Lecture
Handout
Introduction
to programming
Lecture
No. 6
Reading Material
Deitel
& Deitel – C++ How to Program chapter
2
2.7,
2.8, 2.9, 2.20
Summary
- Repetition Structure (Loop)
- Overflow Condition
- Sample Program 1
- Sample Program 2
- Infinite Loop
- Properties of While loop
- Flow Chart
- Sample Program 3
- Tips
Repetition
Structure (Loop)
In
our day to day life, most of the things are repeated. Days and nights repeat
themselves
30
times a month. Four seasons replace each other every year. We can see similar
phenomenon
in the practical life. For example, in the payroll system, some procedures
are
same for all the employees. These are repeatedly applied while dealing with the
employees.
So repetition is very useful structure in the programming.
Let’s
discuss a problem to understand it thoroughly. We have to calculate the sum of
first
10
whole numbers i.e. add the numbers from 1 to 10. Following statement may be one
way
to do it.
cout
<< “Sum of first 10 numbers is = “ << 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
+ 9 + 10;
This
method is perfectly fine as the syntax is right. The answer is also correct.
This
procedure
can also be adopted while calculating the sum of numbers from 1 to 100. We
can
write the above statement adding all the digits from 1 to 100. But this method
will not
be
suitable for computing the sum of numbers from 1 to 1000.The addition of a very
big
number
of digits will result in a very ugly and boring statement. Let’s analyze it
carefully.
Our first integer is 1, is there any other way to find out what is the next
integer?
Yes,
we can add 1 to the integer and get the next integer which is 2. To find the
next
integer
(i.e. 3) we add 1 to the previous integer (i.e. 2) and get the next integer
which is 3.
So
whenever we have to find out the next integer, we have to add 1 to the previous
integer.
We
have to calculate the sum of first 1000 integers by taking a variable sum of
type int. It
is
a good programming practice to initialize the variable before using it. Here,
we
initialize
the variable sum with zero.
int sum = 0;
Now
we get the first integer i.e. 1. We add this to the sum
(sum
becomes 0 + 1 = 1). Now
get
the next integer which can be obtained by adding 1 to the previous integer i.e.
2 and
add
it to the sum
(sum
becomes 1 + 2 = 3). Get the next integer by adding 1 to the
previous
integer and add it to the sum (sum
becomes 3 + 3 = 6) and so on. This way,
we
get
the next integer by adding 1 to the previous integer and the new integer to the
sum. It
is
obvious that we are repeating this procedure again and again i.e. adding 1 to
the
previous
integer and add this new integer to the sum. So we need some repetition
structure
in the programming language. There are many looping constructs in C
Language.
The repetition structure we are discussing in this lecture is 'while loop
structure'.
‘while’ is
also a key word of 'C' so it cannot be used as a variable name.
While
means, 'do it until the condition is true'. The use of while construct can be
helpful
in
repeating a set of instructions under some condition. We can also use curly
braces with
while just like we used with if. If we omit to use the braces with
while construct, then
only
one statement after while will be repeatedly executed. For good programming
practices,
always use braces with while irrespective of the number of
statements in while
block.
The code will also be indented inside the while block as Indentation makes the
code
easy to understand.
The
syntax of while construct is as under:
while ( Logical Expression )
{
statement1;
statement2;
………….
}
The
logical expression contains a logical or relational operator. While this
logical
expression
is true, the statements will be executed repeatedly. When this logical
expression
becomes false, the statements within the while block, will not be executed.
Rather
the next statement in the program after while
block, will be executed.
Let’s
discuss again the same problem i.e. calculation of the sum of first 1000
integers
starting
from 1. For this purpose, we need a variable to store the sum of integers and
declare
a variable named sum.
Always use the self explanatory variable names. The
declaration
of the variable sum in this case is:
int sum = 0;
The
above statement has performed two tasks i.e. it declared the variable sum of
type int
and
also initialized it with zero. As it is good programming practice to initialize
all the
variables
when declared, the above statement can be written as:
int sum;
sum = 0;
Here
we need a variable to store numbers. So we declare a variable number of
type int.
This
variable will be used to store integers.
int number;
As
we have declared another variable of int data
type, so the variables of same data type
can
be declared in one line.
int sum, number;
Going
back to our problem, we need to sum up all the integers from 1 to 1000. Our
first
integer
is 1. The variable number is
to be used to store integers, so we will initialize it by
1
as our first integer is 1:
number = 1;
Now
we have two variables- sum and number. That means we
have two memory
locations
labeled as sum and number which will be used to store sum of integers and
integers
respectively. In the variable sum, we have to add all the integers
from 1 to 1000.
So
we will add the value of variable number into
variable sum, till the time the value of
number
becomes 1000. So when the value of number becomes 1000, we will stop adding
integers
into sum. It
will become the condition of our while loop. We can say sum the
integers
until integer becomes 1000. In C language, this condition can be written as:
while ( number <= 1000 ) {
………Action ………
}
The
above condition means, 'perform the action until the number is 1000 or less
than
1000'.
What will be the Action? Add the number, the
value of number is 1 initially, into
sum.
This is a very simple statement:
sum = sum + number;
Let’s
analyze the above statement carefully. We did not write sum = number; as this
statement
will replace the contents of sum and the previous value of sum will
be wasted
as
this is an assignment statement. What we did? We added the contents of sum and
contents
of number
first (i.e. 0 + 1) and then stored the result of this (i.e. 1) to the sum.
Now
we need to generate next integer and add it to the sum. How can we get the next
integer?
Just by adding 1 to the integer, we will get the next integer. In ‘C’, we will
write
it
as:
number = number + 1;
Similarly
in the above statement, we get the original contents of number (i.e. 1). Add 1
to
them
and then store the result (i.e. 2) into the number. Now we need to add this new
number
into sum:
sum = sum + number;
We
add the contents of sum (i.e. 1) to the contents of number
(i.e. 1) and then store the
result
(i.e. 2) to the sum. Again we need to get the next integer which can be
obtained by
adding
1 to the number. In other words, our action consists of only two statements
i.e.
add
the number to the sum and get the next integer. So our action statements will
be:
sum = sum + number;
number =
number + 1;
Putting
the action statements in while construct:
while ( number <= 1000 ) {
sum = sum + number;
number = number + 1;
}
Let's
analyze the above while loop. Initially the contents of number is
1. The condition in
while loop (i.e. number <= 1000)
will be evaluated as true, contents of sum and
contents
of
number will
be added and the result will be stored into sum. Now
1 will be added to
the
contents of number and number
becomes 2. Again the condition in while loop
will be
evaluated
as true and the contents of sum will be added to the contents of number .The
result
will be stored into sum. Next 1 will be added to the
contents of number and number
becomes
3 and so on. When number
becomes 1000, the condition in while loop evaluates
to
be true, as we have used <= (less than or equal to) in the condition. The
contents of
sum will be added to the contents of number
(i.e. 1000) and the result will be stored into
the
sum.
Next 1 will be added to the contents of number and number
becomes 1001. Now
the
condition in while loop is evaluated to false, as number is
no more less than or equal
to
1000 (i.e. number has
become 1001). When the condition of while loop
becomes false,
loop
is terminated. The control of the program will go to the next statement
following the
ending
brace of the while construct. After the while
construct, we can display the result
using
the cout statement.
cout
<< “ The sum of first 1000 integers starting from 1 is “ << sum;
The
complete code of the program is as follows:
/* This program calculate the sum
of first 1000 integers */
#include <iostream.h>
main()
{
//declaration of variables
int sum, number;
//Initialization of the variables
sum = 0;
number = 1;
// using the while loop to find out the sum of first 1000 integers
starting from 1
while(number <= 1000)
{
// Adding the integer to the contents
of sum
sum = sum + number;
// Generate the next integer by adding
1 to the integer
number = number + 1;
}
cout << "The sum of first 1000 integers starting from 1 is
" << sum;
}
The
output of the program is:
The
sum of first 1000 integers starting from 1 is 500500
While construct is a very elegant and powerful construct. We have seen that
it is very
easy
to sum first 1000 integers just with three statements. Suppose we have to
calculate
the
sum of first 20000 integers. How can we do that? We just have to change the
condition
in the while loop (i.e. number <= 20000).
Overflow
Condition:
We
can change this condition to 10000 or even more. Just try some more numbers.
How
far
can you go with the limit? We know that integers are allocated a fixed space in
memory
(i.e. 32 bits in most PCs) and we can not store a number which requires more
bits
than integer, into a variable of data type, int. If
the sum of integers becomes larger
than
this limit (i.e. sum of integers becomes larger than 32 bits can store), two
things can
happen
here. The program will give an error during execution, compiler can not detect
such
errors. These errors are known as run time errors. The second thing is that 32
bits of
the
result will be stored and extra bits will be wasted, so our result will not be
correct as
we
have wasted the information. This is called overflow. When we try to store
larger
information
in, than a data type can store, overflow condition occurs. When overflow
condition
occurs either a run-time error is generated or wrong value is stored.
Sample
Program 1:
To
calculate the sum of 2000 integers, we will change the program (i.e. the while
condition)
in the editor and compile it and run it again. If we need to calculate the sum
of
first
5000 integers, we will change the program again in the editor and compile and
run it
again.
We are doing this work again in a loop. Change the program in the editor,
compile,
execute
it, again change the program, compile and execute it and so on. Are we doing
this
in
a loop? We can make our program more intelligent so that we don’t need to
change the
condition
every time. We can modify the condition as:
int upperLimit;
while (number <=
upperLimit)
where
upperLimit is a variable of data type int. When the value of upperLimit is
1000,
the
program will calculate the sum of first 1000 integers. When the value of
upperLimit is
5000,
the program will calculate the sum of first 5000 integers. Now we can make it
re-
usable
and more effective by requesting the user to enter the value for upper limit:
cout << “Please enter the
upper limit for which you want the sum ”;
cin >> upperLimit;
We
don’t have to change our program every time when the limit changes. For the sum
of
integers,
this program has become generic. We can calculate the sum of any number of
integers
without changing the program. To make the display statement more
understandable,
we can change our cout statement as:
cout << “ The sum of first “ << upperLimit << “
integers is “ << sum;
Sample
Program 2:
Problem
statement:
Calculate
the sum of even numbers for a given upper limit of integers.
Solution:
We
analyze the problem and know that while statement will be used. We need to sum
even
numbers only. How can we decide that a number is even or not? We know that the
number
that is divisible by 2 is an even number. How can we do this in C language? We
can
say that if a number is divisible by 2, it means its remainder is zero, when
divided by
2.
To get a remainder we can use C’s modulus operator i.e. %. We can say that for
a
number
if the expression (number % 2) results in zero, the number is even. Putting
this in
a
conditional statement:
If ( ( number % 2) == 0 )
The
above conditional statement becomes true, when the number is even and false
when
the
number is odd (A number is either even or odd).
The
complete code of the program is as follows:
/* This program calculates sum
of even numbers for a given upper limit of
integers
*/
#include <iostream.h>
main()
{
//declaration of variables
int sum, number, upperLimit;
sum = 0;
number = 1;
// Prompt the user to enter upper limit of integers
cout << “Please enter the upper limit for which you want the sum ”
;
cin >> upperLimit;
// using the while loop to find out the sum of first 1000 integers
starting from 1
while(number <= upperLimit)
{
// Adding the even integer to the
contents of sum
if ( ( number % 2 ) == 0 )
{
sum = sum + number;
}
// Generate the next integer by adding
1 to the integer
number = number + 1;
}
cout << "The sum of even numbers of first “ <<
upperLimit << “ integers starting
from 1 is " << sum;
}
The
output of the program is:
Please
enter the upper limit for which you want the sum 10
The
sum of even numbers of first 10 integers starting from 1 is 30
Suppose
if we don’t have modulus operator in the C language. Is there any other way to
find
out the even numbers? We know that in C integer division gives the integer
result
and
the decimal portion is truncated. So the expression (2 * (number / 2)) gives
the
number
as a result, if the number is even only. So we can change our condition in if
statement
as:
if ( ( 2 * ( number /2 ) )
== number )
Infinite
Loop:
Consider
the condition in the while structure that is (number <=
upperLimit) and in the
while block the value of number is
changing (number = number + 1) to ensure that the
condition
is tested again next time. If it is true, the while
block is executed and so on. So
in
the while block statements, the variable
used in condition must change its value so that
we
have some definite number of repetitions. What will happen if we do not write
the
statement
number
= number + 1; in our program? The value of number will
not change,
so
the condition in the while loop
will be true always and the loop will be executed
forever.
Such loops in which the condition is always true are known as infinite loops as
there
are infinite repetitions in it.
Property
of while loop:
In
the above example, if the user enters 0, as the value for upper limit. In the
while
condition
we test (number <= upperLimit) i.e. number is less than or equal to
upperLimit
(
0 ), this test return false. The control of the program will go to the next
statement after
the
while
block. The statements in while structure will not be executed
even for a single
time.
So the property of while loop is that it may execute zero or more time.
The
while loop
is terminated, when the condition is tested as false. Make sure that the
loop
test has an adequate exit. Always use braces for the loop structure. If you
forget to
put
the braces, only one statement after the while statement is considered in the while
block.
Flow
Chart:
At
first, we will draw a rectangle and write while in it. Then draw a line to its
right and
use
the decision symbol i.e. diamond diagram. Write the loop condition in the
diamond
and
draw a line down to diamond which represents the flow when the decision is
true. All
the
repeated processes are drawn here using rectangles. Then a line is drawn from
the last
process
going back to the while and decision connection line. We have a line on the
right
side
of diamond which is the exit of while loop. The while loop
terminates, when the loop
condition evaluates to false and the control gets out of while
structure.
Here
is the flow chart for sample program 2:
Sample
Program 3:
Problem
statement:
Calculate
the factorial of a given number.
Solution:
The
factorial of a number N is defined as:
N(N-1)(N-2)………….3.2.1
By
looking at the problem, we can see that there is a repetition of multiplication
of
numbers.
A loop is needed to write a program to solve a factorial of a number. Let's
think
in
terms of writing a generic program to calculate the factorial so that we can
get the
factorial
of any number. We have to multiply the number with the next decremented
number
until the number becomes 1. So the value of number will decrease by 1 in each
repetition.
factorial = 1;
number = 1;
// Prompt the user to enter upper limit of integers
cout << “Please enter the number for factorial ” ;
cin >> number;
// using the while loop to find out the factorial
while(number > 1)
{
factorial = factorial * number;
number = number - 1;
}
cout << "The factorial is
“ << factorial;
}
Exercise:
1) Calculate the sum of odd integers for a given
upper limit. Also draw flow chart of
the
program.
2) Calculate the sum of even and odd integers
separately for a given upper limit
using
only one loop structure. Also draw flow chart of the program.
Tips
• Always use the self explanatory variable names
• Practice a lot. Practice makes a man perfect
• While loop may execute zero or more time
• Make sure that loop test (condition) has an
acceptable exit.
Here
is the code of the program.
/*This
program calculates the factorial of a given number.*/
#include <iostream.h>
main()
{
//declaration of variables
int factorial, number;
//Initialization of the variables
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