.
Examples – A, a, T, x, 9, 3, %, &, @, $
String (char array) represents a sequence of characters
that can be typed through the keyboard. They cannot be used
in arithmetic calculations.
Examples – Year2012, Gone with the wind, email@mailbox,
**(*)**
Constants
•
Constants are the entities that do not change their
values during program execution.
•
There are four types – string, character, floating
point and integer constants.
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12. String Constants Examples
String or character array constants should be always
enclosed in double quotes.
•
•
•
•
“123”
“Abc”
“This is some text.”
“so many $”
(character
(character
(character
(character
array
array
array
array
)
)
)
)
char Constants Examples
Character constants should be always enclosed in single
quotes.
•
•
‘A’
‘z’
(character)
(character)
integer constants Examples
•
•
123
34
(integer)
(integer)
floating point constants Examples
•
•
34.78
5666.778
(floating point)
(floating point)
A sample program using constants
12
13. Variables
•
Variables are actually named memory locations which
can store any value.
•
It is the programmer who assigns the name for a memory
location.
•
Variables are the entities that can change their
values during program execution.
Example
A = 10;
A contains 10 now
A = A + 1;
A contains 11 now
A = 30;
A contains 30 now
•
There can be integer variables which can store integer
values, floating point variables to store floating
point values, character variables to store any
character or string variables to store a set of
characters.
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14. The type (i.e., datatype) of a variable will be covered in
next chapter.
Rules for naming the variables
•
•
A variable name should not contain space or any other
special character.
•
14
All variable names should begin with a letter and
further it can have digits or underscores or letters.
Another rule that you have to consider when naming the
variables is that they cannot match any keyword of the
C++ language. The standard reserved keywords in C++
are:
15. •
The C++ language is a "case sensitive" language. It
means that a variable written in capital letters is
not equivalent to another one with the same name but
written in small letters. Thus, for example, the
RESULT variable is not the same as the result variable
or the Result variable. These are three different
variables.
•
Generally, it is considered a good practice to name
variables according to their purpose/functionality.
Exercise
1. Identify the type of constants
a. 123.45
b. 145.0
c. 1166
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18. Chapter – 4
C++ Datatypes
•
Datatypes are available in a programming language to
represent different forms of data and to determine the
bytes used by each form of data.
•
Datatype technically indicates the amount of memory used
in the form of bytes.
•
•
•
•
•
So we need to understand bits and bytes.
A bit in memory can store either 1 or 0.
8 bits make a byte.
A byte can be used to store a sequence of 0’s and 1’s.
Example – 10110011 or 11110000
Decimal to Binary Conversion
Calculation of the range
•
18
With 1 bit, we can represent 2 numbers (21)
Bit Combination Decimal Value
0
0
19. •
1
1
With 2 bits, we can represent 4 numbers (22)
Bit Combination Decimal Value
00
0
01
1
10
2
11
3
•
With 3 bits, we can represent 8 numbers (23)
Bit Combination Decimal Value
000
0
001
1
010
2
011
3
100
4
101
5
110
6
111
7
•
With 8 bits, we can represent 256 numbers (28). If it
has to cover both +ve and –ve numbers, that 256 has to
be split as -128 to + 127 including 0.
•
With 16 bits, we can represent 65536 numbers (216). If
it has to cover both +ve and –ve numbers, that 65536
has to be split as -32768 to + 32767 including 0.
Fundamental Datatypes
•
There are five basic data types in C++ - int, char,
float, double and void.
•
int datatype is used to represent an integer.
• char datatype is used to represent a single character.
char is internally stored as a number as the system can
only understand numbers, that too binary numbers. So each
character present in the keyboard has a number equivalent,
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20. called ASCII code. The ASCII table for the alphabet and
digits are given below.
Char
ASCII code
A to Z
65 to 90
a to z
97 to 122
0 to 9
48 to 57
Please refer wikipedia for the complete ASCII table (for
all characters in the keyboard).
•
float datatype is used to represent a floating point
number.
•
double datatype is used to represent a larger floating
point number.
•
void datatype is used in C++ functions covered in chapter
11. It means “nothing” stored.
•
There are other data types like long, short, enum, union
etc out of which long is used more often.
•
long datatype is used to represent larger integers.
•
The following table gives the size in bytes and the range
of numbers accommodated by each datatype.
Datatype Size in bytes
(Memory used)
Range
char
1 byte
-128 to 127
int
float
double
long
2
4
8
4
-32768 to 32767
Approx 7 digits
Approx 15 digits
-2147483648 to
2147483647
void
None
bytes
bytes
bytes
bytes
Variable Declaration
20
Example data that can
be stored using the
given datatype
‘a’, ‘A’, ‘$’, ‘1’,
‘0’, ‘%’
(Any single character
on the keyboard)
12, 9489
12.43, 6789.566
56789.66666
45674, 998304
21. •
Any variable should be declared before we use it.
•
Associating a variable name with a datatype is called
declaration.
int a;
float mynumber;
These are two valid declarations of variables.
The first one declares a variable of type int with the
name a. Now a can be used to store any integer value.
It will occupy 2 bytes in memory.
The second one declares a variable of type float with
the name mynumber. Now mynumber can be used to store
any floating point value. It will occupy 4 bytes in
memory.
Similarly, we can declare char, long, double variables.
char ch;
long num;
double d;
•
(ch occupies 1 byte)
(num occupies 4 bytes)
(d occupies 8 bytes)
If you are going to declare more than one variable of
the same type, you can declare all of them in a single
statement by separating their names with commas.
For example:
int a, b, c;
This declares three variables (a, b and c), all of
them of type int, and has exactly the same meaning as:
int a;
int b;
int c;
•
Multiple declaration of a variable is an error.
For example,
int a;
int a = 2; //error as a is already declared in the
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22. previous statement
Variable Initialization
•
Giving an appropriate value for a variable is known as
initialization
•
Examples
int a = 10;
float b = 12.5;
float c;
c = 26.0;
char ch1, ch2;
ch1 = ‘q’;
ch2 = ‘p’;
long t = 40000;
double d = 1189.345;
int x = 9, y = 10; (Multiple initializations)
A sample program using variables
#include <iostream.h>
void main ()
{
int a, b;
int result;
a = 5;
b = 2;
a = a + 1;
// 1 is added to a value and stored in
// a again. a becomes 6 now.
result = a - b; // result contains 4 now
cout << result; // 4 is printed on screen
}
Exercise
1. Write correct declarations for the following
a. A variable to store 13.5
b. A variable to store Hello
c. A variable to store grade (A / B / C / D / E)
d. A variable to store 10000
22
23. 2. Convert the following decimal numbers to binary
a. 1024
b. 255
c. 1189
d. 52
3. Draw the table to show the bit combination and
decimal value for 4 bits binary.
4. Identify the errors in the following
a. int a = ‘a’;
b. float x; y; z;
c. int num = 45678;
d. char ch = “abc”;
e. char c = ‘abc’;
f. int m$ = 10;
g. long m = 90; n = 3400;
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25. Chapter – 5
Operators and Expressions
Two Categories
•
•
Binary operators – operators which take 2 operands.
Examples +, - , > , <, =
Unary operators – operators which take 1 operand.
Examples ++, --, !
Operations and the Operators
Arithmetic Operators +, -, *, /, %
Arithmetic operators are used to perform addition (+),
subtraction (-), multiplication (*) and division (/).
% is a modulo operator which gives the remainder after
performing the division of 2 numbers.
10%3 will give 1.
14%2 will give 0.
27%7 will give 6.
Logical Operators &&, ||, !
Logical operators are used in conditions to combine
expressions.
a > 10 && a < 100
b == 10 || b == 20
!(b == 10)
Relational Operators >, <, >=, <=, !=, ==
Relational operators are used for comparisons.
10 > 2 will give true (1).
12 == 2 will give false (0).
6 <= 12 will give true (1).
7 != 7 will give false (0).
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26. Assignment Operator =
Assignment operator is used to assign a value to a
variable
a = 10;
b = b + 20;
c = b;
There are shortcuts used in assignment.
A = A + B; can be written as A += B;
A = A * B; can be written as A *= B; (similarly for - , /
and % operators)
Increment Operator ++
Decrement Operator -•
•
These two operators increment or decrement the value
of a variable by 1.
There are 2 versions of them – post and pre
Pre increment – First the increment happens and then the
incremented value is used in the expression
Example a = 10, b = 20
C = (++a) + (++b) = 11 + 21 = 32
Post increment – First the current value is used in the
expression and then the values are incremented
Example a = 10, b = 20
C = (a++) + (b++) = 10 + 20 = 30 then a becomes 11 and b
becomes 21
Pre decrement – First the decrement happens and then the
decremented value is used in the expression
Example a = 10, b = 20
C = (--a) + (--b) = 9 + 19 = 28
Post decrement – First the current value is used in the
expression and then the values are decremented
Example a = 10, b = 20
C = (a--) + (b--) = 10 + 20 = 30 then a becomes 9 and b
becomes 19
Sample Problem
int a =
int c =
a = a +
b = c cout <<
26
20 , b = 40;
a++ + ++b;
b++;
--a;
a << b << c;
20
21
61
61
+
+
–
0
41
41
61
61
= 61 (a becomes 21)
= 62 (b becomes 42)
= 0
will be printed
27. Expressions
C++ Expressions are the result of combining constants,
variables and operators. Expressions can be arithmetic or
logical.
Arithmetic expressions use arithmetic operators and
int/float constants and variables.
Examples of arithmetic expression
(a, b, c are integer variables)
a = b + c;
a = a + 20;
b++;
c = c * 13 % 5;
Logical expressions use logical or relational
operators with constants and variables. The result of a
logical expression is always true (any non zero value,
usually 1) or false.
Examples of logical expression
(a, b, c are integer variables)
a > 10 && a < 90
a != b
((c%6 == 1) || (c%6 == 2))
!(a < 100)
The evaluation of the expression generally follows order
of precedence (similar to BODMAS rule)
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28. Expressions with char variable
When char is used in an expression, its ASCII number
equivalent is taken.
Example
char ch = ‘a’;
ch = ch + 5;
cout << ch;
will print ‘f’ on the screen.
char c = ‘S’;
c = c – 2;
cout << c;
will print ‘Q’ on the screen
Exercise
1. Evaluate the following expressions
a. cout << 10 + 5 * 3;
b. int a = 10;
a += 20;
cout << a++;
c. cout << (10 > 5 && 10 < 20);
d. int z = !(10 < 50);
cout << z;
e. int f = 20 % 3 + 2;
cout << f;
2. Write C++ expressions for the following
a. C = A2 + B2
28
29. b.
c.
d.
3. Find
a.
A is greater than 0 and less than 100
Grade is A or B
Increment the value of X by 20
the output for the following code snippets
int a = 39, b = 47;
a = a++ + 20;
b = ++b = 40;
int c = ++a + b++;
cout << c;
b. int x = 23, y = 34;
x = x++ + ++x;
y = y + ++y;
cout << x << “ “ << y;
c. int m = 44, n = 55;
int k = m + --n;
int j = n – m--;
int l = k / j;
cout << l;
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31. Chapter – 6
Input and Output
The output operation is already illustrated in Chapter 2
– “First Program in C++”. A recap of the same is given
below.
Output using cout
cout is used along with the insertion operator, which is
written as << (two "lesser than" signs).
cout << "Output sentence"; // prints Output sentence on screen
cout << 120;
// prints number 120 on screen
cout << x;
// prints the content of x on screen
Notice that the sentence in the first statement is
enclosed between double quotes (") because it is a
constant string of characters. Whenever we want to use
constant strings of characters we must enclose them
between double quotes (") so that they can be clearly
distinguished from variable names. For example, these two
sentences have very different results
cout << "Hello";
cout << Hello;
// prints Hello
// prints the content of Hello variable
The insertion operator (<<) may be used more than once in
a single statement:
cout << "Hello, " << "I am " << "a C++ statement";
This last statement would print the message Hello, I am a
C++ statement on the screen.
We can combine variables, constants and expressions in a
cout statement.
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32. cout << "Hello, I am " << age << " years old and my zipcode is "
<< zipcode;
If we assume the age variable to contain the value 24 and
the zipcode variable to contain 90064 the output of the
previous statement would be:
Hello, I am 24 years old and my zipcode is 90064
It is important to notice that cout does not add a line
break after its output unless we explicitly indicate it,
therefore, the following statements:
cout << "This is a sentence.";
cout << "This is another sentence.";
will be shown on the screen one following the other without
any line break between them:
This is a sentence.This is another sentence.
even though we had written them in two different insertions
into cout. In order to perform a line break on the output
we must explicitly insert a new-line character. In C++ a
new-line character can be specified as n (backslash, n):
cout << "First sentence.n";
cout << "Second sentence.nThird sentence.";
This produces the following output:
First sentence.
Second sentence.
Third sentence.
Additionally, to add a new-line, you may also use
the endl manipulator. For example:
cout << "First sentence." << endl;
cout << "Second sentence." << endl;
would print out:
First sentence.
Second sentence.
32
33. Input using cin
cin is used along with the extraction operator, which is
written as >> (two "greater than" signs).
int age;
cin >> age;
The first statement declares a variable of
type int called age, and the second one waits for an input
from cin (the keyboard) in order to store it in this
integer variable.
You can also use cin to request more than one data input
from the user:
cin >> a >> b;
is equivalent to:
cin >> a;
cin >> b;
Note:
An uninitialized variable has junk value.
Example
int a;
a = a + 10;
cout << a;
The a value is undefined here.
Sample Program showing input and output
#include <iostream.h>
void main ()
{
int i;
cout << "Please enter an integer value: ";
cin >> i;
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34. cout << "The value you entered is " << i;
cout << " and its double is " << i*2;
}
In the above program, an int variable is declared and its
value is taken as input. Similarly, other type variables
(long, char, float, double) can be declared and their
values can be taken as inputs.
Exercise
1. Write programs for the following taking necessary inputs
a. To find and display the area of circle, triangle and
rectangle
b. To calculate the average of 3 numbers
c. To find and display the simple interest
d. To convert the Fahrenheit to Celsius and vice versa
F = 9/5 * C + 32
C = 5/9 (F – 32)
e. To convert the height in feet to inches (1 foot = 12
inches)
34
36. Chapter – 7
Conditions (if and switch)
Null Statement
•
•
•
•
Statements are the instructions given to the computer
to perform any kind of action.
Statements form the smallest executable unit within a
program
Statements are terminated with a ;
The simplest is the null or empty statement ;
Compound Statement
•
•
•
It is a sequence of statements enclosed by a pair of
braces { }
A compound statement is also called a block.
A compound statement is treated as a single unit and
can appear anywhere in the program.
Control Statements
•
•
•
•
•
36
Generally a program executes its statements from the
beginning to the end of main().
But there can be programs where the statements have to
be executed based on a decision or statements that
need to be run repetitively.
There are tools to achieve these scenarios and those
statements which help us doing so are called control
statements
In a program, statements can be executed sequentially,
selectively (based on conditions) or iteratively
(repeatedly in loops).
The sequence construct means the statements are being
executed sequentially, from the first statement of the
main() to the last statement of the main(). This
represents the default flow of statements.
37. Operators used in conditions
•
>, <, >=, <=, ==, !=, &&, ||, !
•
Examples
grade == ‘A’
a > b
!x
x >=2 && x <=10
grade == ‘A’ || grade == ‘B’
•
Please note that == (double equal-to) is used for
comparisons and not = (single equal-to)
•
Single equal-to is used for assignment.
•
AND (&&) and OR(||) can be used to combine conditions
as shown below
•
a > 20 && a < 40
Both the conditions a > 20 and a < 40 should be
satisfied in the above case to get a true.
•
a == 0 || a < 0
Either the condition a == 0 or the condition a < 0
should be satisfied to get a true value
•
•
•
•
Not operator (!) negates or inverses the truth value
of the expression
!true becomes false
!false becomes true
!(10 > 12) becomes true
Note:
a > 20 && < 40 is syntactically wrong.
a == 0 || < 0 is syntactically wrong.
Conditional structure: if
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Priya Kumaraswami
38. •
•
Also called conditional or decision statement
Syntax of if statement
if ( condition )
statement ;
•
•
•
•
Statement can be single or compound statement or null
statement
Condition must be enclosed in parenthesis
If the condition is true, statement is executed. If it
is false, statement is ignored (not executed) and the
program continues right after the conditional
structure.
Example
if (x == 100)
cout << "x is 100";
•
•
If x value is 100, “x is 100” will be printed.
If x value is 95 or 5 or 20 (anything other than 100),
nothing will be printed.
•
If we want more than a single statement to be executed
in case that the condition is true we can specify a
block using braces { }
if (x == 100)
{
cout << "x is ";
cout << x;
}
Note:
if (x == 100)
{
cout << "x is ";
cout << x;
}
•
•
If x value is 100, x is 100 will be printed.
If x value is 95 or 5 or 20 (anything other than 100),
nothing will be printed.
if (x == 100)
cout << "x is ";
cout << x;
•
38
If x value is 100, x is 100 will be printed.
39. •
If x value is 95 (anything other than 100), 95 will be
printed. This is because the if statement includes
only one statement in the absence of brackets. The
second statement is independent of if. Therefore, the
second statement gets executed irrespective of whether
the if condition becomes true or not.
Note:
Any non-zero value is true.
0 is false.
Conditional structure: if and else
We can additionally specify what we want to happen if the
condition is not fulfilled by using the keyword else. It is
used in conjunction with if.
if (condition)
statement1 ;
else
statement2 ;
// Note the tab space
// Note the tab space
Example:
if (x == 100)
cout << "x is 100";
else
cout << "x is not 100";
prints on the screen x is 100 if x has a value of 100, but
if it is not 100, it prints out x is not 100.
Conditional structure: if … else if … else
If there are many conditions to be checked, we can use the
if … else if … else ladder as shown below in the example.
if (x > 0)
cout << "x is positive";
else if (x < 0)
CS WORKBOOK IX
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Priya Kumaraswami
40. cout << "x is negative";
else
cout << "x is 0";
The above syntax can be understood as… if condition1 is
satisfied, do something. Otherwise, check if condition 2 is
satisfied and if so, do something else. Otherwise (else),
do completely something else.
Remember that in case we want more than a single statement
to be executed for each condition, we must group them in a
block by enclosing them in braces { }.
if (x > 0)
{
cout << "x is
cout << “ and
}
else if (x < 0)
{
cout << "x is
cout << “ and
}
else
{
cout << "x is
cout << “ and
}
positive";
its value is “ << x ;
negative";
its absolute value is “ << -x ;
0";
its value is “ << 0 ;
Points to remember
•
•
•
40
If there is only one statement for ‘if’ and ‘else’, no
need to enclose them in curly braces { }
Example
if ( grade == ‘A’)
cout << “Good grade”;
else
cout << “Should improve”;
In an if statement, DO NOT put semicolon in the line
having test condition
if ( y > max) ;
if the condition is true, only
null statement will be executed.
{
max = y;
}
41. Conditional structure: Nested if
•
In a nested if construct, you can have an if...else
if...else construct inside another if...else if
...else construct.
•
Syntax
if (condition 1)
{
if (condition 2)
statement 1;
else
statement 2;
}
else
statement 3;
if (condition 1)
statement 1;
else
{
if (condition 2)
statement 2;
else
statement 3;
}
if (expression 1)
{
if (condition 2)
statement 1;
else
statement 2;
else
{
if (condition 3)
statement 3;
else
statement 4;
}
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Priya Kumaraswami
42. Dangling else problem
if ( ch >= ‘A’)
if(ch <= ‘Z’)
upcase = upcase + 1;
else
•
•
Which if the else belongs to, in the above case?
The else goes with the immediate preceding if
•
The above code can be understood as shown below
if ( ch >= ‘A’)
{
if(ch <= ‘Z’)
upcase = upcase + 1;
else
others = others + 1;
•
If the else has to be matching with the outer if, then
use brackets { } as below
if ( ch >= ‘A’)
{
if(ch <= ‘Z’)
upcase = upcase + 1;
}
Difference between if…else if ladder and multiple
ifs
42
43. int a = 10;
int a = 10;
if( a >= 0)
if( a >= 0)
cout << a;
else if ( a >=
cout << a;
5)
if ( a >=
cout << a + 5;
5)
cout << a + 5;
else if (a >= 10)
if (a >= 10)
cout << a + 10;
cout << a + 10;
else
else
The output here is 10.
The first condition is
satisfied and therefore it
gets inside and prints 10.
The other conditions will
not be checked.
The output here is 101520.
The code here has multiple
ifs which are independent.
The else belongs to the
last if.
Conditional structure: switch
•
•
Switch is also similar to if.
But it tests the value of an expression against a list
of integer or character constants.
•
Syntax
switch (expression)
{
case constant1:
statements;
break;
case constant2:
statements;
break;
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Priya Kumaraswami
44. •
switch evaluates expression and checks if it is
equivalent to constant1, if it is, it executes group
of statements under constant1 until it finds
the break statement. When it finds the
break statement, the program jumps to the end of
the switch structure.
•
If expression was not equal to constant1 it will be
checked against constant2. If it is equal to this, it
will execute group of statements under constant2 until
a break keyword is found, and then will jump to the
end of the switch structure.
•
Finally, if the value of expression did not match any
of the specified constants (you can include as
many case labels as values you want to check), the
program will execute the statements included after
the default: label, if it exists (since it is
optional).
•
If there is no break statement for a case, then it
falls through the next case statement until it
encounters a break statement.
•
A case statement cannot exist outside the switch.
Example of a switch construct with integer constant
int n;
cin >> n;
switch(n)
{
case 1:
cout << “C++”;
break;
case 2:
cout << “Java”;
break;
case 5:
cout << “C#”;
break;
44
45. default:
cout << “Algol”;
break;
}
In the above program, if input is 1 for n, C++ will be
printed.
If input is 2 for n, Java will be printed. If input is 5
for n, C# will be printed. If any other number is given as
input for n, Algol will be printed.
Example of a switch construct with character
constant
char ch;
cin >> ch;
switch(ch)
{
case ‘a’:
cout << “Apple”;
break;
case ‘$’:
cout << “Samsung”;
break;
case ‘3’:
cout << “LG”;
break;
default:
cout << “Nokia”;
break;
}
In the above program, if input is ‘a’ for ch, Apple will be
printed. If input is ‘$’ for ch, Samsung will be printed.
If input is ‘3’ for ch, LG will be printed. If any other
character is given as input for ch, Nokia will be printed.
Scope of a variable
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46. •
A variable can be used only within the block it is
declared.
•
Example1 - the scope of j is within that if block
only, starting brace of if to ending brace of if.
if( a == b)
{
int j = 10;
cout << j;
à valid as j is declared inside
the if block
}
cout << j;
à invalid as j is used outside the
block
•
Example2 - the scope of n is within the main(),
starting brace of main() to ending brace of main()
void main()
{
int n;
cin >> n;
if ( n > 100)
{
cout << n << “is invalid”;
}
cout << “Enter the correct value for n”;
}
Sample Programs
7.1 Program to print pass or fail given the mark
#include <iostream.h>
void main()
{
int x;
cout << “Enter the mark”;
cin >> x;
if (x >= 40)
{
cout << “Pass J” ;
}
else
{
46
47. cout << “Fail L” ;
}
}
7.2 Program to find the grade
#include “iostream.h”
void main()
{
int x;
cout << “Enter the mark”;
cin >> x;
char grade;
if (x >= 80 && x <= 100)
{
grade = ‘A’;
}
else if ( x >= 60 && x < 80)
{
grade = ‘B’;
}
else if ( x >= 45 && x < 60)
{
grade = ‘C’;
}
else if ( x >= 33 && x < 45)
{
grade = ‘D’;
}
else if ( x >= 0 && x < 33)
{
grade = ‘E’;
}
else
{
cout << “Not a valid mark”;
grade = ‘N’;
}
cout << “The grade for your mark is” <<
grade;
}
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48. 7.3 Program to check whether a number is divisible by 5
#include <iostream.h>
void main()
{
int x;
cout << “Enter a number”;
cin >> x;
if (x % 5 == 0)
{
cout << “The number is divisible by 5”;
}
else
{
cout << “The number is not divisible by 5”;
}
}
7.4 Program to check whether a number is positive or
negative
#include <iostream.h>
void main()
{
int i;
cout << “Enter a number”;
cin >> i;
if( i >= 0)
cout << “positive integer”;
else
cout << “negative integer”;
}
7.5 Program to calculate the area and circumference of a
circle using switch
This program takes r (radius) as input and also a choice
n as input.
Depending on the choice entered, it calculates area or
circumference.
It calculates area if the choice is 1. It calculates
circumference if choice is 2.
#include <iostream.h>
void main()
{
int n , r;
48
49. cout << “Enter choice:(1 for area, 2 for
circumference)” ;
cin >> n ;
cout << “Enter radius”;
cin >> r;
float res;
switch(n)
{
case 1:
res = 3.14 * r * r;
cout << “The area is “ << res;
break;
case 2:
res = 3.14 * 2 * r;
cout << “The circumference is “ << res;
break;
default:
cout << “Wrong choice”;
break;
}
}
7.6 Program to calculate the area and circumference of a
circle using switch fall through
This program takes r (radius) as input and also a choice
n as input.
Depending on the choice entered, it calculates area or
(area and circumference).
It calculates area if the choice is 1. It calculates area
and circumference if choice is 2.
#include <iostream.h>
void main()
{
cout << “Enter choice:(1 for area, 2 for both)” ;
int n , r;
cin >> n >> r;
float res;
switch(n)
{
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Priya Kumaraswami
50. case 2:
res = 3.14 * 2 * r;
cout << “The circumference is “ << res;
case 1:
res = 3.14 * r * r;
cout << “The area is “ << res;
break;
default:
cout << “Wrong choice”;
break;
}
}
7.7 Program to write remarks based on grade
#include <iostream.h>
void main()
{
cout << “Enter the grade” ;
char ch;
cin >> ch;
switch(ch)
{
case ‘A’:
cout << “Excellent. Keep it up. “;
break;
case ‘B’:
cout << “Very Good. Try for A Grade”;
break;
case ‘C’:
cout << “Good. Put in more efforts”;
break;
case ‘D’:
cout << “Should work hard for better
results”;
break;
case ‘E’:
cout << “Hard work and focus required”;
break;
default:
cout << “Wrong Grade entered.”;
break;
}
}
50
51. Exercise
1. Find the output for the following code snippets
a. int a , b;
cin >> a >> b;
if( a % b == 0)
cout << a / b;
cout << a % b;
if a is 42 & b is 7 (1st time run)
if a is 45 & b is 7 (2nd time run)
b. int x, y = 4;
switch(x % y)
{
case 0:
cout << x;
case 1:
cout << x * 2;
break;
case 2:
cout << x * 3;
case 3:
cout << x * 4;
default:
cout << 0;
when x is given as 50,
51, 52, 53, 54, 55
}
c. char ch;
cin >> ch;
int val = 2;
switch (ch)
{
case ‘0’:
cout << “converting from giga to mega”;
cout << val * 1024;
break;
case ‘1’:
cout << “converting from giga to kilo”;
cout << val * 1024 * 1024;
break;
case ‘2’:
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Priya Kumaraswami
52. cout << “converting from giga to bytes”;
cout << val * 1024 * 1024 * 1024;
break;
default:
cout << “invalid option”;
break;
}
When ch is given as 1 2 3 a 0
d. int a, b;
cin >> a >> b;
if( ! (a % b == 0))
if( a > 10)
a = a + b;
else
a = a + 1;
else
b = a + b;
cout << a << “ “ << b;
when a = 10 and b = 5
when a = 9 and b = 3
when a = 8 and b = 6
2. Find the mistakes
#include <iostream.h>
void main()
{
int a, b;
if( a > b ) ;
cout << a << “is greater than “;
cout << b;
else if ( b > a)
cout << b << “is greater than “ ;
cout << a
else ( a == b)
cout << a << “is equal to “ <<;
cout << b;
}
3. Convert the following if construct to switch construct
int m1, m2;
if( m1 >= 80)
{
if( m2 >= 80)
cout << “Very Good”;
else if (m2 >= 40)
cout << “Good”;
else
cout << “improve”;
52
53. }
else if( m1 >= 40)
{
if( m2 >= 80)
cout << “Good”;
else if (m2 >= 40)
cout << “Fair”;
else
cout << “improve”;
}
else
{
if( m2 >= 80)
cout << “Good”;
else if (m2 >= 40)
cout << “improve”;
else
cout << “work hard”;
}
4. Write Programs
a. To accept the cost price and selling price and
calculate either the profit percent or loss percent.
b. To accept 3 angles of a triangle and check whether
the triangle is possible or not. If triangle is
possible, check whether it is acute angled or right
angled or obtuse angled.
c. To accept the age of a candidate and check whether
he can vote or not
d. To calculate the electricity bill according to the
given tariffs.
Units consumed
Charges
Upto 100 units
Rs 1.35 / unit
> 100 units and upto Rs 1.50 / unit
200 units
> 200 units
Rs 1.80 / unit
e. To accept a number and check whether the number is
divisible by 2 and 5, divisible by 2 not 5,
divisible by 5 not 2
f. To accept 3 numbers and check whether they are
Pythagorean triplets
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54. g. To accept any value from 1 to 7 and display the
weekdays corresponding to the number entered. (use
switch case)
h. To find the volume of a cube or a cuboid or a sphere
depending on the choice given by the user
Notes
54
55. Chapter – 8
Loops
A loop is a piece of code that repeats itself until a
condition is satisfied. A loop alters the flow of control
of the program. Each repetition of the code is called
iteration.
Parts of a loop
•
•
•
•
Initialization expression – Control / counter / index
variable has to be initialized before entering inside
the loop. This expression is executed only once.
(Initial Value with which the loop will start)
Test expression – This is the condition for the loop
to run. Until this condition is true, the loop
statements get repeated. Once the condition becomes
false, the loop stops and the control goes to the next
statement in the program after the loop.
(Final Value with which the loop will end)
Update expression – This changes the value of the
control / counter / index variable. This is executed
at the end of the loop statements for each iteration
(Step Value to reach the final value from the initial
value)
Body of the loop – Set of statements that needs to be
repeated.
for loop
•
Syntax
for (initialization expr ; test expr ; update expr)
statement;
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56. The statement can be simple or compound.
•
Example 1
In this code, i is the index variable. Its initial
value is 1. Till i value is less than or equal to 10,
the statements will be repeated. So, the code prints 1
to 10. After i=10 and printing it, the index variable
becomes 11, but the condition is not met. So the loop
stops.
Note:
i=i+1 can also be written as i++.
i=i-1 can also be written as i--;
•
Example 2
for (int a= 10; a >= 0; a=a-3)
{
cout << a;
}
This code prints 10 to 0 with a step of
So 10 7 4 1 get printed.
•
Example 3
for (int c= 10; c >= 1; c=c-2)
{
cout << ‘*’;
}
56
-3.
57. This code loops 10 to 1 with a step of -2. But the c
variable is not printed. ‘*’ is printed 5 times.
•
Example 4
for (int i = 0; i < 5; i=i+1)
cout << i * i;
This code prints 0 1 4 9 16. If there is only one
statement to be repeated, no need to enclose it in
brackets.
•
Example 5
for (int c= 10; c >= 1; c=c-2)
cout << c;
cout << c;
This code loops 10 to 1 with a step of -2. So 10 8 6 4
2 get printed. Since the brackets are not there, the
for loop takes only one statement into consideration.
The above code can be interpreted as
for (int c= 10; c >= 1; c=c-2)
{
cout << c;
}
cout << c;
So outside the loop, one more time c gets printed. So
0 gets printed.
10 8 6 4 2 0 is the output from the above code.
Variations in for loop
Null statement in a loop
for (a= 10; a >= 0; a=a-3); Note the semicolon here
cout << a;
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Priya Kumaraswami
58. This means the null statement gets executed repeatedly.
cout << a; is independent.
Multiple initialization and update in a for loop
for (int a= 1, b = 5; a <= 5 && b >= 1; a++, b--)
cout << a << “ “ << b << endl;
The output of the above code will be
1 5
2 4
3 3
4 2
5 1
Note:
for(int
for(int
for(int
for(int
c
c
c
c
=
=
=
=
0;
0;
0;
0;
c
c
c
c
<
<
<
<
5;
5;
5;
5;
c++)
c=c++)
c=c+2)
c+2)
correct
wrong
correct
wrong
The scope rules are applicable for loops as well.
A variable declared inside the body (block) of a for loop
or a while loop is not accessible outside the body (block).
while loop
•
Syntax
Initial expression
while (test expression)
Loop body containing update expression
•
58
Example 1
59. •
In this code, a is the index variable. Its initial
value is 0. Till a is less than or equal to 10, the
statements will be repeated. So, the code prints 0 to
10. After i=10 and printing it, the index variable
becomes 11, but the condition is not met. So the loop
stops.
Example 2
int a= 10;
while (a >= 0)
{
cout << a;
a=a-3;
}
This code prints 10 to 0 with a step of -3.
So 10 7 4 1 get printed.
Variations in while loop
Infinite loop
•
Example 1
int a= 10;
while (a >= 0); Note the semicolon here
{
cout << a;
a=a-3;
}
This means the null statement gets executed repeatedly. It
becomes infinite loop as the update does not happen within
the loop.
• Example 2
int a= 10;
while (a >= 0)
{
cout << a;
}
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Priya Kumaraswami
60. There is no update statement here. So this also becomes
infinite loop as it does not reach the final value.
Multiple initialization and update in a while loop
int a= 1, b = 5;
while (a <= 5 && b >= 1)
{
cout << a << “ “ << b << endl;
a++;
b--;
}
The output of the above code will be
1 5
2 4
3 3
4 2
5 1
Quick Comparison of for and
while
for loop syntax
while loop syntax
60
61. Nested Loops
•
•
A loop may contain one or more loops inside its body.
It is called Nested Loop.
Example 1
for(int i= 1; i<=3; i++)
{
for(int j= 2; j<=4; j++)
{
cout << i * j << endl;
}
}
Here, the i loop has j loop inside. For each i value,
the j value ranges from 2 to 4.
i value
i = 1
i = 2
i = 3
j
j
j
j
j
j
j
j
j
j
value
= 2
= 3
= 4
= 2
= 3
= 4
= 2
= 3
= 4
output
1
3
4
4
6
8
6
9
12
• Example 2
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Priya Kumaraswami
62. for(int i= 1; i<=2; i++)
{
for(int j= 3; j<=4; j++)
{
for(int k= 5; k<=6; k++)
{
cout << i * j * k<< endl;
}
}
}
Here, the i loop has j loop inside which has k loop
inside it. For each i value, the j value ranges from 2
to 4. For each value of j, k value ranges from 5 to 6.
i value
i = 1
j value
j = 3
j = 4
i = 2
j = 3
j = 4
•
k
k
k
k
k
k
k
k
k
value
= 5
= 6
= 5
= 6
= 5
= 6
= 5
= 6
output
15
18
20
24
30
36
40
48
Example 3
for(int i= 1; i<=2; i++)
{
for(int j= 3; j<=4; j++)
{
cout << i * j << endl;
}
for(int k= 5; k<=6; k++)
{
cout << i * k<< endl;
}
}
Here, the i loop has j loop and k loop inside it. But
j loop is independent of k loop. For each i value, the
j value ranges from 2 to 4. For each value of i, k
value ranges from 5 to 6.
i value
62
j value
k value
output
63. i = 1
j = 3
j = 4
k = 5
k = 6
i = 2
j = 3
j = 4
k = 5
k = 6
•
•
3
4
5
6
6
8
10
12
In the syllabus, we learn to use nested loops for
printing
Tables
Patterns
Jump Statements
We will be learning 2 jump statements – break and return.
Basically, jump statements are used to transfer the
control. The control jumps from one place of the code to
another place. We’ve already seen break statement in
switch. It breaks out of the switch construct. break can be
used inside a loop too to break out of the loop if the
condition is met.
Example
for(int c = 9; c < 30; c = c + 2)
{
if(c%7 == 0)
break;
}
cout << c;
c starts from 9 and goes till 29 with a step of 2.
If in between any c value is divisible by 7, it breaks out
of the loop and prints that value.
The output will be 21.
return statement is used to return control from a function
to the calling code. If return statement is present in main
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Priya Kumaraswami
64. function, the control will be given to the OS (which is the
calling code) and the program will stop.
Example
#include <iostream.h>
void main()
{
for(int i=0; i<10;i++)
{
cout << i;
if(i==4)
return;
}
}
The above program will print 0 1 2 3 4 and stop.
Sample Programs
8.1 Program to print multiplication tables from 1 to 10
upto x 20 using nested for loops.
#include <iostream.h>
void main()
{
for (int i = 1 ; i <= 10; i=i+1)
{
cout << i << “ table starts ”;
for (int j = 1 ; j <= 20; j=j+1) // this for loop
has only one statement
to repeat
cout << i << “ x “<< j << “ = “<< i * j<< endl ;
cout << i << “ table ends” ;
cout << endl;
}
}
8.2 Program to print multiplication tables from 1 to 10
upto x 20 using nested while loops.
#include <iostream.h>
void main()
{
int i = 1;
64
65. while (i <= 10)
{
cout << i << “table starts” ;
int j = 1;
while (j <= 20)
{
cout<<i << “ x “<< j << “ = “<< i * j<<endl;
j = j + 1;
}
cout << i << “table ends” ;
cout << endl;
i = i + 1;
}
}
*
8.3 Program to print the
**
pattern.
***
#include <iostream.h>
****
void main()
{
for (int i = 1 ; i <= 5; i++)
{
for (int j = 1 ; j <= i; j++)
cout << ‘*’;
cout << endl;
}
}
8.4 Program to print the pattern.
1
12
#include <iostream.h>
123
void main()
1234
{
for (int i = 1 ; i <= 5; i++)
{
for (int j = 1 ; j <= i; j++)
cout << j;
cout << endl;
}
CS WORKBOOK IX
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Priya Kumaraswami
66. }
8.5 Program to print the pattern.
#include <iostream.h>
void main()
{
for (char i = ‘A’ ; i <= ‘E’; i++)
{
for (char j = ‘A’ ; j <= i; j++)
cout << j;
cout << endl;
}
}
8.6 Program to print the pattern.
#include <iostream.h>
void main()
{
}
66
*
***
*****
*******
A
AB
ABC
ABCD
67. 8.7 Program to print the pattern.
*******
*****
#include <iostream.h>
void main()
{
***
}
8.8 Program to print the pattern.
*******
*
#include <iostream.h>
void main()
{
*
* *
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68. }
8.9 Program to print the pattern.
1
121
#include <iostream.h>
void main()
{
12321
1234321
}
8.10 Program to print the pattern.
#include <iostream.h>
void main()
{
68
2
242
24642
2468642
69. }
8.11 Program to print the pattern.
2468642
#include <iostream.h>
void main()
{
24642
242
}
8.12 Program to print the pattern.
$$$$$
$
$
$
$
#include <iostream.h>
void main()
{
$
$
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Priya Kumaraswami
70. }
8.13 Program to print the pattern.
$$$$$
$&&&$
#include <iostream.h>
void main()
{
}
Exercise
1. Find the outputs
70
$&&&$
$&&&$
71. a. int a = 33, b = 44;
for(int i = 1; i < 3; i++)
{
cout << a + 2 << b++ << endl;
cout << ++a << b – 2<< endl;
}
b. long num = 64325;
int d, r = 0;
while(num != 0)
{
d = num % 10;
r = r * 10 + d;
num = num / 10;
}
cout << r;
2. Convert the following nested for loops to nested while
loops and guess the output
for(int x = 10; x < 20; x=x+4)
{
for(int y = 5; y <= 50; y=y*5)
{
cout << x << endl;
}
cout << y << endl;
}
3. Write programs
a. To print the odd number series till 999
b. To print the sum of natural numbers from 1 to 100
c. To print the factorial of a given number
d. To check whether a number is prime or not
e. to generate the Fibonacci and Tribonacci series
f. to accept 10 numbers and print the minimum out of
them
g. to print the following patterns
1
55555
12345
54321
1
1
1357531
22
4444
1234
4321
21
131
13531
123
321
321
13531
131
333
CS WORKBOOK 333
IX
4444 22
%%%%%%%
1
%
%
123
%
%
12
12345
%%%%%%%
1234567
71 21
4321
1357531
Priya Kumaraswami
1
74. Chapter – 9
Number Arrays
•
•
•
An array represents continuous memory locations having
a name which can store data of a single data type.
An array is stored in contiguous memory locations.
Lowest address having the first element and highest
address having the last element.
Arrays can be single dimensional or multi dimensional.
Single Dimensional Array
Two Dimensional Array (Table format)
We learn only single dimensional arrays in class 9.
74
75. Declaration of an array
•
•
•
•
•
•
•
•
•
datatype arrayname [size];
Example
int a [6];
The datatype indicates the datatype of the elements
that can be stored in an array.
We give the arrayname and it has to follow the rules
of naming the variables.
An array has size. The size denotes the number of
elements that can be stored in the array. Size should
be positive integer constant.
In the above example, a is an array that can store 6
integers.
Each array element is referenced by the index number.
The index number always starts from zero.
So, An array A [6] will have the following elements.
A [0], A [1], A [2], A [3], A [4]and A[5].
The index number ranges from 0 to 5
In this example, we have used an array of size 6.
int a[6];
a[0] = 25;
a[1] = 36;
a[2] = 92;
a[3] = 45;
a[4] = 17;
a[5] = 63;
cout << a[4]; will print 17 on the screen.
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76. Calculation of bytes
The memory occupied by an array is calculated as
size of the element x number of elements
For example, int a [15]; will take up 2 (size of an int) x
15 (number of elements) = 30 bytes.
float arr[20]; will take up 4 (size of a float) x 20
(number of elements) = 80 bytes
Loops for input, processing and
output
Usually, we use loops to take array elements as input, to
display array elements as output.
Example
int a[6];
for(int i = 0; i < 6; i++)
cin >> a[i];
// input loop
for(i = 0; i < 6; i++)
a[i] = a[i] * 2;
// processing loop
for(i = 0; i < 6; i++)
cout << a[i];
// output loop
Combining the input and processing loops
int a[6];
for(int i = 0; i < 6; i++)
// input & processing
loop
{
cin >> a[i];
a[i] = a[i] * 2;
}
for(i = 0; i < 6; i++)
cout << a[i];
Note:
76
// output loop
77. Generally we avoid combining input and output of array
elements in a single loop because it mixes up the display
on the screen.
Taking the size of the array from the user
int a[50];
int size;
cin >> size;
for(int i = 0; i < size; i++)
// input & processing
loop
{
cin >> a[i];
a[i] = a[i] * 2;
}
for(i = 0; i < size; i++)
cout << a[i];
// output loop
Operations
The following operations are covered in class 9.
• Count of elements
• Sum of elements
• Minimum and Maximum in an array
• Replacing an element
• Reversing an array
• Swapping the first half with the second half
• Swapping the adjacent elements
• Searching for an element
Sample Programs
9.1 Program to count the odd elements.
#include <iostream.h>
void main()
{
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78. }
9.2 Program to sum up all the elements.
#include <iostream.h>
void main()
{
78
79. }
9.3 Program to find the max and min in the given array.
#include <iostream.h>
void main()
{
}
9.4 Program to replace an element.
#include <iostream.h>
void main()
{
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80. }
9.5 Program to reverse the array.
#include <iostream.h>
void main()
{
80
81. }
9.6 Program to swap the first half with the second half.
#include <iostream.h>
void main()
{
}
9.7 Program to swap the adjacent elements.
#include <iostream.h>
void main()
{
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82. }
9.7 Program to search for a given element using flag.
#include <iostream.h>
void main()
{
82
83. }
Exercise
1. Fill in the blanks
The following program searches for a given element using
count. Some portions of the program are missing. Complete
them.
#include <iostream.h>
void main()
{
int a [ 10];
for (
)
//input for array
;
int num;
cout << “Enter the number to find”;
cin >> num;
int count = 0;
for(__________________________)
{
if(____________)
count++;
}
if(count ______)
cout << “found in the array”;
else
cout << “not found in the array”;
}
2. Find the number of bytes required in memory to store
a. A double array of size 20.
b. A char array of size 30
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84. c. A float array of size 50
3. Write Programs
a. To count the number of elements divisible by 4 in
the given array of size 20
b. To sum up the even elements in the array of size 10
c. To combine elements from arrays A and B into array C
Notes
84
85. Chapter – 10
Char Arrays
Char arrays are similar to number arrays. The storage in
memory is similar to that of number arrays.
The difference comes when the char array is treated as a
single string. In that case, the input and output need not
use a loop. At the end, a null character 0 has to be
present. Null character marks the end of the string.
In the below diagram, though 8 bytes are allocated, we use
only 6 bytes. So the presence of null character next tells
us that the string has ended.
The above diagram shows a char array of size 8. But all the
8 bytes are not used since the array has to store only
“class9” which has length 6, one additional byte to store
the null character.
Declaration of char array
char arrayname [size];
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86. •
•
•
Example
char A [8];
We give the arrayname and it has to follow the rules
of naming the variables.
An array has size. The size denotes the number of
elements that can be stored in the array. Size should
be positive integer constant.
In the above example, A is an array that can store 8
characters including the null character.
•
•
Each array element is referenced by the index number.
The index number always starts from zero.
•
•
•
So, An array A [8] will have the following elements.
A [0], A [1], A [2], A [3], A [4], A[5], A[6] and A[7]
The index number ranges from 0 to 7
char array as a single string
Input Operation
char str [20];
gets (str);
where gets is a function to get a string from the user.
Here, the null character is automatically added.
Output Operation
puts (str);
86
87. where puts is a function to display a string on the
screen. This will display the string till the null
character.
To use gets() and puts() function, we need to
#include <stdio.h>
char array not as a string
Input Operation
char str [20];
for(int i=0; i<20;i++)
cin>> str[i];
Here, the null character is not added and it will take
all the 20 characters.
Output Operation
for(int i=0; i<20;i++)
cout << str[i];
This will display all the 20 characters on screen.
Library Functions
We use two libraries ctype.h and string.h to do the char
and string manipulation. ctype.h functions work at
character level. string.h functions work at string level
considering the sequence of characters as a single
string.
ctype.h functions
•
•
•
islower() to check if the char is lowercase
isupper() to check if the char is uppercase
tolower() to convert to lowercase
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88. •
•
•
•
toupper()
isalpha()
isdigit()
isalnum()
to
to
to
to
convert to uppercase
check if the char is alphabet
check if the char is digit
check if the char is alphanumeric
Usage of ctype.h functions
char ch;
cin >> ch;
if(islower(ch))
if(isupper(ch))
if(isalpha(ch))
if(isdigit(ch))
if(isalnum(ch))
ch = toupper(ch);
ch = tolower(ch);
string.h
•
•
•
•
functions
strlen()
strcmp()
strcpy()
strcat()
Usage of string.h functions
char s[30];
gets(s);
char c[30];
gets(c);
int len = strlen(s); //Calculates the actual length
if(strcmp(s, c) == 0) // if s and c are equal, it
will give 0
strcpy(s, c);
//copies c to s
strcat(s, c);
//Joins s and c and stores in s
char manipulation inside the
loop
All character manipulations take place inside the loop.
Example
88
89. char arr[40];
// declare char array of required size
gets (arr);
// take input for char array
int len = strlen(arr);
// find the exact length
for(int i=0; i<len; i++)
// loop to process
{
//arr[i] can be modified or processed here
//arr[i] denotes each character in the char array
}
Operations
The following operations are covered in class 9.
• Count of a char in an array
• Conversion of cases
• Replacing a char
• Reversing an array
• Checking for palindrome
• Searching for a char
Sample Programs
10.1
Program to take a string as input and change their
cases. For example, if “I am Good” is given as input,
the program should change it to “i AM gOOD”
#include “iostream.h”
#include “string.h”
#include “stdio.h”
#include “ctype.h”
void main()
{
//for
//for
//for
//for
cout and cin
strlen()
gets() and puts()
char functions(isupper()etc)
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90. char arr [51];
gets(arr);
int len = strlen(arr);
for(i = 0; i < len; i= i+1)
{
if( isupper(arr[i]))
arr[i] = tolower(arr[i]);
else if( islower(arr[i]))
arr[i] = toupper(arr[i]);
}
puts(arr);
}
10.2
Program to take a string as input and count the number
of spaces, vowels and consonants
10.3
Program to take a string as input and replace a given
char with another char
90
91. 10.4
Program to take a string as input, reverse it and
check if it’s a palindrome
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92. 10.5
Program to take a string as input and search for a char
Exercise
1. Find the output
92
93. a. char ch = ‘&’;
char st[20]="BpEaCeFAvourEr";
for(i=0;st[i] != ‘0’;i++)
{
if(st[i]>='D' && st[i]<='J')
st[i]=tolower(st[i]);
else if(st[i]=='A' || st[i]=='a'|| st[i]=='B'
|| st[i]=='b')
st[i]=ch;
else if(i%2!=0)
st[i]= st[i]+1;
else
st[i]=st[i-1];
}
puts(st);
b. char msg[20] = “WELCOME”;
for (int i = strlen (msg) - 1; i > 0; i--)
{
if (islower(msg[i]))
msg[i] = toupper (msg[i]);
else
if (isupper(msg[i]))
if( i % 2 != 0)
msg[i] = tolower (msg[i-1]);
else
msg[i]=msg[i-1];
}
cout << msg << endl;
2. Write Programs
a. To convert the vowels into upper case in a string
b. To replace all ‘a’s with ‘e’s (consider case)
c. To count the number of digits and alphabet in a
string
d. To find the longest out of 3 strings
Notes
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94. Chapter – 11
Functions
•
•
•
•
•
Large programs are difficult to manage and maintain.
A large program is broken down into smaller units
called functions.
A function is a named unit consisting of a set of
statements.
This unit can be invoked from other parts of the
program.
Two types
• Built in – They are part of the libraries which
come along with the compiler like strlen( ).
• User defined – They are created by the programmer.
Parts of a function in a Program
•
•
•
Function Prototype or Declaration (just after the
iostream.h)
Function Definition (below the main function)
Function Call (in the main function )
Example
#include “iostream.h”
int calculateresult (int, int);
int main( )
{
int a, b, c, d, e, f;
cin >> a;
cin >> b;
c = calculateresult (a, b);
cin >> d;
cin >> e ;
f = calculateresult (d, e);
cout << f;
return 0;
}
94
Function Prototype
Function call 1
Function call 2
95. int calculateresult (int p, int q)
{
int r;
r = p + q + (p * q);
return r;
}
Function Definition
Working of a function
Function Prototype is just to announce the presence of a
function later in the program.
Function call in the main() (or some other function) makes
the control go to the required function definition and do
the work.
The argument values in the function call are copied to the
argument values in the function definition.
Function Definition is the code which does the actual work
and may return a result.
The control goes back to the main() and the result can be
copied to a variable or printed or used in an expression.
Function Prototype Syntax
•
•
Function prototype is the declaration of the function
that tells the program about the type of the value
returned by the function and the number and type of
arguments.
It enables a compiler to carefully compare each use of
the function with the prototype to check whether the
function is invoked properly i.e.,
• the right arguments are passed (number of
arguments and type)
• The right type is expected as return type
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96. ReturnDatatype functionname ( datatype arg1, datatype arg2,
……) ;
ReturnDatatype denotes the datatype of the return value
from the function, if any. If the function is not
returning any value, then void should be given. A
function can return only one value.
•
•
void data type specifies an empty value
It is used as the return type of functions which do
not return a value.
Functionname denotes the name of the function and it has to
follow the rules of naming the variables.
Within brackets (), the argument list is present. For each
argument, the datatype and name should be written. There
can be zero or more arguments. If there are more arguments,
they should be comma separated.
Example
1. float func ( int a, int b );
func is the function name, a and b are the arguments with
int datatype, float is the return datatype
2. int demo ( float x);
demo is the function name, x is the argument with float
datatype, int is the return datatype
3. void test ( float m, int n, char p);
test is the function name, m is the first argument with
float datatype, n is the second argument with int datatype,
p is the third argument with char datatype, void is the
return datatype
4. int example ()
example is the function name, there are no arguments, int
is the return datatype
Function Definition Syntax
•
96
Function definition is just like Prototype declaration
except that it has body and it does not have a
semicolon
97. •
The function prototype and the function definition
must agree EXACTLY on the return type, function name
and argument / parameter list (number and type)
if the ReturnDatatype is not void
ReturnDatatype functionname ( datatype arg1, datatype arg2,
……)
{
return ____ ;
}
if the ReturnDatatype is void
void functionname ( datatype arg1, datatype arg2, ……)
{
return;
// This is optional
}
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98. Example
1. float func ( int a, int b )
{
float k = 0.5 * a * b;
return k;
}
2. int demo ( float x)
{
int n = x / 2;
return n;
}
3. void test ( float m, int n, char p)
{
cout << (m + n)* p;
}
4. int example ()
{
int k;
cin >> k;
k = k * 2;
return k;
98
99. }
Function call Syntax
if the ReturnDatatype is not void
ReturnDatatype variable = Functionname (arg1, arg2 …);
if the ReturnDatatype is void
Functionname (arg1, arg2 …);
Note that the arguments do not have their datatypes.
The returned value can be stored in a variable, printed or
used in an expression as shown below
•
•
•
int c = calculateresult (a, b );
returned value stored in a variable
cout << calculateresult (a, b);
returned value printed directly
int d = calculateresult (a, b) + m * n;
returned value used in expression
Example
1.
2.
3.
4.
float ans = func (a, b );
int res = demo (x);
test (m, n, p);
int val = example ();
Possible Function Styles
1.
2.
3.
4.
Void function with no arguments
Void function with some arguments
Non void function with no arguments
Non void function with some arguments
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100. Function Scope
•
Variables declared inside a function are available to
that function only
Example
Sample Programs
11.1 Program to calculate power ab using a function (returns
a value)
#include <iostream.h>
long power (int a, int b);
//Function Prototype
void main()
{
int n1, n2;
cin >> n1 >> n2;
long p = power (n1, n2); //Function Call
cout << p;
}
100
101. long power (int a, int b)
//Function Definition
{
long pow = 1;
for (int i = 1; i <= b, i++)
pow = pow * a;
return pow;
}
11.2 Program to calculate factorial n! using a function
(void)
#include <iostream.h>
void factorial (int n);
//Function Prototype
void main()
{
int a;
cin >> a;
factorial (a); //Function Call
}
void factorial (int m)
//Function Definition
{
long fact = 1;
for (int i = 1; i <= m, i++)
fact = fact * i;
cout << fact;
}
11.3 Program to calculate area of a triangle using a
function (area = ½ x b x h)
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Priya Kumaraswami
102. 11.4 Program to calculate volume and surface area of a
sphere using 2 separate functions (volume = 4/3 pi r3 ,
surface area = 4 pi r2)
11.5 Program to print the multiplication table of a given
number upto x 20 using a function
102
103. 11.6 Program to check whether a given number is prime using
a function
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104. 11.7 Program to find the max in an int array of size 20
using a function
Exercise
1. Find the output
a. #include <iostream.h>
int calc ( int x, int y);
void main()
{
int m = 60, n = 44;
int p = calc (m, n);
104
105. m = m + p;
p = calc (n, m);
n = n + p;
cout << m << “ “ << n << “ “ << p;
}
int calc ( int x, int y)
{
x = x + 10;
y = y – 10;
int z = x % y;
return z;
}
b. #include <iostream.h>
int calc ( int x[6], int s);
void main()
{
int arr[6] = { 12, 15, 3, 9, 20, 18 };
calc (arr, 6);
}
int calc ( int x[6], int y)
{
for(int i=0; i<6; i=i+2)
{
switch(i)
{
case 0:
case 1:
cout << x[i] * (i+1);
break;
case 2:
case 3:
cout << x[i] * 3;
case 4:
case 5:
cout << x[i] * 5;
break;
}
}
}
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106. 2. Rewrite the following program after correcting
syntactical errors
#include <iostream.h>
void demo ( int a ; int b)
void main()
{
float p, q;
int r = demo(p , q);
cout << r;
}
void demo ( int a ; int b);
{
p = p + 10;
q = q – 10;
r = p + q;
return r;
}
3. Write Programs using functions
a. To find a3 + b3 given a and b
b. To convert the height of a person given in inches to
feet
c. To print the sum of odd numbers in a given int array
d. To convert the cases of a char array
106
108. Worksheet - 1
1.
(C++ Datatypes, Constants, Variables, Operators, Input and Output)
Match the following
int a;
a. An integer constant
“c++”
b. A floating point
constant
C = 20;
c. A char constant
9.5
d. A string (char array)
constant
‘n’
e. A variable declaration
100
f. A variable
initialization
2. Give examples of your own for
a.
b. An integer constant
c. A floating point
constant
d. A char constant
e. A string (char array)
f.
g.
constant
A variable declaration
A variable
initialization
4.
5.
3.
List the basic data types in C++ and explain their ranges.
Identify constants, variables and data types in the following
statements.
a.
b. int a = 10;
e. cin >> d;
c. char c = ‘d’;
f. float f = 3.45;
d. cout << c << “alphabet”;
g. double d = m;
6.
Find the mistakes in the following C++ statements and write correct
statements.
a.
b. int a = 10,000;
f. cout << the result is
<< “k”;
c. char c = “f”;
g. cout << a(b+c);
d. char ch = ‘go’;
h. float f$ = 5.49;
e. cin >> “g”;
i. int k = 40,129;
8.
Identify the valid and invalid variables from the following giving
reasons for invalid ones
a.
b. int My num;
f. int num1;
c. int my_num;
g. int main;
d. int my-num;
h. int Main;
10.
e. int 1num;
Write C++ expressions (applying BODMAS)
7.
9.
11.
108
109. a.
b.
c.
d.
e.
a 2 + 2ab
n−b
g. k =
1
h. z = ab + 3 (x – y)2
a2b
2ab
pnr/100
a2 + 2ab + b2
f. sum =
n(n + 1)
2
12.
13. Write
a.
b.
c.
d.
e.
f.
g.
14.
15.
16.
C++ statements for the following
Initialize a variable c with value 10
Take an integer as input in variable k
Print a floating point variable f
Print a string constant “Programming is fun”
Print a string constant “Result” and an integer constant 10
Print a string constant “Result” and an integer variable x
Take 2 numbers as input and find their squares and print them
with proper messages
Write C++ programs for the following with proper messages wherever
necessary
a. Print the area and perimeter of a square whose side is 50
b. Print the sum, difference, product and quotient of any 2
numbers
c. Interchange the values of 2 variables using a third variable
d. To accept temperature in degree Celsius and convert it to
degree Fahrenheit (F = 9/5 * C + 32)
Evaluate the following if a = 88, b = 101, c = 34
a. b = (b++) + c;
a = a – (--b);
c = (++c) + (a--);
b. a * b / c + 10
c. a / 2 + b * b + 3 * c
d. a * b – a / b + c * 2
e. a = b + 20;
c = (a++) + 10;
b = b – (--a)
Find the output of the following
#include <iostream.h>
#include <iostream.h>
void main()
void main()
{
{
int x = 9, y = 99, z = 199;
int p = 20;
cout << x + 7 << endl;
int q = p;
x = x
CS WORKBOOK IX + y;
q = q + 3;
109
Priya Kumaraswami
110. 17.
#include <iostream.h>
#include <iostream.h>
void main
void main()
{
{
int k;
int s = 75;
k = k + 10;
int ____ = 100;
p = p + 20;
____ t = 85;
cin >> “k”;
int tot = ___ + m + t;
cin >> 10;
int a = tot / 3;
Find the mistakes in the following code sample
18. Fill in the blanks
19. List the types of operators.
20. Give examples of unary and binary operators.
110
111. Worksheet – 2 (if..else)
1. Write ‘if’ statements for the following
a. To check whether the value of int variable a is equal to 100
b. To check whether the value of int variable k is between 40
2.
and 50 (including 40 and 50)
c. To check whether the value of int variable k is between 40
and 50 excluding 40 and 50)
d. To check whether the value of int variable s is not equal
to 50
e. To check whether the value of a char variable ch is equal
to ‘lowercase a’ or ‘uppercase a’
Find the output of the following
a.
int i = j = 10;
if ( a < 100)
if( b > 50)
i = i + 1;
else
j = j + 1;
cout << i << “ “ << j ;
Value supplied for a and b a = 30, b = 30
Value supplied for a and b a = 60, b = 70
b.
int i = j = 10;
if ( a < 100)
{
if( b > 50)
i = i + 1;
}
else
j = j + 1;
cout << i << “ “ << j ;
Value supplied for a and b a = 30, b = 30
Value supplied for a and b a = 60, b = 70
c.
if (!3)
{
cout << “Tricky”;
}
cout << “Yes”;
d.
if ( 3 )
cout << “Tricky again”;
else
CS WORKBOOK IX
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Priya Kumaraswami
112. cout << “Am I right?”;
cout << “No??”;
e.
if ( 0 )
cout << “Third Time Tricky”;
cout << “Am I right?”;
f.
if ( !0 )
cout << “Fourth Time Tricky”;
cout << “No??”;
g.
if ( 0 )
cout << “Not again”;
else
cout << “Last time”;
cout << “Thank God”;
3. Find the mistakes and correct them.
int b;
a)
if (b = 10);
cout << “Number of bats = 10” ;
cout << “10 bats for 11 players… Not sufficient!”;
else if ( b = 15)
cout << “Number of bats = 15”;
cout << “Cool… Bats provided for the substitutes too..” ;
else (b = 20)
cout << “Number of bats = 20” ;
4. Write complete C++ programs using if construct
a. To find the largest and smallest of the given 3 numbers A,
B, C
b. To find whether the number is odd or even, if its even
number check whether it is divisible by 4.
c. To find whether a year given as input is leap or not
d. To convert the Fahrenheit to Celsius or vice-versa
depending on the user’s choice
e. To create a four function calculator ( +, -, /, *)
f. To calculate the commission rate for the salesman. The
commission is calculated according to the following rates
Sales
Commission Rate
112
113. 5.
30001 onwards
15%
22001 – 30000
10%
12001 – 22000
7%
5001 – 12000
3%
0 – 5000
0%
Illustrate Nested If with examples
Worksheet – 3 (switch)
1. Convert the following ‘if’ construct to ‘switch’ construct
2.
3.
char ch;
cin >> ch;
if( ch == ‘A’)
cout << “Excellent. Keep it up.”;
else if (ch == ‘B’)
cout << “Good job. Try to get A grade next time”;
else if (ch == ‘C’)
cout << “Fair. Should work hard to get good grades”;
else if (ch == ‘D’)
cout << “Preparation not enough. Should work very hard”;
else
cout << “Invalid grade”;
Find the output of the following
int ua = 0, ub = 0, uc = 0, fail = 0, c;
cin >> c;
switch(c)
{
case 1:
case 2: ua = ua + 1;
case 3:
case 4: ub = ub + ua;
case 5: uc = uc + ub;
default: fail = fail + uc;
}
cout << ua << “-“ << ub << “-“ << uc << “-“ << fail ;
The above code is executed 6 times and in each execution, the
value of c is supplied as 0, 1, 2, 3, 4 and 5.
Find the mistakes and correct them.
CS WORKBOOK IX
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Priya Kumaraswami
118. int b;
cin >> b;
switch (b)
{
case ‘10’;
cout <<
“Number of bats = 10” ;
break;
case ‘10’:
cout << “Number of bats = 15” ;
4. Write complete C++ programs using switch construct
a. To display the day depending upon the number (example, if 1
5.
118
is given, “Sunday” should be printed, if 7 is given,
“Saturday” should be printed.
b. To display the digit in words for digits 0 to 9
c. To calculate the area of a circle, a rectangle or a
triangle depending upon user’s choice
Convert the following ‘if’ construct to ‘switch’ construct
int a; char b;
cin >> a; cin >> b;
if( a == 1)
{
cout << “Engineering”;
if(b == ‘a’)
cout << “Mechanical”;
else if ( b == ‘b’)
cout << “Computer Science”;
else if ( b == ‘c’)
cout << “Civil”;
}
else if (a == 2)
{
cout << “Medicine”;
if(b == ‘a’)
cout << “Pathology”;
else if ( b == ‘b’)
cout << “Cardiology”;
else if ( b == ‘c’)
cout << “Neurology”;
}
else if (a == 2)
{
cout << “Business”;
if(b == ‘a’)
cout << “Finance”;
else if ( b == ‘b’)
cout << “Human Resources”;
else if ( b == ‘c’)
cout << “Marketing”;
}
119. 6. Find the output of the following if a gets values 0, 1 and 2 in
7.
three consecutive runs.
int a; cin >> a;
switch (a)
{
default:
cout << 'd';
case 0:
cout << 0;
case 1:
cout << 1;
}
Find the output of the following if a gets values 0, 1 and 2 in
three consecutive runs.
int a; cin >> a;
switch (a > 100)
{
default:
cout << 'd';
break;
case 0:
cout << 0;
break;
case 1:
cout << 1;
break;
}
CS WORKBOOK IX
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Priya Kumaraswami
120. Worksheet – 4 (Loops)
1. Write a program to print tables of 3,6,9,….n upto x15
2.
( multiplied by 15)
Write Programs to print the following patterns using nested loops
1
2
23
242
456
24642
2468642
(*****)
1
1
97531
&
(***)
13
31
7531
&$&
(*)
135
531
531
&$$$&
1357
7531
31
&$$$$$&
3. How many times “hello” will be printed in the following code
fragment:
for (i=0; i<5; i++)
for (j=0; j<4; j++)
cout<< “hello”;
4. Find the output
a. #include <iostream.h>
void main()
{
int A = 5, B = 10;
for (int c = 1; c <= 2 ; c++)
{
cout << “Line 1 “ << A++ << “ & “ << --B << endl;
cout << “Line 2 “ << B + 3 << “ & “ << A + 5 << endl;
}
}
b. #include <iostream.h>
void main()
{
int m = -3, n = 1;
while( m > -7 )
{
m = m - 1;
n = n * m;
cout << n;;
}
}
5. What is wrong with the following while loops?
a. int counter = 1;
b.
int counter = 1;
while(counter < 100)
while(counter < 100)
{
cout << counter;
cout << counter);
counter ++;
counter --;
}
6. What will be the output of the following code? Explain.
for (int
{
120
c = 0; c <= 10; c++)
121. if(c == 4) break;
else cout << c;
7.
8.
}
cout << c;
Convert the following nested for loops to nested while loops and
find the output
for(int i=0; i<5;i++)
{
for(int j=1;j<3;j++)
{
if( i != j)
cout << i * j;
}
}
Explain the steps and find the
output.
int a = 5, b = 15;
int num = 12345;
if ( (a + b) % 2 == 0)
int d;
{
int s = 0;
b = b % 10;
while (num != 0)
a = a + b;
{
cout << a << “ “ << b << endl;
d = num % 10;
}
cout << d << endl;
if ( a % 10 == 0)
s = s + d;
{
num = num / 10;
switch(a)
9. What will be the output?
a. for (int c =
cout << c;
cout << c;
b. for (int c =
cout << c;
cout << c;
c. for (int c =
{
cout << c;
cout << c;
}
d. for (int c =
{
cout << c;
}
cout << c;
{
0; c <= 10; c++);
0; c <= 10; c++)
0; c <= 10; c++)
0; c <= 10; c++)
CS WORKBOOK IX
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Priya Kumaraswami
122. 10. Which out of the following will execute 5 times?
a. for ( int j = 0; j <= 5; j++)
b. for ( int j = 1; j < 5; j++)
c. for ( int j = 1; j < 6; j++)
d. for ( int j = 0; j < 6; j++)
e. for ( int j = 0; j < 5; j++)
11. What is a nested loop?
12. What are the parts of a loop?
122
123. Worksheet – 5
Function related Programs
1. Write a C++ program with a function to find the volume of a cone.
The function has the following prototype:- float vol (int radius,
int height); volume of a cone = 1/3 pi r2 h
2. Write a C++ program with a function to find the simple interest.
The function has the following prototype:- void calcinterest (int
p, int q, float r);
Char Array Programs
3. Write a C++ program to convert the string given as input as
specified below
a. All capital letters should be replaced with the next letter
(for example – A should be changed as B, B should be
changed as C …. Z remains Z etc)
b. All small letters should be replaced with the previous
letter (for example – a remains a, b should be changed as
a, c should be changed as b etc)
c. Example Input Computer
Example Output Dnlotsdq
Integer / Float Array Program
4. Write a program in C++ which takes single dimensional array and
size of array and find sum of elements which are positive.
If 1D array is
10 , 2 , −3 , −4 , 5 , −16 , −17 , 23
Then positive numbers in above array is 10, 2, 5, 23
Sum = 10 + 2 + 5 + 23 = 40
Output is 40
5. Write a program in C++ to combine the contents of two equi-sized
arrays A and B by computing their corresponding elements with the
fornula 2 *A[i]+3*B[i], where value I varies from 0 to N-1 and
transfer the resultant content in the third same sized array.
6. Write a program in C++ which accepts an integer array and its
size and replaces elements having even values with its half and
elements having odd values with twice its value .
eg:
if the array contains
3, 4, 5, 16, 9
then the array should be rearranged as
6, 2,10,8, 18
Find the output of the following code snippets
7. void main()
{
int AY[5]={5,10,15,20,25};
int loop = 5;
for(int m=0; m<loop ;m++)
CS WORKBOOK IX
123
Priya Kumaraswami
124. {
}
switch (m)
{
case
case
case
case
}
0:
4: cout<<AY[m]*5;
2:
1: cout<<AY[m]<<endl;
}
8. void main()
{
char Text[ ]= “Mind@Work#”;
for (int I=0; Text[I] != ‘0’; I++)
{
if ( ! isalpha(Text[I]))
Text[I]=’*’;
else if (isupper (Text[I]))
Text[I]=Text[I]+1;
else
Text[i]=Text[I+1];
}
puts(a);
}
9. int stock[ ]={ 10,22,15,12,18};
10.
11.
124
int total=0;
for(int I=0; I<5; I++)
{
if(stock[I]>15)
total+=stock[I];
}
cout<<total;
void execute(int x,int y)
{
int temp = x+y;
x = x + temp;
if(y!=200)
cout << temp << ” “ << x << ” “ << y;
}
void main( )
{
int a=50, b=20;
execute (b, 200);
cout <<a << b << ”n”;
execute (a,b);
cout <<a << b << ”n”;
}
void arm(int n)
{
int number, sum=0,dg,dgg,digit;
number=n;
while(n>0)
{
dg=n/10;
dgg=dg*10;