Topic 8 Programing

7/27/2019 Topic 8 Programing

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1

CHAPTER 8:

PROGRAMMING

8.1 Introduction to Programming

sanggagakputih.blogspot.com



2


8.1.1 Definition8.1.2 Types of Programming Languages

8.1.3 Programming Language Paradigm8.1.4 Translator




3

LEARNING OUTCOME:

At the end of this topic, students shouldbe able to:

1. Define programming language.2. Differentiate between Low-level and

High-level Language.

3. State paradigms of programming language.4. Identify the functions of translators.





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What is Programming ???




5

A programming is a series of instructions thatdirects a computer to perform tasks.

– created by a programmer using a

programming language.- instructions must be written in a way thecomputer can understand.

Programming




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Computer Program - Example




7

What is a ProgrammingLanguage ???




8

Programming Language

Set of words, abbreviations, and symbols

that enables a programmer to communicate

instructions to a computer.

- Discovering Computer 2011 -




9

Programming Language

Languages that are used for…

Writing games

Writing application programs (like Excel)




10

10

8.1.2 Types of Programming Language

Low-levelLanguage

High-levelLanguage

Two Types




11

(1) Low-level Language

Programming language that is machinedependent that runs on only oneparticular type of computer.

These programs are not easily portableto other types of computer.




12

12

MachineLanguage

AssemblyLanguage

Two Types

Low-level Language




13

Low-level Language – Machine

Language

The FIRST generation of programminglanguage

The ONLY language that computer directly recognizes. Machine language instructions use aseries of binary digits

(1s and 0s) or combination of numbersand letters that represent binary digits.




14

Low-level Language – Machine

Language

Disadvantages:

their instructions are extremelycryptic and difficult to learn programs are lengthy Machine-dependent not portable.




15

Low-level Language – Assembly

Language

The SECOND generation of programminglanguage

Using symbolic instruction codes (meaningful abbreviat ions)

- A → addition- C → compare

- L → load- M → multiply




16


Language

Disadvantages:

Difficult to learn. (Programs must be convertto machine language before the computer can execute, or run the program.)Programmers must use assembler to convert

assembly language to machine language.




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Language

Disadvantages:programs are lengthy.Machine-dependentnot portable.




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Language




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(2) High-level Language

• instructions are quite English-like

• easier to learn than machine or assembly language

• a single instruction corresponds to

many operations at the machinelevel




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High-level Language

have to be converted to machine-languagebefore they can be executed

Portable

examples : FORTRAN, COBOL, BASIC,

Pascal, Ada, C and C++




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• Computer languages are categorized accordingto the approach they use to solve a problem.

• A paradigm, is a way in whichcomputer language looks at the problem to be

solved. (style/architecture of programming

language)

21

8.1.3 Programming LanguageParadigm




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FORTRAN

COBOL

BASIC

C

Pascal

Ada

Smaltalk

C++

Visual Basic

Java

C#

F#

LISP

Scheme

Prolog

22




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Object-Oriented - Java




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Translator Program -- the program thattranslates the source program into the objectprogram.

Source Program -- program written in a high-level programming language.

Object Program -- the source program after it

has been translated into machine language.

8.1.4 Translator




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Assembler Interpreter Compiler

convertsprograms writtenin assembly

language tomachine code

translates onehigh-levelprogram

instruction at atime into machinecode

convert all of thesource programinto object

program

25

8.1.4 Translator

Three Types


8 1 Introduction to Programming



26

Definition of programming languageDifferences between low-level and high-

level languageDifferences of the language paradigm inprogrammingFunctions of translator

26

SUMMARY





27

CHAPTER 8:

PROGRAMMING

8.2 Approach in Problem Solving




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8.2.1 Input Process Output (IPO) Analysis8.2.2 Algorithm8.2.3 Pseudocode

8.2.4 Flow Chart8.2.5 Control Structure

8.2.5.1 Sequence8.2.5.2 Selection

8.2.5.3 Looping


8 2 Approach in Problem Solving



29

LEARNING OUTCOME:


1. Identify input, process and output from a give

problem.

2. Define algorithm.

3. Solve problem using pseudocode.

4. Solve problem using flow chart.5. a) explain the purpose of each control structure.

b) apply appropriate control structure in problem

solving.



P bl S l i



30

Problem Solving – Everyday in Li fe




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Problem Solving in Programming

Programming is a process of problem

solving

Problem solving techniques

Analyze the problem

Outline the problem requirements

Specify what the solution should doDesign steps, called an algorithm, to

solve the problem (the general

solution)sanggagakputih.blogspot.com

8 2 1 Input Process Output (IPO) Analysis



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8.2.1 Input Process Output (IPO) Analysis

• The IPO is used to analyze problems anddevelop algorithms

• Used to organize and summarize theresults of a problem analysis

• It shows where in the solution theprocessing takes place





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Input Processing Output

Processing items:

Algorithm:

• It can also be represent using IPO chart






34

To do the IPO Analysis, start with:

1- Output (Display the results)

2- Input (Read the data)3- Process (Perform the computation)

Input :

Process :

Output :

Identify






35

Output should answer the following question:

What does the user want to see pr in ted

on the pr inter, displayed on the sc reen, or

sto red in a f i le?






36

Input should answer the following question:

What info rmat ion w i l l the compu ter need

to know to pr int , disp lay, or s tore the outpu t i tems?






37

Processing item:

An in termediate value that the algo r i thm uses

when processing the inpu t into the output



Analyze the Problem



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Analyze the Problem

• What is it you are trying to accomplish?

• What outcome you are trying to arrive at?

• List the Inputs and Outputs

• Often you work backwards from the

Output

• List the Outputs, and then figure out what

Inputs you need in order to arrive at the Outputsanggagakputih.blogspot.com

Analyze the Problem



39

Examp le 1 Problem statement:

Calcu late the area o f a rectang le

We can summarize the information contained

in the problem statement as follows

width

height

Analyze the Problem


Analyze the Problem



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Prob lem Analys is :


1 – Output

2 – Input

3 – Process

Analyze the Problem

Input :

Process :

Output :

Identify


Analyze the Problem



41

Problem Analys is :

Input: width, height

Process: area = width x height

Output : area of rectangle

Analyze the Problem


IPO Chart



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The Problem Analysis, can also be represent

using IPO (Input, Processing, Output) Chart

O C a t


width,height

Processing items:

area = width xheight

Algorithm:

area of rectangle


Analyze the Problem



43

Example 2

Problem statement:

Find the average of three numbers input by

user.

We can summarize the information contained inthe problem statement as follows...

y




Analyze the Problem



45

Problem Analys is :

Input : number1, number2, number3

Process:

average = (number1 + number2 +

number3) / 3

Output : average

y


IPO Chart



46


number1, number2,number3

Processing items:

average = (number1 +number2 + number3) /

3

Algorithm:

average


Analyze the Problem



47

Example 3

Problem statement:

Determ ine the total cost o f dur ians, given

the number of k i los o f dur ians purchased

and the cost of dur ians per k i lo .

y


Analyze the Problem



48



1 – Output

2 – Input

3 – Process

Input :

Process :

Output :

Identify


Analyze the Problem



49

Problem Analys is :

Input : Number of kilos of durians,

Cost of durians per kilo

Process: Total cost = Number of kilos of durians ×


Output :

Total cost of durians


IPO Chart



50


Number of kilos of durians,


Processing items:

Total cost = Number of kilos of durians x


Algorithm:



Analyze the Problem



51

Example 4 Problem statement:

JKL Corpo rat ion gives a bonus to Ir fan based

on his sales. The bonus is 5 percent o f the sales. Calcu late Irfan's bonus.


Analyze the Problem



52



1 – Output

2 – Input

3 – Process

Input :

Process :

Output :

Identify




53

Problem Analys is :

Input : Irfan's sales,

Bonus rate of 5%

Process: Bonus = Sales x Bonusrate 5%

Output : Irfan's bonus


IPO Chart



54


Irfan's sales,Bonus rate of

5%

Processing items:

Bonus = Sales xBonus rate 5%

Algorithm:

Irfan's bonus


8.2.2 Algorithm



55

• The algorithm is the abstract idea of

solving a problem.

• An algo r i thm is a step-by -step

inst ruct ions that will transform input into

output.

• It can be represent using pseudo code or

flow chart.sanggagakputih.blogspot.com

From Algorithms to Programs



56

Problem

C++ Program

Problem Analysis

Algorithm


Algorithm in everyday‟s life



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g y y

How to make a mug of hot coffee ??





58

g y y

How to make a mug of hot coffee:1. Start

2. Boil water

3.Prepare a mug4. Put a tea spoon of coffee & sugar

5. Pour hot water

6. Stir

7. End


8.2.2 Algorithm



59

can be described using three control structures

(Bohm and Jacopini-1966);

•A Sequence – is a series of statements that execu te one after

another

•A Selection – (branch) – statement is used to determine which of two

different statements to execute depending on

cer tain co ndi t ions

•A Looping – (repetition) – statement is used to repeat s tatements whi le certain

condi t ions are met


Planning the Algorithm



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2. Transfer to pseudo code or f low char t

3. a. Must start with a start

b. Must close with an end

1. Do Problem Analysis

Input :

Process :Output :

Identify


TECHNIQUES TO REPRESENTTHE ALGORITHM



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THE ALGORITHM

Pseudo code Flow chart

• Artificial, informal

language used todevelop algorithms

• Similar to everyday

English

• Graphical representation

of an algorithm

• Special-purpose

symbols connected byarrows (flow lines)

1.Pseudo code

2.Flow Chart





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Calculate the area of a circle


Start:Input radiusCalculate circle areacircle area = 3.142 x radius

x radius

Print the circle areaEnd:

ample of Pseudocode & Flow char t

radius

circle area = 3.142 x radius x radius

circle area

start

end

THE ALGORITHM


SUMMARY



63

SUMMARY

1. Identify input, process and output from a givenProblem.



CHAPTER 8:



64

CHAPTER 8:

PROGRAMMING



LEARNING OUTCOME:




65

LEARNING OUTCOME:


1. Identify input, process and output from a givenproblem.


3. Solve problem using pseudo code.

4. Solve problem using flow chart.


8.2.2 Algorithm



66

• The algorithm is the abstract idea of

solving a problem.

• An algo r i thm is a step-by -step

inst ruct ions that will transform input into

output.

• It can be represent using pseudo code or

flow chart.sanggagakputih.blogspot.com




67

Problem

C++ Program

1) Problem Analysis

2) Algorithm





68

How to make a mug of hot coffee ??





69

How to make a mug of hot coffee:1. Start

2. Boil water

3. Prepare a mug

4. Put a tea spoon of coffee & sugar

5. Pour hot water

6. Stir

7. End


8.2.2 Algorithm



70

can be described using three control structures

(Bohm and Jacopini-1966);

•A Sequence – is a series of statements that execu te one after

another

•A Selection – (branch) – statement is used to determine which of two

different statements to execute depending on

cer tain co ndi t ions

•A Looping – (repetition) – statement is used to repeat s tatements whi le certain

condi t ions are met


Planning the Algorithm



71

2. Transfer to pseudo code or f low char t

3. a. Must start with a start

b. Must close with an end

1. Do Problem Analysis

Input :

Process :Output :

Identify







73

Calculate the area of a circle


Start:1. Input radius2. Calculate circle area:

circle area = 3.142 xradius x radius

3. Print the circle areaEnd:

ample of Pseudocode & Flow char t

radius

circle area = 3.142 x radius x radius

circle area

start

endsanggagakputih.blogspot.com

8.2.3 Pseudo Code



74

• Artificial, informal language used to develop

algorithms.

•Similar to everyday English.


8.2.3 Pseudo Code



75

Pseudocode Format

start

statement 1statement 2

end

1

3

2

The statement

refers to any

i nput , ou tpu t &

process involved


Example 1



76

Problem statement:

Calcu late the area of a rectangle

width

height




77

Problem Analys is :





Pseudo Code – Examp le 1



78

start

1. Input width, height

2. Compute area:area = width x height

3. Print area of rectangle

end


P bl t t t

Example 2



79

Problem statement:

Find the average of th ree numbers input

by user.


Example 2



80

Problem Analys is :


Process: average = (number1 + number2 +

number3) / 3

Output : average





81

start

1. Input number1, number2, number3

2. Compute average:

average = (number1 + number2+ number3) /3

3. Print average

end


P bl t t t

Example 3



82

Problem s tatement:

Determ ine the total cos t of du rians , given

the number of ki los o f dur ians purchased



P bl A l i

Example 3



83

Problem Analys is :



Process:

Total cost = Number of kilos of durians ×


Output :






84

start 1. Input number of kilos of durians,

cost of durian per kilo.

2. Compute total cost of durian:

total cost = number of kilos of durians x cost of

durian per kilo

3. Print total cost of durianend


Example 4



85

Problem statement:


on his sales. The bonus is 5 percent o f the

sales. Calcu late Ir fan's bonus .


Example 4



86

Problem Analys is :

Input : Irfan's sales,

Bonus rate of 5%

Process: Bonus = Sales x Bonus

rate 5%

Output : Irfan's bonus





87

start

1. Input sales, bonus rate

2. Compute bonus:bonus = sales x bonus rate

3. Print bonus

end


8.2.4 Flow Chart



88

• Graphical

representation of

an algorithm.

• Special-purpose

symbols connected

by arrows (flow

lines).


Graphic Name Meaning

Flow Chart Symbols



Symbol

Terminal Symbol

(ellipse)

indicates the beginning and end

points of an algorithm

Process Symbol

(rectangle)

shows an instruction other than

input, output or selection

Input-Output

Symbol

(parallelogram)

shows an input or output operation

89

89sanggagakputih.blogspot.com

G hi S b l N M i

Flow Chart Symbols



Graphic Symbol Name Meaning

Disk Storage Input-

Output Symbol

(cylinder)

indicates input from or output to disk

storage

Printer Output

Symbolshows hardcopy printer output

Selection Symbol(diamond)

shows a selection process for two-wayselection

90


Graphic Name Meaning

Flow Chart Symbols



Graphic

Symbol

Name Meaning

Flow Lines

(arrow)

indicates the logical sequence of

execution steps in the algorithm

Off-Page

Connector

provides continuation of a logical path

on another page

On-Page

Connector

(circle)

provides continuation of a logical path at

another point in the same page

91


Flow Chart - Format



start

Statement 1

Statement 2

end

The

statement

refers to

any input , outpu t &

process

involved

92


P bl t t t

Example 1



93

Problem statement:

Calcu late the area of a rectangle

width

height




94

Problem Analys is :





start

Flow chart - Example 1



area = width x height

end

width, height

area

95


Problem statement:

Example 2



96

Find the average of th ree numbers input

by user.


Problem Analys is :

Example 2



97

Problem Analys is :


Process: average = (number1 + number2 +

number3) / 3

Output : average


start

Flow chart - Examp le 2



average = (number1 + number2 + number3) / 3

end

number1, number2, number3

average

98


Problem s tatement:

Example 3



99

Determ ine the total cos t of du rians , given

the number of ki los o f dur ians purchased



Problem Analys is :

Example 3



100

y



Process:

Total cost = Number of kilos of durians ×


Output :



start




total cost = number of kilos of durians x costof durian per kilo

end

number of kilos of durians,

cost of durian per kilo

total cost

101


Problem statement:

Example 4



102

Problem statement:


on his sales. The bonus is 5 percent o f the

sales. Calcu late Ir fan's bonus .


Problem Analys is :

Example 4



103

Problem Analys is :

Input : Sales,

Bonus rate 5%

Process: Bonus = Sales x Bonus

rate 5%

Output : Bonus


start




Bonus = Sales x Bonus rate 5%

end

Sales, Bonus rate 5%

Bonus

104




105

EXERCISE


Problem statement:

Exerc ise 1



106

Problem statement:

Pustaka Mukmin Bookstore offers 15%

discount for each revision book. Calculate and

display the price that customer should paid.

Prepare IPO analysis, pseudo code & flow

chart




Problem Analys is :



108

Problem Analys is :

Input : Price

Process: New Price = Price x 85%

Output : New Price


Pseudo Code



109

start 1. Price

2. Compute Discount:

Discount = Price x 15%

3. Compute New Price:New Price = Price – Discount

4. Print New Price

end


start

Flow chart



Discount = Price x 15%New Price = Price - Discount

end

Price

New Price

110


SUMMARY1. Identify input, process and output from a given

problem.



111

2. Define algorithm.3. Solve problem using pseudo code.4. Solve problem using flow chart.


CHAPTER 8:



112

PROGRAMMING


8.2.5 Control Structure


LEARNING OUTCOME:




113


1. Explain the purpose of each control structure.

2. Apply appropriate control structure in

Problem solving.



Three (3) control structures



114

A Sequence

is a series of statements that execute one after ano ther

A Selection – (branch)

statement is used to determine whichof two different statements to execute

depend ing on certain cond i tions


8.2.5.1 Sequence



115

Control Structures

SELECTION LOOPINGSEQUENCE


What do you do after wake up in the

8.2.5.1 Sequence(Analogy)



116

morning?

Brush teethWash faceTake a bathComb hair Smile in front of mirror Having breakfastGo to class


8.2.5.1 Sequence



117

• The simplest programs consist justsequence of statements :

– no loops, no selections amongstalternative actions.


8.2.5.1 Sequence



118

• Instruction in sequence programming are executed sequentially one by one.

• The sequence structure directs the

computer to process the instructions,one after another, in the order listed in

the program.

first instruction → last instruction. sanggagakputih.blogspot.com

Problem statement:

Example 1



119

Find the produc t of two numbers.


IPO Analys is:

Example 1



120

Input : num1, num2

Process: product = num1 x num2

Output : product


Example 1 – Pseudo Code



121

start

1. Input num1, num2

2. Compute product:

product = num1 x num2

3. Print product

end


start

Examp le 1 - Flow chart



product = num1 x num2

end

num1, num2

product

122


Problem statement:

Example 2



123

Find a volum e for a cone.

Formula :


IPO Analys is:

Example 2



124

Input : radius, height

Process:

volume = (1/3) x 3.142 x radius x

radius x height

Output : volume


Example 2 – Pseudo Code



125

start

1. Input radius, height

2. Compute volume:

volume = (1/3) x 3.142 x radius xradius x height

3. Print volume

end


start

Flow chart - Examp le 2



volume = (1/3) x 3.142 x radius x radius x height

end

radius , height

volume

126


C t l St t

8.2.5.2 Selection



127

Control Structures

SELECTION LOOPINGSEQUENCE


Analogy

8.2.5.2 Selection



128

You need to choose to make

“Plain Coffee” or “Coffee Milk”


What is Select ion Struc ture?

8.2.5.2 Selection



129

• The selection structure allows instructions

to be executed non-sequentially.

It compares a condition / expression, and

based on the result of the condition, it takes

a certain course of action.sanggagakputih.blogspot.com

8.2.5.2 Selection

Types of selection structures



130

if-else nested if-else switch

Two-way

selectionMulti-way

selectionMulti-way

selection

if

One-way

selection


The if construct has the general

Selection - (1) if



131

• The if construct has the generalform

if (expression)statement


• Expression is evaluated FIRST.

If i l l

Selection - (1) if



132

• If it evaluates to a non-zero value

(TRUE), then statement is executed.

• If it evaluates to zero (FALSE), the

statement is terminated.

• This structure has ONE entry point

and ONE exit point.


Selection - (1) if

Pseudo Code Format:



133

start

1. inpu t

2. if (express ion)

Statement

end


Selection - (1) if

Flow chart Format:



134

expression

statement

True

False


Selection - (1) if [example]

Problem statement:



135

Problem statement:

Print message “a is bigger than b” if

the value for a is bigger than b.


Selection - (1) if

IPO Analysis:

Input: a b



136

Input: a, b

Process:i f

( a > b)print “a is bigger than b”

Output: message “a is bigger than b”


Pseudocode:

Selection - (1) if



137

start

input a , bif (a > b)

print “a is bigger than b”

end


start

Flow chart

Selection - (1) if



138end

a, b

“a is bigger than b”

a > b

true

false


Selection - (1) if [example 2]

Problem statement:



139

Problem statement:

Print message “PASS” if marks for

computer science test is greater than or equal to 60.


Selection - (1) if-

IPO Analysis:

Input: marks



140

Input: marks

Process:i f ( marks >= 60)

print “ PASS ”

Output: message “ PASS ”


Pseudocode:

Selection - (1) if



141

start

input marks

if (marks >= 60)

print “ PASS ”

end


start

Flow chart

Selection - (1) if



142end

marks

“ PASS ”

marks >= 60

true

false


• The if-else construct has thel f

Selection - (2) if-else



143

The if else construct has thegeneral form

if (expression)statement 1

else

statement 2sanggagakputih.blogspot.com

• Expression is evaluated FIRST.

• If it evaluates to a non-zero value




144

If it evaluates to a non zero value

(TRUE), statement 1 is executed.

• If it evaluates to zero (FALSE),

statement 2 is executed.

• Either statement 1 is executed or

statement 2 BUT NOT BOTH.



Pseudo Code Format:



145

start

1. input

2. if (exp ress ion ) Statement 1

else

Statement 2

end



Flow chart Format:



146

expression

statement 1

True

False

statement 2



• Example 1:

Print “Excellent!” when user enter marks



147

greater than or equal to 80, else print

“Sorry, try again”.



IPO Analysis:

Input: marks



148

p

Process:i f (marks >= 80)

print “Excellent” else

print “Sorry, try again”

Output: “Excellent!” or “Sorry, try again” sanggagakputih.blogspot.com

start

Pseudocode:




149

start

input marks

if (marks >= 80)

print “Excellent!”

else


end sanggagakputih.blogspot.com

start

Flow chart




150end

marks

“Excellent!”

marks >= 80

true “Sorry, try

again”

false



• Example 2:

A hi h h l i i i



151

A high school poetry competition is open

only for students above 15 years old.

Display “Eligible” if the students meetthe requirement, else display “Not

eligible” if otherwise.



IPO Analysis:

Input: age



152

Process:i f (age > 15)

print “Eligible” else

print “Not eligible”

Output: “Eligible” or “Not eligible” sanggagakputih.blogspot.com

start

Pseudocode:




153

start

input age

if (age > 15)

print “Eligible”

else

print “Not eligible”


start

Flow chart




154end

age

“Eligible”

age > 15

true “Not eligible”

false



• Example 3:



155

If x is greater than y, display “x is bigger

than y” else display “x is smaller than y”.



IPO Analysis:

Input: x, y



156

Process:i f ( x > y)

print “x is bigger than y” else

print “x is smaller than y”

Output: “x is bigger than y” or “x is smaller than y” sanggagakputih.blogspot.com

start

Pseudocode:




157

input x, y

if (x > y)

print “x is bigger than y”

else

print “x is smaller than y”


start

Flow chart




158end

x , y

“x is bigger than y”

x > y

true “x is smaller than

y”

false


• The nested if-else structure can

be nested to any depth.

Selection - (3) nested if-else



159

if (expression 1)

statement 1else if (expression 2)

statement 2

else if (expression n)

statement nelse

last statementsanggagakputih.blogspot.com

• In this nested form, expression 1 is

evaluated first. If it evaluates to non-ero (TRUE) statement 1 is




160

zero (TRUE), statement 1 is

executed and the entire statement is

terminated.• If not, control goes to the second if,

where expression 2 is evaluated. If it

evaluates to non-zero (TRUE),statement 2 is executed.sanggagakputih.blogspot.com

• If not, expression n is evaluated. If it

evaluates to non-zero (TRUE),statement n is executed




161

statement n is executed.

• Otherwise, last statement will be

executed.

• Only ONE of the statements will be

executed.sanggagakputih.blogspot.com

start 1. input

Pseudo Code Format:




162

p

2. if (exp ress ion 1)

statement 1

else if (expression 2)statement 2

else if (expression n)

statement n

elselast statement


Flow chart Format:Selection - (3) nested if-else

expression 1

True

False

statement 1

statement 1

statement 1

statement 1



163

expression 2

expression n

True

TrueFalse

statement 2

statement n

last statement

False


• Example 1:

if student‟s grade is greater than or equal to 80Print “A”

f




164

else if student‟s grade is greater than or equal

to 60 Print “B”

else if student‟s grade is greater than or equal

to 50 Print “C”

else

Print “Failed”


IPO Analysis:

Input: marksProcess: i f (marks >= 80)




165

Process: i f (marks 80)

print “A”

else if (marks >= 60)

print “B” else if (marks >= 50)

pr int “C”

else

pr in t “Failed”

Output: “A” or “B” or “C” or “Failed” sanggagakputih.blogspot.com

start 1. input marks

Pseudocode:




166

2. if (marks >= 80)

print “A”


print “B”


pr in t “C”

else

pr in t “Failed” end


Flow chart


k 80

True“A”

marks

start



167

marks >=80

False

marks >=60

marks >=50

True

TrueFalse

False

“A”

“B”

“C”

“Failed”

end



• Example 2:

if (x > 0)



168

( )

display "x is positive"

else if (x < 0)display "x is negative"

else

display "x is 0"sanggagakputih.blogspot.com

IPO Analysis:

Input: x Process: i f (x > 0)




169


else if (x < 0)

display "x is negative"else

display "x is 0"

Output: “x is positive” or “x is negative” or “x is 0” sanggagakputih.blogspot.com

start

1 input x

Pseudocode:




170

1. input x

2. if (x > 0)


else if (x < 0)

display "x is negative"

else

display "x is 0"

end


Flow chart


X > 0

True “x is positive”

x

start



171

X > 0

False

X < 0 True

False

x is positive

“x is negative”

“x is 0”

end


• The switch structure is a more

elegant alternative to the complex

nested if-else structure.

Selection - (4) switch



172

• The switch structure lets user chooseONE of several alternatives.

• Also known as mult ip le-choice statement. sanggagakputih.blogspot.com

• Switch structure has the general form:

switch (expression)

{




173

case expression 1: statement block

break


break

case expression n: statement block

break

default : statement block

} sanggagakputih.blogspot.com

• expression must evaluate to an integer

or a character constant.




174

• Each alternative (except the default

alternative) in the switch block is normallyterminated with a break statement.


• The switch statement is executed as

follows:




175

– The switch expression is evaluated

FIRST. If it evaluates to expression 1,the statement block for that case is

executed and break statement in the

block terminates the entire switch

statement.sanggagakputih.blogspot.com

• If the switch expression evaluates to

expression 2, the statement block for thatcase is executed and break terminates

th it h t t t




176

the switch statement.

• If the switch expression DOES NOT

EVALUATE to any of the case

expressions, the statement block for the

default case is executed and thenterminates the switch statement.sanggagakputih.blogspot.com

• A break is NOT necessary for the default

case since its execution will terminate theswitch statement.





start

1. input

2. sw itch (exp ression )

{

Pseudo Code Format:Selection - (4) switch



178


break

case expression 2: statement blockbreak

case expression 'n': statement block

break

default : statement block}


Flow chart Format:Selection - (4) switch




• Example 1:


x Print

1 X is 1



180

1 X is 1

2 X is 2

Other value Value of x unknown


IPO Analysis:

Input: x Process: sw i tch (x)

{




181

{

case 1: Print "x is 1”

breakcase 2: print "x is 2“

break

default: Print "value of x unknown”}

Output: "x is 1" or "x is 2" or "value of x unknown” sanggagakputih.blogspot.com

Pseudocode:


start 1. Input x 2. switch (x )



182

{case 1: Print "x is 1“

breakcase 2: Print "x is 2"

breakdefault: Print "value of x unknown”

}end


Flow chartSelection - (4) switch




EXERCISE




• Exercise 1:

If student's grade is greater than or



185

equal to 50, print “Passed” else print

“Failed”.

Do IPO analysis, pseudo code and flow

chart.sanggagakputih.blogspot.com



CHAPTER 8:PROGRAMMING

8.3 Introduction to C++




8.3 Introduction to C++



188

8.3.1 Program

Structure

8.3.2 Using Compiler


LEARNING OUTCOME:




189

8.3.1 Program Structure

Identify components of C++.


Edit, compile, link and execute program.



Components of C++:

i. Comment



190

ii. Preprocessor directive

iii. Function

iv.Body

v. Return statementsanggagakputih.blogspot.com

8.3.1 Program Structure• A computer program consists a list of

instructions written in a computer language.

//This is my first C++ program //It prints a line of text

#include <iostream>

i. comment

ii preprocessor



191

#include <iostream> using namespace std;

int main()

{cout << "My first C++ program";

cin.get();

return 0;}

ii.preprocessor

directive

iii. function

iv. body

v. returnstatement


8.3.1 Program Structure// rectangleAREA.cpp // Find Area of rectangle


int main()

{

i. comment

ii.preprocessor

directive

iii. function



192

{int width,height,area;

cout << “Enter value for width: ”; cin >> width;cout << “Enter value for height: “;

cin >> height;area = width * height;cout << “Area: “ << area;

cin.get();return 0;

}

iv. body

v. return

statementsanggagakputih.blogspot.com


i. Comment

Style to insert a comment in C++.

i. Text begin with two double slash //

normally used for single line comment



193

- normally used for single line comment.

// Th is is my firs t C++ program

// It p r in ts a line o f tex t


i. Comment





i. Comment

Style to insert a comment in C++.

ii.Text begin with /* and ends with */

( ibl t i i li )



195

(possibly containing many lines)

/* Th is is my firs t C++ program

It pr ints a l ine of text */


i. Comment




8.3.1 Program Structurei. Comment

• The purpose to insert a comment: – To document a program

– To improve program readability → help



197

To improve program readability helpother people read and understand a

program.

• Remember – comment DO NOT cause thecomputer to perform any action when the

program is run.sanggagakputih.blogspot.com


ii. Preprocessor directive

- Preprocessor directive is a general instructionto the C++ compiler.

Sh i ( #) d b



198

- Sharp sign ( #) :are processed by

preprocessor before program is compiled.- #include <iostream> → tells the

preprocessor to include in the program the

contents of the input/output stream workproperly.sanggagakputih.blogspot.com

8.3.1 Program Structureii. Preprocessor directive

- header file (with the file extension.h) is stored in

the C++ library.

- each header file contains a set of related



199

functions.



iii. Function

int main() – All C++ programs must have the main()

function



200

function.

–

The keyword void before main() indicatesthat main() will not return any value to the

operating system.

– C++ programs normally begin executing

at function main.sanggagakputih.blogspot.com

iii. Function





iv. Body Function

- All the statements in a C++ function

(main) must appear between a pair of



202

( ) pp p

curly braces { }.- The group of statements enclosed

between the braces is called a function

block.sanggagakputih.blogspot.com


iv. Body Function

- The body of main may consist of:



203

a) variable declaration and reserved worde.g. : int width,height,area;

reserved wordvariables


8.3.1 Program Structureiv. Body Function


b) input / output console

input console: use standard input stream object



204

p p j

cin and input operator, >>, to allow user type

in a value (or values).

e.g. : cin >> width;



iv. Body Function


b) input / outpu t conso le

output console: use standard output stream



205

p p

object - cout and the output operator, <<, to

output the message.

e.g. : cout << “Area: “ << area;



iv. Body Function

- The body of main may consist of:c) C++ statement

Every statement must end with semicolon



206

- Every statement must end with semicolon

( ; )e.g. :

i. cout << "My first C++ program”;

ii. area = width * height;sanggagakputih.blogspot.com


v. return statement

return 0;

- is included at the end of main function.

0



207

- value 0 indicates that the program has

terminated successfully.



program.cpp

(C++ source code)

C++

compiler

program.obj

(object code)

Linker



208

C++ Library

program.exe

(executable program)

Fig. 1: Building a C++ Programsanggagakputih.blogspot.com


• Edit – enter the program statement

• To write a C++ program, you need to enter theprogram statements by using:

– Text editor (e.g. Notepad, Microsoft Word, etc.)



209

– Integrated Development Environment (IDE)(e.g.

Dev C++, Borland C++, Visual C++, etc)

• The complete program statements called

“source code”.



• Compile – translating C++ into machine code

(also called “object code”) – Compiler determines the syntax error.



210

• Link – Run the linker – combine the machine

code with code from C++ library; after compiles

is successful.



• Execute Program – an application can be run.

–Finally, the computer, under the control of itsCPU, executes the program one instruction

at a time.



211

–Compiler detects grammatical (syntax) error

but NOT the program-logic error.



Editor

Compiler Compile

EditSource code

Tool Step Product



212

Linker

Run

Link

Object code

Executable image

Result / Outputsanggagakputih.blogspot.com

SUMMARY1. Identify components of C++.

2. Edit, compile, link and execute program.





8.4 Data, Operator and Expression






215

8.4.4


At the end of the lesson student should

be able to:

Id tif i bl d d d

8.4.1 Variables & Reserved Word



216

Identify variables and reserved word.








gross_income

city_tax

student_age

It 10

• Student age

• Continue

• 17throw

• Principal+Interest

Valid vs. Invalid

Iden t i f iers / User def ined words



220

Item10

count

Number_of_characters

_abc

ABC123Z7

X_1

• Principal+Interest

• 12

• 3X

• %change

• Data-1

• Myfirst.csanggagakputih.blogspot.com

• A variable is an identifier that is usedto hold a value.

Variables



221

• Must be declared before they can beused.

• The general form of a declaration isdataType variable_list;


Dec lar ing variables:

#include <iostream>

using namespace std;

int main ()

{



222

// declaring variables:char name;

int age, home_address;

float weight, height;

return 0;

}


• Keywords that identify languageentities.

• Have special meanings to the

Keyword / Reserved Words



223

Have special meanings to the

compiler.

• Must appear in the correct location

in a program, be typed correctly, and

be used in the right context.

• Must be typed fully in lowercase.


Keyword / Reserved Words





be able to:

Identify various basic data types.

8.4.2 Data Type



225

Differentiate basic data type and their usage.


A set of data values and a set of operations on those

values. Five classes:

character , char

integer , int

Bas ic Data Type and Dec larat ion

http://localhost/var/www/apps/conversion/tmp/scratch_2/#Slide%20120




226

floating-point, floatdouble floating-point, double

valueless, void


Data Type: (i) char

Used to declare character variables

(hold character value).

8.4.2 Data Type





227

( )

For example:

char choice;


Used to declare numeric program variables of

integer type. (range: -32768 to 32767)

8.4.2 Data Type

Data Type: (ii) int



228

Example:

int counter;


Use to declare a floating-point variable.

(approximately 6 digits of precision)

8.4.2 Data Type

Data Type: (iii) float



229

( pp y g p )

For example:

float gross_salary;sanggagakputih.blogspot.com





Type Typical Size in Bits Minimal Range

char 8 0 to 255

int 16 or 32 -32 768 to 32 767

float 32 6 digits of precision

A ll Data Types Defined by th e

ANSI/ISO C Standard



232

double 64 12 digits of precision

ANSI – Am erican National Standards Inst i tu te for C prog ramm ing langu age.

ISO – Internat ional Organization for Standardizat ion sanggagakputih.blogspot.com



Positive or negative whole numbers with

no fractional part. Consists of an optional plus (+) or a

minus ( –) sign and decimal digits.

In teger Cons tants




Must begin with a non zero decimal digit

Can contain decimal digit values of 0 - 9.

If the sign is missing, the computer assumes

positive.

RULES for Decimal Integer Constants



235

Commas are not allowed in integer constant.


Example :

-15

0

0179

1F8

Decimal Integer Constants

Valid Invalid



236

+250

7550

const int num = -15;

1,756


Character enclosed in single quotation

marks. Examples:

–'F'

Character Constants



237

–'1'

–'*'

char genderFemale = 'F';


A sequence of any number of characters

surrounded by doub le quotat ion m arks . Useful in prompting users for input.

Can contain any printable characters.

Str ing Constants




const char name[] = “Hello World!”;

Str ing Constants - declarat ion




Positive or negative decimal numbers

with an integer part, a decimal point,and a fractional part.

Can be represented either in

Float ing-Point Cons tants



240

convent ional or scient i f ic notation.


Consists of six (6) parts:

A sign A decimal integer (optional )

A decimal point

A decimal fraction (optional )

Scient i f ic Notat ion



241

( p )

The letter e or E followed by a signed or

unsigned integer exponent (optional )

Type suffix f, F, l, or L (optional )


Commas are not allowed.

Three types: –float

–double

Scien t i f ic Notat ion



242

–long double


The following are examples of floating-

point literals:

Scien t i f ic Notat ion

Floating Point Value

5.3876e4 53,876



243

4e-11 0.000000000041e+5 100000

7.321E-3 0.007321

0.5e-6 0.0000005

0.45 0.45


The escape character and any character that

follows. E.g:

Code Meaning

\n New line

\t Horizontal tab

\”

Escape Sequence / Backslash



244

\” Double quote

\’ Single quote

\\ Backslash

\? Question marksanggagakputih.blogspot.com

Escape Sequence / Backs lash : e.g




Escape Sequence / Backslash :

output





be able to:

clarify operations on variables &

operators that can be used in a

8.4.3 Operator



247

program.

Identify precedence of operators


C++ uses a set of built-in operators.

Operators act on operands.

There are several operators:

8.4.3 Operator




Operator Meaning

-Subtraction, (also unary minus )

+Addition

*Multiplication

/Division

Arithmetic Operators



249

/

%Modulus Division (produce the remainder

of an integer division)

--Decrement

++Incrementsanggagakputih.blogspot.com

Rules for Evaluat ion of Ar i thm et ic

Expressions:

Evaluate all parenthesized subexpressions,

beginning with the innermost subexpressions in

case they are nested.

e g:

8.4.3 Operator



250

e.g:

bill = 60 + ((hours – 24) * 5);


C++ provides two special operators '++' and '-

-' for incrementing and decrementing the value

of a variable by 1.

The increment/decrement operator can be

used with any type of variable BUT it cannot

be used with any constant

Inc rement and Decrement



251

be used with any constant. Increment and decrement operators each have

two forms, pre and post.



Increment Decrement

++x or x++ --x or x--

x = x + 1 x = x - 1

x += 1 x = 1



252

x += 1 x -= 1

Pre: ++x Pre: --x

Post: x++ Post: x--


Expression Meaning

x-- decrease 1 after use in expression

--x decrease 1 before use in expression

x++ increase 1 after use in expression




253

x++ increase 1 after use in expression

++x increase 1 before use in expression


++x

x = 10;

z = ++x;

Here

x++

x = 10;

z = x++;

Here




254

Here,

z is equal to 11

x is equal to 11

Here,

z is equal to 10

x is equal to 11


--x

x = 10;

z = --x;

Here

x--

x = 10;

z = x--;

Here




255

Here,

z is equal to 9

x is equal to 9

Here,

z is equal to 10

x is equal to 9


int x,y;

x = 5;

y = x++ - 2;

int x,y;

x = 5;

y = ++x – 2;

Inc rement and Decrement – e.g : 1



256

Answer:

y = 3x = 6

Answer:

y = 4x = 6


int y,z;

y = 8;

z = y-- + 3;

int y,z;

y = 8;

z = --y + 3;

Inc rement and Decrement – e.g : 2



257

Answer:

y = 7z = 11

Answer:

y = 7z = 10


int x,y;

x = 9;

y = x++ + 8;

int x,y;

x = 9;

y = ++x + 8;

Inc rement and Decrement – exercise 1



258

Answer:

y = 3x = 6

Answer:

y = 4x = 6


int x,y;

x = 9;

y = x++ + 8;

int x,y;

x = 9;

y = ++x + 8;




259

Answer:

y = 17x = 10

Answer:

y = 18x = 10


int y,z;

y = 3;

z = y-- - 1;

int y,z;

y = 3;

z = --y - 1;




260

Answer:

y = 7z = 11

Answer:

y = 7z = 10


int y,z;

y = 3;

z = y-- - 1;

int y,z;

y = 3;

z = --y - 1;




261

Answer:

y = 2z = 2

Answer:

y = 2z = 1


int a,b,n;

a = 5;

b = 3;




262

n = ++a * ++b;

Answer: n =

a =


int a,b,n;

a = 5;

b = 3;




263

n = ++a * ++b;

Answer: n = 24

a = 6


int a,b,n;

a = 7;

b = 1;




264

n = a++ * --b;

Answer: n =

a =


int a,b,n;

a = 7;

b = 1;




265

n = a++ * --b;

Answer: n = 0

a = 8


EXERCISE Write the output for the program fragment

given.

1)

int z = 7;

cout << z << endl;

cout << --z << endl;



266

;

cout << z++ << endl;


2) int p,q;

p = 9;

q = --p + 5;

cout << --q << “ \t”; cout << --p << endl;

cout << p++ << endl;




3) int p,q;

p = 7;

q = p++ - 3;

cout << p << “ \t”; cout << ++p << endl;

cout << q++ << “ \t”;

cout << q << endl;



268

q ;


4) int a = 3,b = 2, c;

c = 3 * ((5/a) + (4*(a-b)) % (a+b-2));

cout << ++c;




ANSWER

7

6

6

1

2

3

12 77

7

8

9

4

5

4



270

7


The relational operators are used to test the

relation between two values.

All relational operators are binary operators

and therefore require two operands.

A relational expression returns zero (0) when

the relation is FALSE and a non-zero (1)

Relat ional Operato r



271

when it is TRUE.


Operator Meaning

> Greater than

>= Greater than or equal

<Less than

L h l

Relat ional Operator



272

<= Less than or equal

== Equal

!= Not equalsanggagakputih.blogspot.com

Relat ional Operato r - examples




Logical refers to the ways these relationships

can be connected.

Operator Meaning

&& AND

|| OR

Logical Operator



274

||

! NOT


P Q P&&Q P||Q !P

0 0 0 0 1

0 1 0 1 1

1 0 0 1 0

1 1 1 1 0

Logical Operator – Tru th Tab le



275

1 1 1 1 0


Logical & Relat ional Operator – Example

i. ( 5 < 3 ) && (6 <= 6)

Result: 0 (False)

ii.( 1 < 3 ) || (5 != 6) iii. ! (7 >=8)

TF



276Result: 1 (True) Result: 1 (True)

T T F


The assignment operator '=' is used for

assigning a variable to a value. This operator takes the expression on its right-

hand-side and places it into the variable on its

left-hand-side.

For example:

Ass ignment Operator



277

m = 5;

bill = 200;sanggagakputih.blogspot.com

Other Examples:

a) sumProduct = x * y;

b) bonus = 5 * salary;

c) nett_income = gross – deduction;

d) modulus = a % b;

) BMI idth / (h i ht * h i ht)

Ass ignment Operator



278

e) BMI = width / (height * height);


Precedence Level Operator Associativity

1

(highest)

( ) left to right

2 Unary + unary –

++ --

right to left

right to left

3 * / % left to right

4 + - left to right

5 < <= > >= left to right

Precedence o f A ri thm etic and Relat ional

Operators



279

6 == !== left to right

7

(lowest)

= += -= *= /= %= right to left


Assume a = 10, b = 3, c = 7

1) a + b >= 3 * c

Presedence: Examp le 1

12 3



280

13 >= 21


Assume a = 10, b = 3, c = 7

a != 2 * c + b

10 != 17


1 23



281

Output : 1 (true)


int a;

a = 8 + 5 * 1 % 2 * 4

Output : 12


1 2 34





be able to:

Write expression in correct syntax

8.4.4 Expression






Here are some examples of expressions:

(i) nett_salary = (basic_pay + hours * rate) – (socso +

premium + loan);

(ii) (b * b – 4 * a * c) > 0

8.4.4 Expression



285

(iii) (gender=='M' || gender=='F') && age >= 21


8.4.4 Expression




Mixed mode expression happened when

Constants and variables of different data

types are combined together.

Type casting is used to force an expression to

be of a specific data type.

Mixed Mode Expression




General forms:

(data_type) expression

OR

Type Casting



288

data_type (expression)


float a, b;

float addition, multiplication; float modulus;

addition = a + b;

multiplication = a * b;

modulus = (int) a % (int) b;

Type Casting - examples




You can add tabs and spaces to expressions

to make them easier to read.

For example, the following two expressions arethe same:

(i) x=10/y*(127/x);

Space and Parentheses



290

(ii) x = 10 / y * (127 / x); sanggagakputih.blogspot.com

Redundant or additional parentheses do not

cause errors or slow down the execution of an

expression.

You should use parentheses to clarify the

exact order of evaluation, both for yourself andfor others.





Which of the following two expressions is

easier to read?

x = y/3-34*temp+127;

x = (y/3) - (34*temp) + 127;





SUMMARY

1. Identify variables and reserved word.2. Identify various basic data types.3. Differentiate basic data type and their usage.

4. Clarify operations on variables & operatorsthat can be used in a program.

5. Identify precedence of operators.6. Write expression in correct syntax.





8.5 Use of Control Structure





be able to:

Write a program segment by using

appropriate control structure.





Control structure refers to the order of execution of instructions in a program.

Types of control structure: – Sequence –

Selection – Looping

Control Structure





Problem

- Problem Analysis (IPO)

- Algorithm (Pseudo code

+ Flow chart)



297C++ Programsanggagakputih.blogspot.com

Series of statements that execu te one after ano ther.

(1) Sequence




(1) Sequence – example 1

Calcu late and

display area of

a c irc le.

IPO Analysis :

I - radiu s

P - area = 3.142 x radius x radius

O - area





Pseudo code:

Start

Input radius

Calcu late area:

area = 3.142 x radius x radius

Disp lay area



300

Disp lay area

End sanggagakputih.blogspot.com

(1) Sequence – examp le 1


int main()

{ float radius, area;

cin >> radius;

area = 3.142 * radius *radius;

Pseudo c ode:

Start

Input radius Calculate area:

area = 3.142 x radius

x radius

Display area End

D

I

P

O



301

cout << area;

cin.get();return 0;

}NOT USER FRIENDLY !!!


Why??

Because there is NO prompt message for user...

Assume that radius entered by user is 8.7

Sample of ou tput :






int main(){

float radius, area;

cout << "Enter radius:";

cin >> radius;

D

I

P

O



303

cin >> radius;

area = 3.142 * radius *radius;sanggagakputih.blogspot.com


Sample of ou tpu t :





Calcu late and display to tal for three (3)

qu izzes below :

Quiz1 → 78.5, Quiz2 → 67.5, Quiz3 → 90.2

IPO Analysis :

I - -

P - to tal = Qu iz1 + Qu iz2 + Qu iz3



305

O - total




int main(){float Quiz1 = 78.5, Quiz2 =

67.5;

float Quiz3 = 90.2, total;

P

D

O



306

total = Quiz1 + Quiz2 +Quiz3; sanggagakputih.blogspot.com







Calcu late and d isp lay p rice of lap top after

15% discoun t.

IPO Analys is:

I - or i pr ice

P - new pr ice = or i pr ice x 0.85



308

O - new pr ice




int main(){float ori_price, new_price;

cout << “Insert original price: RM “;

i >> i i



309

cin >> ori_price;

new_price = ori_price *0.85;







The following C++ code contains errors.

Correct the error and rewrite the code.

#include (iostream.h)#include (stdlib.h)

int main(){char name[25];

cout >> 'Enter yourname: ';



311

name: ;cin << name

cout << "Your name is "sanggagakputih.blogspot.com

Errors:

#include (iostream.h)#include (stdlib.h)


cout >> 'Enter yourname: ';



312

name: ;cin << name


Answer:

#include <iostream.h> #include <stdlib.h>


cout << “Enter yourname: “;



313

name: ; cin >> name;


Sample of output :




The program may contains error(s). Correct

the error and rewrite the program.


void main(){

double dist_km

double dist_mile;

cout << "\n Enter the distance in



315

cout << "\n Enter the distance in\"KM\:";

CIN >> dist_km;sanggagakputih.blogspot.com

Errors:


void main(){

double dist_km

double dist_mile;

cout << “\n Enter the



316

cout << “\n Enter thedistance in \”KM \:”;

CIN >> dist_km;sanggagakputih.blogspot.com

Answer:


void main(){

double dist_km;

double dist_mile;

cout << “\n Enter the



317

cout << \n Enter thedistance in \”KM \”:”;

cin >> dist_km;sanggagakputih.blogspot.com

Study the C++ program below. Using problem

analysis, determine input, output and process

involve in this program.


int main(){

int feet, yards;

cout << “Enter value (yards)”;



318

;

cin >> yards;feet = 3 * yards;


Answer:

IPO Analys is:

I - yards

P - feet = 3 * yards

O - feet



319

O - feet


Exercise:

(1) Wri te a complete C++ prog ram that w i l l calculate salary earned when user enter number of days

worked and rate per day. Your output should

disp lay number of days worked, rate per day and

amount earned.

(2) Wri te a complete C++ prog ram that w i l l calculate

amount paid for an i tem when u ser enter quant i ty

and pr ice of the item . The ou tput m ust d isplay the

quant i ty, pr ice of the i tem and amount paid.

(3) Write a C++ program that wil l calculate and



320

ou tput the average of two f loat ing-po int numbers

entered from the keyboard.sanggagakputih.blogspot.com

Answer:

(1) Wri te a complete C++ prog ram that w i l l calculate salary earned when user enter number of days

worked and rate per day. Your output should

disp lay number of days worked, rate per day and

amount earned.IPO Analys is:

I - days worked, rate

P - salary = days worked x rate



321O - days worked, rate, salary sanggagakputih.blogspot.com

Program:




Sample of output :




Answer:

(2) Wri te a complete C++ prog ram that w i l l calculate amount paid for an i tem when user enter quant ity

and pr ice of the item . The ou tput m ust d isplay the

quant i ty, pr ice of the i tem and amount paid

IPO Analys is:

I - quanti ty, p rice

P - amount paid = quant i ty x pr ice



324O - quantit y, pr ice, amoun t paid sanggagakputih.blogspot.com

Program:




Sample of output :




Answer:

(3) Write a C++ program that wil l calculate and ou tput the average of two f loat ing-po int numbers

entered from the keyboard

IPO Analys is:

I - num1, num2

P - average = (num1 + num2) / 2



327

O - average sanggagakputih.blogspot.com

Program:




Sample of output :




(2) Selection

• The selection structure allows instructions tobe executed non-sequentially.

It compares a condition / expression, andbased on the result of the condition, it takes a

certain course of action.




Types of selection structures

if-else nested if-else switch

Two-wayselection

Multi-wayselection

Multi-wayselection

if

One-wayselection

(2) Selection




(2) Selection [if] – example 1

Print “Passed!” when user enter marks greater

than or equal to 60.

IPO Analysis:

Input: marks

Process: i f (marks >= 60)

print “Passed!”

Output: message “Passed!”






int main(){

float marks;

cout << "Enter marks: ";cin >> marks;

if ( k > 60)



333if (marks >= 60)

cout << "\nPassed!";sanggagakputih.blogspot.com


Sample of ou tput :





Assign car allowance = 200 and housing

allowance = 800 if job code for a worker is

equal to '1'.

IPO Analysis:

Input: job code

Process: i f (job code == '1')

car allowance = 200housing allowance = 800

Output: -






int main(){

char job_code;float car_allowance,

housing_allowance;

cout << "Enter your jobcode: ";



336

code: ;

cin >> job_code;sanggagakputih.blogspot.com


Sample of ou tput :




(2) Selection [if-else] – example 1

Print “Excellent!” when user enter marks

greater than or equal to 80, else print “Sorry,

try again”.

IPO Analysis:

Input: marks

Process: i f (marks >= 80)

print “Excellent” else


O t t “E ll t!” “S t i ”



338Output: “Excellent!” or “Sorry, try again”




int main(){

float marks;

cout << "Enter marks: ";cin >> marks;

if (marks >= 80)



339if (marks >= 80)

cout << "\nExcellent";else



Sample of ou tput :





Calculate price of car rental. Customer will charged for RM 60 if total hours less or equal to 24. Otherwise, RM

5 will be charged for next every hour. Display the price

of car rental.

IPO Analysis:

Input: hours

Process: i f (hours <= 24)

amount paid = 60

print amount paidelse

amount paid = 60 + ((hours – 24) x 5 )

print amount paid



341

Output: amount paidsanggagakputih.blogspot.com



int main(){

int hours;float amount_paid;

cout << "Enter totalhours: ";

cin >> hours;



342

cin >> hours;

if (hours <= 24)







Sample of ou tput :




(2) Selection [nested if-else] – example

1

This program prompts the user to enter a

number. If that number is greater than zero, it

tells the user that a positive number was

entered. If the number is less than zero, ittells the user that a negative number was

entered. If both are not true, it tells the user

that the number is equal to zero.





1

IPO Analysis:

Input: num

Process: i f (num > 0)

print “Positive Number” else if (num < 0)

print “Negative Number”

else

print “Number is equal to zero”

Output: message “Positive Number” OR

“Negative Number” OR

“Number is equal to zero”





int main(){

int num;

cout << "Enter anynumber: ";

cin >> num;


1



347

if (num > 0)sanggagakputih.blogspot.com

348


1

k ih bl




349

Samples of output :


1

k tih bl t




350


2

Computes and prints the product or total of

three numbers based on the user request. If

user request is 1, the program will count the

product of three numbers. If it is 2, theprogram will count the total of three numbers.

If the number is not 1 or 2, display this

message; “You make a wrong choice”.

sanggagakputih blogspot com




351


2

IPO Analysis:

Input: user request, num1, num2, num3

Process: i f (user request == 1)

product = num1 x num2 x num3print product

else if (user request == 2)

total = num1 + num2 + num3

print total

else print “You make a wrong choice!”

Output: product OR total

OR message “You make a wrong choice!” sanggagakputih blogspot com




352


2

sanggagakputih blogspot com




353

Samples of output :


2





354


2




gg g p g p

355

Samples of output :


2




gg g p g p

356

(2) Selection [switch] – example 1

Computes and prints the product or total of

three numbers based on the user request. If

user request is 1, the program will count the

product of three numbers. If it is 2, theprogram will count the total of three numbers.

If the number is not 1 or 2, display this

message; “You make a wrong choice”.




357


IPO Analysis:

Input: user request, num1, num2, num3

Process: swi tch (user request)

{

case 1 : product = num1 x num2 x num3

print product

break

case 2 : total = num1 + num2 + num3

print total

break

default: print “You make a wrong choice!”

}

Output: product OR totalsanggagakputih.blogspot.com



OR message “You make a wrong choice!”

358





359

Switch Nested if-elseequivalent

switch (user_request)

{

case 1 : product = num1 * num2 *

num3;

cout << "\nProduct of threenumber is: " << product;

break;

case 2 : total = num1 + num2 + num3;

cout << "\nTotal of three number

is: " << total;

Break;

default: cout << "\nYou make a wrong

choice!";

`if (user_request == 1)

{

product = num1 * num2 * num3;

cout << "\nProduct of three number is:

" << product;

}

else if (user_request == 2)

{

total = num1 + num2 + num3;

cout << "\nTotal of three number is: "<< total;

}

else

cout << "\nYou make a wrongsanggagakputih.blogspot.com



} choice!";

360


This program needs user to key in a number to

make a choice for their favorite's holiday

destination in Malaysia. If the choice is equal

to 1, it will display “Pulau Langkawi”. If thechoice is equal to 2, it will display “Cameron

Highland”. If the choice is equal to 3, it will

display “Genting Highland”. If the choice is

equal to 4, it will display “Bukit Merah”. Other than that, display “Your choice is not in the

list”.sanggagakputih.blogspot.com



361


IPO Analysis:

Input: user choice

Process: swi tch (user choice)

{

case 1 : print “Pulau Langkawi”

breakcase 2 : print “Cameron Highland”

break

case 3 : print “Genting Highland”

break

case 4 : print “Bukit Merah” break

default: print “Your choice is not in the list”

}

Output: “Pulau Langkawi” or “Cameron Highland” or

“Genting Highland” or “Bukit Merah” or “Yousanggagakputih.blogspot.com



g g

make a wrong choice!”

362





363


Samples of output :




364

Switch Nested if-elseequivalent

switch (user_choice)

{

case 1 : cout << "\nPulau Langkawi ";

break;

case 2 : cout << "\nCameron Highland ";break;

case 3 : cout << "\nGenting Highland ";

break;

case 4 : cout << "\nBukit Merah ";break;

default: cout << "\nYour choice is not in

the list";

}

`if (user_choice == 1)

cout << "\nPulau Langkawi ";

else if (user_choice == 2)

cout << "\nCameron Highland ";


cout << "\nGenting Highland ";


cout << "\nBukit Merah ";

else

cout << "\nYour choice is not in the list";sanggagakputih.blogspot.com



365

What is the output for this program if color value is 'R'

?




366

Output :




367

What is the output for this program code?




368

Output :




369

Study the following fragment code.

int n;

cout << "Enter integer: " <<endl;

cin >> n;

if (n < 5)cout << "\nan integer entered

less than 5 ";

else if (n > 5)cout << "\nan integer entered

greater than 5 ";

(i) What will be the output if user enters the integer value is 0

(ii) What values for n will cause the output “an integer entered not

interesting” ? sanggagakputih.blogspot.com



else

370

(i) What will be the output if user enters

the integer value is 0

Answer : an in teger entered less than

5

(ii) What values for n will cause the

output “an integer entered not

interesting” ?

Answer : n = 5 sanggagakputih.blogspot.com



371

(3) Looping

• The looping (or repetition) structure allows

a sequence of instructions to be executed

repeatedly until a certain condition is

reached.

• The looping structure has three forms:

• while• do-while

• for sanggagakputih.blogspot.com



372

Display “Hello” for five (5) times

IPO Analys is:

I - none

P - none

O - “Hello” for five (5) times

Looping – (1) while – example 1Looping – (1) while – example 1




373

Looping – (1) while – example 1

start

1. ini t ial ize coun ter

i = 1

2. wh ile (i <= 5)

{ pr int “Hello”

3. coun ter increment

i = i + 1

}

4. repeat un ti l i > 5

end

start

end

“Hello”

i <= 5

T

F

i = 1

i = i + 1




374





375


Output :




376

Find the average of th ree num bers fo r three

(3) times

IPO Analys is:

I - num1, num2, num3

P - average = (num1 + num2 + num3) / 3

O - average (3 times)





377





378


Sample of outpu t :




379

Calcu late area of a circ le two fo r (2) t imes

IPO Analys is:

I - rad ius


O - area (2 times)





380





381


Sample of outpu t :




382

Looping – (1) while – t race outpu t 1




383

Looping – (1) while – answer




384

Looping – (1) while – t race outpu t 2




385

Looping – (1) while – answer




386

Display “C++” for seven (7) times

IPO Analys is:

I - none

P - none

O - “C++” (7 times)

Looping – (2) do-while – example 1




387

start

1. ini t ial ize coun ter

i = 1

2. do

{ pr int “C++”

3. coun ter increment

i = i + 1

} 4. wh ile ( i <=7)

repeat wh i le i <= 7

end

start

end

“C++”

i <= 7T

F

i = 1

i = i + 1





388





389


Output :




390



(3) times

IPO Analys is:



O - average (3 times) sanggagakputih.blogspot.com



391





392


IPO Analys is:

I - rad ius


O - area (2 times)





393





394

Display “I Love KMM” for ten (10) times

IPO Analys is:

I - none

P - none

O - “I Love KMM” (10 times)

Looping – (3) for – example 1




395





396


Output :




397



(3) times

IPO Analys is:



O - average (3 times) sanggagakputih.blogspot.com



398





399


IPO Analys is:

I - rad ius


O - area (2 times)





400





401

EXERCISE (1)

Change the C++ segment code below into for & do-while statement.

int i = 7;

while ( i <= 77)

{ cout << i << endl;

i = i + 7;

}




402

EXERCISE (2)

Change the C++ segment code below intowhile & do-while statement.

int j;

for (j=20; j >= 2; j-=2)

{

cout << j << endl;

}




403

EXERCISE (3)

The following C++ codes containe errors.

Correct the errors and rewrite the code.#include <iostream.h>

#include <stdlib.h>

int main[]

{ int Num, Power, bil;

cout << “Please key in any number: “;

cin >> Num;

for (bil == 1, bil <= 12; bil++;)

{ Power = bil x num;

cout << bil << “ x ” << num << “ = “ << Power << endl;

}

cin.get();




404

EXERCISE (4)

a) Write a program segment that will display

value 1 to 100 in increments of 1.(while, do-while, for)

b) Write a program segment that will displayvalue 100 to 1 in decrements of 1.

(while, do-while, for)




405

EXERCISE (1) - Answer

f or statement.int i;

for (i=7; i <= 77; i=i+7)

{

cout << i << endl;

}




406


do-while statement.int i = 7;

do

{ cout << i << endl;

i = i + 7;

} while(i <= 77);




407


while statement.

int j = 20;

while (j >= 2)

{ cout << j << endl;

j-=2;

}sanggagakputih.blogspot.com



408


do-while statement.

int j = 20;

do


j-=2;

} while (j >= 2);sanggagakputih.blogspot.com

EXERCISE (3)



409


#include <iostream.h>

#include <stdlib.h>

int main( )

{ int Num, Power, bil;

cout << “Please key in any number: “;

cin >> Num;

for (bil = 1; bil <= 12; bil++)

{ Power = bil * Num;cout << bil << “ x ” << Num << “ = “ << Power << endl;

}

cin.get();

return 0;}


EXERCISE (4)



410


a) Write a program segment that will display

value 1 to 100 in increments of 1.(while, do-while, for)

int j = 1;

while (j <= 100)


j++;

} sanggagakputih.blogspot.com

EXERCISE (4) A



411


b) Write a program segment that will display

value 100 to 1 in decrements of 1.(while, do-while, for)

int j = 100;

do


j--;

} while (j >= 1); sanggagakputih.blogspot.com





413

Expression for accumulating: (e.g:) sum = sum + num

total = total + marks




414

Example 1:

Calculate and display sum of odd numbers from1 to 10

IPO Analys is:

I - none

P - sum = sum + counter

O - sum sanggagakputih.blogspot.com



415

int counter = 1;

int sum = 0;

while ( counter <= 10 )

{sum = sum + counter;

counter = counter + 2;

}

cout << “Sum of ODD numbers is: “<< sum;

Example 1: while




416

Example 1: while




417

int counter = 1;

int sum = 0;

do

{

sum = sum + counter;counter = counter + 2;

}

while ( counter <= 10 );


Example 1: do-while




418

Example 1: do-while




419

int counter;

int sum = 0;

for (counter = 1; counter <= 10; counter = counter + 2){

sum = sum + counter;

}


Example 1: for




420

Example 1: for


E l 1



421

Output :

Example 1:




422

Example 2:

Calculate and display sum of EVEN numbersfrom 2 to 20

IPO Analys is:

I - none

P - sum = sum + counter

O - sum sanggagakputih.blogspot.com



423

int counter = 2;

int sum = 0;


{sum = sum + counter;

counter = counter + 2;

}

cout << “Sum of EVEN numbers is: “<< sum;

Example 2: while




424

Example 2: while




425

int counter = 2;

int sum = 0;

do

{

sum = sum + counter;counter = counter + 2;

}

while ( counter <= 20 );


Example 2: do-while


E l 2 d hil



426

Example 2: do-while


E l 2 f



427

int counter;

int sum = 0;

for (counter = 2; counter <= 20; counter = counter + 2){

sum = sum + counter;

}


Example 2: for




Example 2:



429

Output :

Example 2:


E ample 3



430

Example 3:

Calculate and display average of three (3)quizzes.

IPO Analys is:

I - mark

P - to tal = to tal + mark

average = total / 3

O - average sanggagakputih.blogspot.com

Example 3: while



431

int counter = 1;

float total = 0, mark, average;


{

cout << “Enter mark: “ ; cin >> mark;

total = total + mark;

counter++;

}average = total / 3;

cout << “Average for 3 quizzes: “<< average;

Example 3: while


Example 3: while



432

Example 3: while


Example 3: do while



433

Example 3: do-while

int counter = 1;


do

{ cout << “Enter mark: “ ;

cin >> mark;total = total + mark;

counter++;

} while ( counter <= 3 );

average = total / 3;cout << “Average for 3 quizzes: “<< average;


Example 3: do while



434

Example 3: do-while


Example 3: for



435

Example 3: for

int counter;


for (counter=1; counter <= 3; counter++)

{ cout << “Enter mark: “ ;

cin >> mark;total = total + mark;

}

average = total / 3;

cout << “Average for 3 quizzes: “<< average;


Example 3: for



436

Example 3: for


Example 3:



437

Sample of Output :

p


Exercise 1



438

Compute the product of all the numbersfrom 1 to 8.

(Hint: Use a variable

called product instead of sum and initialize product to 1)


Exercise 2



439

Find the sum of the square of all thenumbers from 1 to 5.

(i.e. 1*1 + 2*2 + 3*3 +... )


Exercise 1 - Answer




Exercise 1 - Output




Exercise 2 - Answer




Exercise 2 - Output







444

Write a program segment by usingappropriate control structure.

SUMMARY




445

CHAPTER 8:

PROGRAMMING

8.6 Header File and Functions

8.6.1 Standard Header file

8.6.2 Functions


8 6 Header file and Functions



446


be able to:

1) use standard header file in a program

2) use appropriate functions in a program

8.6 Header file and Functions


Introduction



447

Introduction

Header file is a text file containing smallbits of program code, which is used todescribe the contents of the main body of code to other modules.

Also known as pre-processor directive.

The header file tells the compiler what isavailable in the library.


8.6.1 Standard header file



448

•

The following header files are suppliedwith the standard libraries:

. iostream.h

. stdlib.h OR cstdlib.h

. time.h OR ctime.h

. math.h

http://www.cplusplus.com/reference/clibrary/


i) iostream.h – input outpu t stream






449

)

Declares objects that control reading fromand writing to the standard streams.

The only header files to perform input and output in a C++ program.



i) iostream.h – input outpu t stream



450

Defines global objects:

)


i) iostream.h – example



451

# include <iostream.h>

int main()

{

cout << “My first C++ program”; cout <<

endl;

cin.get();

return 0;

}





452


int main()

{

char name [30];

cout << “Enter Your Name: ”;cout << endl;

cin >> name;

cin.get();

return 0;

}






ii) cstdlib.h (stdlib.h)



454

Used for testing and converting characters.

This header defines several general purposefunctions, including dynamic memory

management, random number generation,communication with the environment, integer arithmetic, searching, sorting and converting.

#include <stdlib.h>









456

#include <stdlib.h>


#include <time.h>

int main( )

{

unsigned masa = time(NULL) ;cout<<“masa = “<< masa <<endl ;

srand(masa); //Initialize random number generator

for (int i=0 ; i<8 ; i++)cout << rand( ) << endl;

return 0;

}





Sample of outpu t :



458

p p






iii) ctime.h (time.h)



460

This header file contains definitions of

functions to get and manipulate date and timeinformation.

Declares the structure tm, which includes atleast the following objects:

int tm_sec seconds [0,61] int tm_min minutes [0,59] int tm_hour hour [0,23] int tm_mday day of month [1,31] int tm_mon month of year [0,11] int tm_year years since 1900sanggagakputih.blogspot.com








462

Output:sanggagakputih.blogspot.com

iv) cmath (math.h)



463

cmath declares a set of functions tocompute common mathematical

operations and transformations.

#include <math.h>


iv) cmath (math.h)




iv) cmath (math.h)




iv) cmath (math.h) - example



466

Output:sanggagakputih.blogspot.com

8.6.2 Functions



467

A function is a sub-program that programcan call to perform a task.

When you have a piece of code that isrepeated often you should put it into afunction and call that function instead of repeating the code.


8.6.2 Functions -Creating your own function



468

Declare a function in a similar way as createthe main function.

Here is a function that prints Hello.

void PrintHello()

{

cout << "Hello\n";}


8.6.2 Functions -Calling a function



469

Once you have created the function you must call it

from the main program. Here is an example of how to call the PrintHello

function.

void PrintHello()

{

cout << "Hello\n";

}

int main()

{

PrintHello();

return 0;

}


8.6.2 Functions -Calling a function




Summary



471

1) use standard header file in a program2) use appropriate functions in a program


CHAPTER 8:



472

CHAPTER 8:

PROGRAMMING

8.7 Simple C++ Program


8.7 Simple C++ Program



473

At the end of the lesson student shouldbe able to:

1) Write a simple C++ program.


Documents

Topic 8 Programing