computer programming all chapters

LOGO

Prepaid By: Eng. Ibrahim Elewah

Higher Technological Institute 10th of Ramadan City6th of October Branch

Electrical and Computer Engineering Department

Main Reference

HTI Student Book and “C For Dummies”

by Dan Gookin 2nd Edition

Lecture Notes in

1

Course Contents

Introduction

C Programming Higher Technological Institute2

1

Program Development2

The Essential of C Programs3

Manipulating Data with Operators 4

Reading from and Writing to Standard I/O5

Decision6

Iteration7

Arrays8

C Functions9

Course Contents

Introduction

Higher Technological Institute3

1

Computer Program

Programming Language

Machine Language

Assembly Language

High Level Language

The C Programming Language

Interpreter, Compiler and Assembler

C Programming

Computer Program

A Computer Program is a sequence of instructions written to perform a specified task with a computer. The program has an executable form that the computer can use directly to execute the instructions.

The Executable program in its human-readable form is called thesource code. Computer source code is often written by computerprogrammers. It is written in a programming language.

Source Code may be converted into an executable file ( anexecutable program or a binary) by a compiler and later executed by a central processing unit. Alternatively, computer programs may be executed with the aid of an interpreter, or may be embedded directly into hardware.

Higher Technological Institute4C Programming

Programming Language

A Programming language is an artificial language designed to communicate instructions to a computer.

Most programming language are purely textual; they use sequences of text including words, numbers and punctuation.

The programming languages are divided into three groups:

1. Machine Language

2. Assembly Language

3. High Level Language


Machine Language

Essentially, computers really understand only one language, which consists of zeroes and ones, also known as machine language.

Buto Difficult and take a Long Time to Read and Write

o Difficult to Detect and Correct the Errors

o It is also called Binary CodeC Programming

Assembly Language


o It uses some Mnemonic English codes instead of Binary Codes.

o Assembly language programs are easier to Readand Write than machine language programs.

o It Needs an Assembler to translate Mnemonic codes into Machine Language.

ExampleADD 5,9

SUB 7,3

C Programming

Assembly Language


o It uses some Mnemonic English codes instead of Binary Codes.

o Assembly language programs are easier to Readand Write than machine language programs.

o It Needs an Assembler to translate Mnemonic codes into Machine Language.

ExampleADD 5,9

SUB 7,3

But in Large ProgramsWriting a program in assemblylanguage can prove extremelyTime-consuming, and complicated.

C Programming

High Level Language


Because writing machine- or assembly-language

programs was so Difficult and Confusing, peopledeveloped additional languages that look more likehuman languages

With names such as JAVA, BASIC, and C.

o High Level Language needs Compiler or Interpreter.

C Programming

High Level Language

Saving your programs into library and invoke them in next programs


Readability

Maintainability

Portability

Reusability

Programs are easy to Read

Programs are easy to Maintain

Programs are easy to port across Different computer Platforms

C Programming

C Programming Language

1. High Level Language.

2. C allows you to get control of computer Hardware.

3. Many other high level language developed based on C.

4. The American National Standard Institute ANSIdefine the standard for the C programming language.

5. Your HTI Book focuses on the functions defined in ANSI

See Page 3


Interpreter, Compiler and Assembler


Interpreter

Compiler

Assembler

High Level

Language

Lower Level Language

Assembly Language

Binary Language

High Level

Language Lower Level Language

Computer

Instruction Binary 0’s and 1’s

Read more Pages 3 and 4

C Programming

Your First C Program

/* This is my First C Program */

#include<stdio.h>

int main ()

{

printf(" Hello World! \n ") ;

return 0 ;

}Higher Technological Institute13C Programming

Comments


o The compiler ignore everything between two slashes and asterisks.

o Just to help you document your program.

o C compilers allows to write comments like


/*This comment does not increase the size of the executable file

(Binary Code), nor does it affect the performance speed

*/

C Programming



#include<stdio.h>

int main ()

{


return 0 ;


Header Files

#include<stdio.h>o This Header Files required by include directive

o Header Files extensions .h means header

o The stdio.h header file numerous prototypes and macros to perform input or output I/O for C Program


stdio.h Header File

C Programming



#include<stdio.h>

int main ()

{


return 0 ;


Main Function

int main ()

{

return 0 ;

}Higher Technological Institute18

o Very special function

o Each C Program MUST have a main() function

o The main() function syntax as follows

C Programming



#include<stdio.h>

int main ()

{


return 0 ;


Printing On Screen



o To print a text or numbers printf is used.

o Syntax printf(“ ”);

o The new line character \n

o To print Hello World! write the following statement

C Programming



#include<stdio.h>

int main ()

{


return 0 ;



References

C Programming

References

Logic Design A Higher Technological Institute23

Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

Course Contents



Phases of Computer Program Development

Problem Definition and Analysis

Algorithms and Flowcharts

Program Coding

Program Execute and Testing

Exercise

C Programming


o The problem which you write a program for should becompletely understood by gathering informationabout it.

o This information is the program input.

o How information is processed to give the requiredoutput.

oExample


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

C Programming


o The General Quadratic Equation


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

C Programming



𝒙 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

C Programming



𝒙 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂

𝒙𝟏 =−𝒃 + 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂𝒂𝒏𝒅 𝒙𝟐 =

−𝒃 − 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

C Programming



𝒙 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂

𝒙𝟏 =−𝒃 + 𝒃𝟐 − 𝟒𝒂𝒄


−𝒃 − 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

Discriminator

C Programming



𝒙 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂

𝒙𝟏 =−𝒃 + 𝒃𝟐 − 𝟒𝒂𝒄


−𝒃 − 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄𝒊𝒇 𝑫 < 𝟎 𝒙𝟏 𝒂𝒏𝒅 𝒙𝟐 𝒂𝒓𝒆 𝑰𝒎𝒂𝒈𝒊𝒏𝒂𝒓𝒚𝒊𝒇 𝑫 > 𝟎 𝒙𝟏 𝒂𝒏𝒅 𝒙𝟐 𝒄𝒂𝒏 𝒃𝒆 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

Discriminator

C Programming


o How information is processed to give the requiredoutput ?

𝒂 , 𝒃 𝒂𝒏𝒅 𝒄 𝒂𝒓𝒆 𝑮𝒊𝒗𝒆𝒏 = 𝑰𝒏𝒑𝒖𝒕𝒔

𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄 Question D > 0 ?

YES 𝒙𝟏 =−𝒃+ 𝒃𝟐−𝟒𝒂𝒄


−𝒃− 𝒃𝟐−𝟒𝒂𝒄

𝟐𝒂

NO 𝒙𝟏 𝒂𝒏𝒅 𝒙𝟐 𝒂𝒓𝒆 𝑰𝒎𝒂𝒈𝒊𝒏𝒂𝒓𝒚


𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎

C Programming

Algorithms and Flowcharts


Flowchart


Preparation

Start, Stop ( Begin , End)

Input , Output

Processing Program Instruction

Decision

Connector

Comment( Instruction that change the program)

C Programming

Flowcharts


𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄

𝒙𝟏,𝟐 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂

C Programming

Flowcharts


𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄

𝒙𝟏,𝟐 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄

𝟐𝒂

YES NO

C Programming

Flowcharts Examples


Example 1

Draw a flow chart to

read two numbers and

compare between both

if they are equal or not.

C Programming

Flowcharts Examples


Example 1





C Programming

Flowcharts Examples


Example 2

Draw a flow chart to generate

a table of squares and cubes

of integers from 1 to 9.

C Programming

Flowcharts Examples


Example 2




C Programming

Flowcharts Examples


Example 3


determine the minimum

value of three input

numbers.

C Programming

Flowcharts Examples


Example 3




numbers.

C Programming

Flowcharts Examples


Example 4

Draw a flow chart to generates the famous sequence

of numbers called the Fibonacci series.

Here are the first few terms of the series:

0 1 1 2 3 5 8 13 21 34 55

C Programming

Flowcharts Examples


Example 4




0 1 1 2 3 5 8 13 21 34 55

Each term is found by adding the two previous

ones:

1+1 is 2, 1+2 is 3, 2+3 is 5, 3+5 is 8, and so on.

C Programming

Flowcharts Examples


Example 4




0 1 1 2 3 5 8 13 21 34 55

Each term is found by adding the two previous

ones:

1+1 is 2, 1+2 is 3, 2+3 is 5, 3+5 is 8, and so on.

𝒇𝟎 = 𝟎 , 𝒇𝟏 = 𝟏 𝒂𝒏𝒅 𝒇𝒏 = 𝒇𝒏−𝟏 + 𝒇𝒏−𝟐

Terminate the program after calculation the first 10

numbers of the Fibonacci series and print them.

C Programming

Flowcharts Examples


Example 4

𝒇𝟎 = 𝟎𝒇𝟏 = 𝟏𝒇𝒏 = 𝒇𝒏−𝟏 + 𝒇𝒏−𝟐

0

1

1

2

3

5

8

13

21

34

55

C Programming

Flowcharts Examples


Example 5

It is desired to compute income Tax for a given

income as per the following table:

Income Amount of tax

Up to Rs 35000 Nil

From Rs 35001

To Rs 60000

20% of income

in Excess of Rs 35000

From Rs 60001

To Rs 120000

Rs 5000 + 30% of income in Excess of Rs 60000

More than Rs 120000Rs 23000+ 40% of income in

Excess of Rs 120000

C Programming

Flowcharts Examples


Flowcharts Examples


Example 6

It is desired to sum 20 values read from a data

statement. This can be done by means of a loop. It

uses the following algorithm for summation.

Sj+1 = Sj + A

Where: j = 1,2,3,..20Sj+1 : The new sum

Sj : The old sum

A : The variable whose values are to be added

C Programming

Flowcharts Examples


Flowcharts Examples


Example 7

Draw a flowchart of Program

that Calculate and Display

The Factorial of n

C Programming

Flowcharts Examples


Example 7



The Factorial of n𝒏! = 𝒏 × 𝒏 − 𝟏 × 𝒏 − 𝟐 … × 𝟑 × 𝟐 × 𝟏

C Programming

Flowcharts Examples


Example 7



The Factorial of n

𝒏! = 𝒏 × 𝒏 − 𝟏 × 𝒏 − 𝟐 … × 𝟑 × 𝟐 × 𝟏

C Programming

Flowcharts Examples


Example 8

Draw a flowchart to compute all possible products of

X and Y

X and Y varying from 0 through 9

(100 products)

Home Work ..!

0x0=0 1x0=0 … 9x0=0

0x1=0 1x1=1 … 9x1=9

0x2=0 1x2=2 … 9x2=18

0x3=0 1x3=3 … 9x3=27

. .

. .

0x9=0 1x9=9 … 9x9=81

Flowcharts Examples


Example 8

0x0=0 1x0=0 … 9x0=0

0x1=0 1x1=1 … 9x1=9

0x2=0 1x2=2 … 9x2=18

0x3=0 1x3=3 … 9x3=27

. .

. .

0x9=0 1x9=9 … 9x9=81

Algorithms


Program Coding

o In this phase, the algorithm ( or Flowchart ) is transferred into program using codes of one of programming language.

o Machine Language ( Binary Code ).

o Assembly Language.

o High Level Language.


Program Executing and Testing


Editing

Compiling

Linking

Executing


Means to use any editingprogram to convert the program intoelectronic form and save it with asuitable and valid name with theextension (.c)

The program becomes an electronic source code


Editing


Means to use a compilerprogram matches with theprogramming language the sourceprogram written with, to detect thelinguistic errors and convert thesource code program into a binarycodes.

The program becomes an objectprogram, saved with the source codename but with the extension (.0).


Compiling


The program when it is writtencontains some functions, as printf( )function which is saved in the library ofthe compiler and may be someexternal functions written to do certainjobs by the program after linking withlibraries functions is converted to anexecutable program, with the samename entered by the user but with theextension .exe (dot exe).


Linking


The program in this step i.e.,after linking is ready to be run ordebugged on the computer.


Executing



Editing

Compiling

Linking

Executing

Steps of Obtaining an Executable Program


Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

Course Contents



Constants and variables

Expressions

Arithmetic operators

Statements

Statement blocks

Data Types and Names in C

Naming a Variable

Expressions

o An expression is a combination of constants,variables, and operators that are used todenote computations

1. Taking the value contained in the drawer(variable) A

2. Multiply this value by 2

3. Subtract 1 from result obtained from 2

4. The value contained in drawer A is omitted, thenputting the result obtained from 3 into drawer A.


𝑨 = 𝟐 ∗ 𝑨 − 𝟏

Arithmetic Operations


Addition

DivisionMultiplication

Subtraction

Reminder

The Reminder operator % is used to obtainthe reminder of the first operand divided bythe second operand



Addition


Subtraction

Reminder

𝟔 % 𝟒 = 𝟐 𝟏𝟎𝟎𝟏%𝟐 = 𝟏 𝟒𝟖%𝟓 =

Example



Addition


Subtraction

Reminder

𝟔 % 𝟒 = 𝟐 𝟏𝟎𝟎𝟏%𝟐 = 𝟏 𝟒𝟖%𝟓 = 𝟑

Example

Constants and Variables

o As its name implies, a constant is a value that neverchanges.

o A variable, on the other hand, can be used to presentdifferent values.

o For instance, consider the following:

𝒊 = 𝟏 ;o Where the symbol i is a constant because it always has

the same value (1) and the symbol i is assigned theconstant 1.

o In other words, i contains the value of 1 after thestatement is executed.


Data Types and Names

oThe C language reserves some keywordswords that have special meanings to thelanguage.

oThose reserved words should not be usedas variables, constants, or function namesin your program.

oAll C keywords must be written inlowercase letters, for instance INT will notbe treated as a keyword, it must be writtenas int.C Programming Higher Technological Institute76

The computer list of C keywords

auto break case char

const continue default do

double else enum extern

float for goto if

int long register return

short signed sizeof static

struct switch typedef union

unsigned void volatile while


Naming a Variable

o Characters A through Z and a through z

o Digit characters 0 through 9, which can be used in any position except the first of a variable name.

o The underscore character _

Examplesstop_sign loop3 and_pause


Valid Variable Name Can Use

Naming a Variable

o A variable name can’t contain any C arithmetic signs.

o A variable name can’t contain any dots.

o A variable name can’t contain any apostrophes.

o A variable name can’t contain any other special symbols such as *, @, #, and so on.

Examples4flags sum-result method*4

return what_size? ahmed.ali


Invalid Variable Name Can NOT be Used

Data Types


C Data Type

char

a, B, $, #

int

5, 17, 128

float

2.5 , 0.3

double

23433.3455

Declaration Statement


Typechar

int

float

double

Namec1

N1

F1

d1

Value‘&’

100

32/10

5e3

char c1 = ‘&’ ;

int n1 = 100 ;



Typechar

int

float

double

Namec1

N1

F1

d1

Value‘&’

100

32/10

5e3

char c1 ;

int n1 ;

c1 = ‘&’ ;

n1 = 100 ;



Typechar

int

float

double

Namec1

n1

f1

d1

Value‘&’

100

32/10

5e3

float f1= 32/100 ;

double d1=5e3 ;



Typechar

int

float

double

Namec1

n1

f1

d1

Value‘&’

100

32/10

5e3

float f1 ;

double d1 ;

f1 = 32/100 ;

d1 = 5e3 ;

Some special characters in C

\b BackspaceMoves the cursor to the left one character

\f Form feedGoes to the top of a new page

\n New lineCarriage return and line feeds

\r ReturnReturns to the beginning of the current line

\t TabAdvances to the next tab stop


Examples

/* Example1 : Printing out characters */

# include <stdio.h>

/* the header file for the printf () function */

main ( )

{ /* the main function body till line 13 */

char c1; /* declaration of the character variable c1 */


c1 = ‘A’; /* assigning c1 with the character constant A */

c2 = ‘a’; /* assigning c2 with the character constant a */

return 0;

}


Examples /* Example1 : Printing out characters */

# include <stdio.h>


main ( )






printf ( “ The character c1 is : %c \n ”, c1);

printf ( “ The character c1 is : %c ” , c1);

printf ( “ while the character c2 is : %c \n”, c2);

return 0;

}



# include <stdio.h>


main ( )









return 0;

}




# include <stdio.h>


main ( )









return 0;

}



New Line

Specifier

for

Character Data Type


# include <stdio.h>


main ( )









return 0;

}


You Have To Know about Data Types


Type

char

int

float

double

Name

c1

n1

f1

d1

Value

‘&’

100

32/10

5e3

Specifier

%c

%d

%f

%e,%E

Examples /* The arithmetic operations on integers */

# include<stdio.h>


main ( )


int m = 3;

int n=2;

printf ( "The summation of %d and %d is : %d.\n", m, n, m+n);

printf ( "The difference between %d and %d is : %d.\n", m, n, m-n);

printf ( "The multiplication of %d by %d is : %d.\n", m, n, m*n);

printf ( "The division of %d by %d is : %d.\n", m, n, m/n);

printf ( "The remainder of division of %d by %d is : %d.\n", m, n, m%n);

return 0 ;

}


Operators Precedence

Operator Sign Name

( ) Brackets

/ Division

* Multiplication

+ Addition

- Subtraction


Logic Operators


Operator Sign Name

| | OR

&& AND

!= NOT

Examples /* Example3 : Integer vs. floating point divisions */

# include <stdio.h> /* the header file for the printf ()

function */

main ( )

{

int n1,n2,n3;

float m1,m2,m3;

n1 = 32/10; m1= 32/10;

n2 = 32.0/10; m2= 32.0/10;

n3 = 32/10.0; m3 = 32/10.0;

return 0;

}


Examples /* Example3 : Integer vs. floating point divisions */

# include <stdio.h> /* the header file for the printf () function */

main ( )

{

int n1,n2,n3; float m1,m2,m3;

n1 = 32/10; m1= 32/10;

n2 = 32.0/10; m2= 32.0/10;

n3 = 32/10.0; m3 = 32/10.0;

printf ( “ The integer division of 32/10 is : %d \n”, n1);

printf ( “The floating point division of 32/10 is : %f \n”, m1);

printf ( “The integer division of 32.0/10 is : %d \n”, n2);

printf ( “The floating point division of 32.0/10 is : %f \n”, m2);

printf ( “The integer division of 32/10.0 is : %d \n”, n3);

printf ( “The floating point division of 32/10.0 is : %f \n”, m3);

return 0;

}C Programming Higher Technological Institute98

Double Data Type

o Here are two examples:

[mantissa] e [exponent]

[mantissa] E [exponent]

Example

5000 5e3.

-300 -3e2

0.0025 2.5e-3.


Specifier %e or %E

With printf ( )

Precedence Example

# include<stdio.h>

main ( )

{

int A,B,C,D,E, X, Y, Z;

A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;

printf ( “ X= %d \n Y= %d \n Z=%d \n”, X,Y,Z);

return 0 ;

}


Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


X = E * D – B / C + A ;

Precedence Example

X = E * D – B / C + A ;


X = E * D – B / C + A ;

Precedence Example


X = E * D – B / C + A ;

Precedence Example


X = E * D – B / C + A ;

A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example


X = E * D – B / C + A ;

2 * 10 – 6 / 3 + 20 A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example


X = E * D – B / C + A ;

2 * 10 – 6 / 3 + 20

20 – 2 + 20

A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example


X = E * D – B / C + A ;

2 * 10 – 6 / 3 + 20

20 – 2 + 20

38

A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


Y = ( E * D ) – ( B / C ) + A ;

Precedence Example

.

.

.


Y = ( E * D ) – ( B / C ) + A ;

Precedence Example

.

.

.

38


Y = ( E * D ) – ( B / C ) + A ;

Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


Z = A * ( D – B ) / ( C + E ) ;

Precedence Example


Z = A * ( D – B ) / ( C + E ) ;

A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example


Z = A * ( D – B ) / ( C + E ) ;

20 * 4 / 5A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example


Z = A * ( D – B ) / ( C + E ) ;

20 * 4 / 5

16

A = 20

B = 6

C = 3

D = 10

E = 2

( )

*

/

+ -

Precedence Example

# include<stdio.h>

main ( )

{


A = 20 ; B = 6 ; C = 3 ; D = 10 ; E = 2 ;

X = E * D – B / C + A ;

Y = ( E * D ) – ( B / C ) + A ;

Z = A * ( D – B ) / ( C + E ) ;


return 0 ;

}


Home Work ..!!Exercises 3 Page 44


Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

Course Contents



Course Contents


Errors Types

Arithmetic Assignment Operators

Unary Minus Operator

The Cast Operator


Error Types


Syntax

Execution

Logic

Error Types

They are errors that youhave made in the form

Or syntax of the language

o Spelling a keyword incorrectly

o Errors that are detectedduring running of a programare shown in a dialog box thatappears and the error ishighlighted in the program.


Syntax

Example

Error Types

If the program has no syntax errors. The computer can execute it.

During execution, execution errors may be detected.

o If a number is divided by zerocauses an execution error.

Whenever an execution erroris detected, the computer displays adialog box with an error messageand stops executing the program.


Execution

Example

Error Types

If the output of the programdoes not agree with what isexpected, this is logic error.

The computer cannot detectsuch an error because it does notknow the logic of the programshould be.

So, it is your responsibility todetect logic errors in the program.


Logic

Error Types


Syntax

Execution

Logic

Arithmetic Assignment Operators

o Here the statement causes the value of the right-hand-operand to be assigned to the memory location of the left-hand-operand.

o Statement writes the value of the right-hand-operand (5) into the memory location of the integer variable a (which is the left-hand-operand in this case).


left-hand-operand = right-hand-operand

a=5 ;

b=a=5 ;

Assignment and Arithmetic Operators

Operator Description

+ = Addition Assignment Operator

- = Subtraction Assignment Operator

* = Multiplication Assignment Operator

/ = Division Assignment Operator

% = Remainder Assignment Operator


Equivalence of Statements

Statement Equivalence

x+ = y; x = x + y;

x- = y; x = x - y;

x* = y; x = x * y;

x/ = y; x = x / y;

x% = y; x = x % y;


z = z * x + y ; z *= x + y ; ?

Equivalence of Statements

Statement Equivalence

x+ = y; x = x + y;

x- = y; x = x - y;

x* = y; x = x * y;

x/ = y; x = x / y;

x% = y; x = x % y;


z = z * x + y ; z *= x + y ; ≠

Example# include <stdio.h> main ( ){ int x, y, z; x =1; y =3; z= 10;printf (“Given x = %d, y = %d, and z = %d,\n”, x, y, z);x= x + y;printf ( “ x= x +y assigns %d to x;\n”, x);x = 1; x+= y ; printf ( “ x+= y assigns %d to x;\n”, x);x = 1; z = z* x + y;printf ( “z = z*x+y assigns %d to z;\n”, z); z = 10; z = z* (x + y);printf ( “z = z*( x+ y) assigns %d to z;\n”, z);z = 10; z *= x + y;printf ( “z *= x+ y assigns %d to z;\n”, z);return 0; }


Unary Minus Operator

o Given an integer, you can get its negation by changing the sign of the integer by using, -, the minus operator, which is called the unary minus operator.

o Differ between the unary minus operator and the subtraction operator.

o The first – symbol is used as the subtraction operator while the second –symbol is the unary minus operator.


x = 1.234 ; -x equals -1.234

z = x- -y; OR z = x- (-y);

Incrementing or Decrementing by ONE


X++ ++X

X-- --X

IncrementP

ost-

Decrement

Pre

-

Example# include <stdio.h> main ( ){int x=5; printf (“ X++ %d ,\n”, x++);printf (“++X %d ,\n”, ++x);printf (“ X-- %d ,\n”, x--);printf (“ --X %d ,\n”, --x);return 0; }


Relational Operators

Operator Description

== Equal to

!= Not equal to

> Greater than

< Less than

>= Greater than or equal to

<= Less than or equal to


oAll relational expressions produce a result of either 0 or 1.

oGiven x = 3 and y = 5, for instance, the relational expression x < y gives a result of 1


x * y < z + 3 (x * y) < ( z + 3 )


Example# include <stdio.h> main ( ){ int x, y;

double z;

x = 7; y = 25; z = 24.46;

printf (“ Given x = %d, y = %d and z = %2f, \n”, x, y, z);printf ( “ x > = y produces: %d \n”, x > =y);printf ( “x = = y produces: %d \n”, x = =y);printf ( “x < y produces: %d \n”, x < y); printf ( “x > y produces: %d \n”, x > y);printf ( “x != y - 18 produces: %d \n”, x ! =y -18); printf ( “x + y ! = z produces: %d \n”, x +y ! = z); return 0; }


This is algebraically True and is supposed to return 1.

o However, the expression returns 0 which means that the equal to relationship does not hold.

o This is because the truncation of the integer division – that is , 1 /2 – produces 0 not 0.5.


1 / 2 + 1 / 2 = = 1 1 Or 0


The Cast Operator

o You can convert one data type to a different oneby prefixing and cast operator to the operand.

o The general form of the cast operator is

o Here data-type specifies the data type you wantto convert to x is a variable ( or, expression)that contains the value of the current data type.

o You have to include the parentheses (and) tomake up a cast operator.


(data-type) x

Example# include <stdio.h> main ( ){int x, y;

x = 7;

y = 5;

printf ( “ Given x = %d, y = %d, \n”, x, y);

printf ( “ x / y produces: %d \n”, x /y);

printf ( “(float) x / y produces: %f \n”, (float) x / y);

return 0 ;

}


Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

Course Contents



scanf ( ) functiono The scanf ( ) function is a very important function in C.

o It is used to read data from the standard input device; the keyboard.

o The syntax for the scanf ( ) function is :

o Example : to input two variables integer x and float y


scanf(“ % % ”, & & );

scanf(“ %d %f ”, &x &y );

scanf ( ) functiono The scanf ( ) function is a very important function in C.

o It is used to read data from the standard input device; the keyboard.

o One thing you need to be aware of; is that the scanf ( ) function doesn’t actually start reading the input untilthe Enter key is pressed.

o Data entered from the keyboard is placed in an inputbuffer.

o When the Enter key is pressed, the scanf () function looks for its input in the buffer.


printf( ) functiono The printf ( ) function is a very important function in C.

o It is used to print out messages on the screen.

o The syntax for the printf ( ) function is

o Example : to print two variables integer x and float y


printf(“ % % ”, );

printf(“ %d %f ”, x y );

printf( ) functiono The printf ( ) function is a very important function in C.

o It is used to print out messages on the screen.

o The number of expressions is determined by thenumber of the format specifiers inside the firstargument.

o The format specifiers and the expressions arematched in order from left to right, and you shoulduse exactly the same number of expressions as thenumber of format specifiers within the format string.


Adding the Minimum Field Width

o A integer is added between the percent sign (%) and the letter in a format specifier to specify the minimum field width and ensures that the output reaches the minimum width.

o For example in %10f

10 is a minimum field width specifier thatensures that the output is at least 10 characterspacers wide.


printf(“ %10f ”, x );

Example


# include <stdio.h>

main ( )

{

int x=12; int y=12345;

printf (“%d \n”, x);

printf (“%d \n”, y);

printf (“%5d \n”, x);


printf (“%2d \n”, y);

return 0;

}

Example


# include <stdio.h>

main ( )

{

int x=12; int y=12345;

printf (“%d \n”, x);

printf (“%d \n”, y);



printf (“%2d \n”, y);

return 0;

}

12

12345

12

00012

12345

Aligning Output

o By default, all output is placed on the Right edge ofthe field, as long as the field width is longer than thewidth of the output.

o You can change this and force output to be left-justified.

o To do so, you need to prefix the minimum fieldspecifier with the minus sign (-).

o Specifies the minimum field width as 12, and justifiesthe output from the left edge of the field.


%-12d

Example


# include <stdio.h>

main ( )

{

int x,y,z,m,n; x=1; y=12; z=123;

m=1234; n=12345;

printf (“%8d %-8d \n”, x, x );

printf (“%8d %-8d \n”, y, y );

printf (“%8d %-8d \n”, z, z );

printf (“%8d %-8d \n”, m, m );

printf (“%8d %-8d \n”, n, n );

return 0;

}

Example


# include <stdio.h>

main ( )

{

int x,y,z,m,n; x=1; y=12; z=123;

m=1234; n=12345;

printf (“%8d %-8d \n”, x, x );

printf (“%8d %-8d \n”, y, y );

printf (“%8d %-8d \n”, z, z );

printf (“%8d %-8d \n”, m, m );

printf (“%8d %-8d \n”, n, n );

return 0;

}1 1

12 12

123 123

1234 1234

12345 12345

The Precision Specifier

o You can put a period (.) and an integer right after theminimum field width specifier.

o The combination of the period and the integer makeup a precision specifier.

o The precision specifier is another important specifieryou can use to determine the number of decimalplaces for floating-point numbers, or to specify themaximum field width for integers or strings.

o The minimum field width length is specified as 10character long, and the number of decimal places isset to 3.


%10.3f

The Precision Specifier

o You can put a period (.) and an integer right after theminimum field width specifier.

o The combination of the period and the integer makeup a precision specifier.

o The precision specifier is another important specifieryou can use to determine the number of decimalplaces for floating-point numbers, or to specify themaximum field width for integers or strings.

o Remember, the default number of decimal places is 6.


Example


# include <stdio.h>

main ( )

{

int x=123; float y=123.456789;

printf (“Default integer format \n %d \n”, x );

printf (“With precision specifier \n %2.8d \n”, x );

printf (“Default float format \n %f \n”, y );

printf (“With precision specifier \n %10.2f \n”, y );

printf (“With precision specifier \n %-10.2f \n”, y );

return 0;

}

Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

Course Contents


Decision6

Course Contents


The if statement

Flowchart of if statement

The if else statement and flowchart

Nested if statements and their flowcharts

The switch statement

The break and continue statement

The goto statements

Decision6

The if statemento The if statement is the most popular conditional

branching statement.

o It can be used to evaluate the conditions as well as tomake the decision whether the block of codecontrolled by the statement is going to be executed.

o Here if condition is logical TRUE, the statements insidethe braces are executed.

o If condition is logical FALSE, then the statements areskipped.


The if statement Flowchart


…..if ( Test Expression) {statement1;statement2;…}….

IF Statement Syntax

The if statement Flowchart


…if ( Test Expression) statement;…

IF Statement Syntax

Example

if ( x>5) printf( “ X > 5 ” );…

Example


# include <stdio.h>

main ( )

{

int i;

printf (“Integers that can be divided by 3 \n ” );

printf (“Enter a positive number : \n ”, x );

scanf(“%d”, &i );

if ( i %3 == 0 )

printf (“The number %d is divisible by 3 \n ”, i );

return 0;

}

Output


Integers that can be divided by 3Enter a positive number : 12The number 12 is divisible by 3

Integers that can be divided by 3Enter a positive number : 7

The if else statement


…if ( Test Expression) {statement1;statement2;…}else {statement_Astatement_B…}…

if else Statement Syntax

The if else statement


…if ( Test Expression) statement1;elsestatementA;…

if else Statement Syntax

if ( X > 5) printf( “ X > 5 ” );elseprintf( “ X < 5 ” );

Example

if else Example







# include <stdio.h>

main ( )

{

int x,y;

printf (“Please Enter X and Y \n ” );

scanf(“%d %d”, &x , &y );

if ( x == y )

printf (“ X = Y \n ” );

else

printf (“ X != Y \n ” );

return 0; }

if else Example

Nested if Flowchart


if ( Test Expression1) { if ( Test Expression2)

{ statement1;statement2;

} else { statementA;

statementB;}

}else {statement X ;}

Nest IF Statement Syntax

Nested if Flowchart





numbers.


# include <stdio.h>

main ( )

{

int A,B,C;

printf (“Please Enter A,B and C \n ” );

scanf(“%d %d %d”, &A , &B , &C );

if ( A<B )

{ if ( A<C )

printf (“ A The is the Smallest \n ” );

else

printf (“ C The is the Smallest \n ” );

} else if ( B<C )

printf (“ B The is the Smallest \n ” );

else

printf (“ C The is the Smallest \n ” );

return 0; }

Nested if Example

The switch statement


o The nested if statements will become very complex ifthere are many decisions that need to be made.

o The switch statement, can be used to make unlimiteddecisions or choices based on the value of a conditionalexpression and specified cases.

o The conditional expression is evaluated first, if thereturn value of the conditional expression is equal tothe constant expression expression 1, the statementstatement 1 is executed.

o the value of the conditional expression is not equal toany values of the constant expressions labeled by thecase keyword, the statement (statement-default)following by the default keyword is executed.

The Switch Statement Syntax


switch ( Conditional Expression ) {

case expression1;statment1;case expression2;statment2;…default;statement-default;

}

The Switch Statement Syntax


switch ( Conditional Expression ) {case expression1;statment1;case expression2;statment2;…default;statement-default; }

o You have to use the case keyword to label each case.

o The default keyword is recommended to be used forthe default case.

o Note that no constant expressions are identical in theswitch statement.

The break statemento An important feature of the switch statement is that

the computer continues to execute the statementsfollowing the selected case until the end of the switchstatement.

o You can add a break statement at the end of thestatement list following every case label, if you wantto exit the switch construct after the statementswithin a selected case are executed


The Break Statement Syntax


switch ( Conditional Expression ) {

case expression1;statment1;break ;case expression2;statment2;break;…default;statement-default;

}

Example

Total Score Letter Grade

85 <= T <= 100 A

80 <= T < 85 B+

75 <= T < 80 B

70 <= T < 75 C+

65 <= T < 70 C

55 <= T < 65 D+

50 <= T < 55 D

T < 50 F


Write Program that compute the grade for a student, the

grade is based on Total Score T

( from 0 to 100) The Grading Scale Is Illustrated In The

Following Table:

# include <stdio.h>

main ( )

{ int score;

printf (" Enter Scorre: \n");

scanf ("%d", &score);

switch (score / 10 ) {

case 10:

printf ("your grade is A \n");

break;

case 9:

printf ("your grade is A \n");

break;

case 8:

printf ("your grade is B \n");

break;

case 7:

printf ("your grade is C \n");

break;


case 6:

printf ("your grade is D \n");

break;

case 5:

printf ("your grade is D \n");

break;

case 4:

printf ("your grade is F \n");

break;

case 3:


break;

case 2:


break;

case 1:


break;

case 0:


break;

default:

printf ( "illegal input \n");

break;

}

printf ( "GOODBYE\n" );

return 0;

}

Example

The Continue statemento There are times when you want to stay in a loop

but skip over some statements within the loop.o To do this, the continue statement causes

execution to jump to the top of the loopimmediately.

o The statement can be used within a while,do…while or for statement to terminate thecurrent iteration of the loop and begin thenext.


The goto statemento Do not use the goto statement unless it’s absolutely

necessary because its usage may make the C programunreliable and hard to debug.

o The Syntax of goto Statement:


…lablename :

statment1;statment2;..

goto lablename ;…


# include <stdio.h>

main ( )

{

int x;

guess_lbl:

printf (“Guess integer number in the range 1 to 10 \n”);

scanf (“%d”, &x );

if ( x == 7)

goto rght_lbl;

printf (“NO, wrong guess, try again \n”);

goto guess_lbl;

rght_lbl:

printf (“WELL DONE your guess = %d is right\n” , x);

return 0;

}

goto Example


# include <stdio.h>

main ( )

{

int x;

guess_lbl:

printf (“Guess integer number in the range 1 to 10 \n”);

scanf (“%d”, &x );

if ( x == 7) goto rght_lbl;

printf (“NO, wrong guess, try again \n”);

goto guess_lbl;

rght_lbl:

printf (“WELL DONE your guess = %d is right\n” , x);

return 0;

}

goto Example

Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

Course Contents


Iteration 7

Course Contents


The for statement

The Null Statement

The while statement

The infinite while Loop

The do-while Statement

The Nested Loop

Examples

Iteration7

Iterationo Iteration (looping) is used in programming

to perform the same set of statementsover and over until certain specifiedconditions are met.


Repeat the same statement(s) until certain condition

Iteration


Example




Iteration


Example




From

1 to 9

Step 1

The for statement


for ( initialization ; Test Expression ; Step) statement;…

for Statement Syntax One Statement

for ( initialization ; Test Expression ; Step) {statement1;statement2;…}

for Statement Syntax Multiple Statements

The for statement


for ( i=1 ; i <= 10 ; i++ )

printf (“ Hello \n ”);

printf (“ Goodbye \n ”);

Will Print Hello 10 Times Then PrintGoodbye

for ( i=1 ; i <= 10 ; i++ )

{



}

Will Print Hello Goodbye10 Times

Example


# include <stdio.h>

main ( )

{

int i=1; int j,k;

for ( i=1 ; i <= 9 ; i++ )

{

j=i*i; k=j*i;

printf (“ i = %d ”, i ) ;

printf (“ j = %d ”, j ) ;

printf (“ k = %d \n”, k ) ;

}

return 0;

}

Null statement


for ( i=1 ; i <= 10 ; i++ ) ;



The while statement


Initializationwhile ( Test Expression ){…Increment Or Decrement ;}

while Statement Syntax

The while statement


int w = 1;

while ( w <= 10 )

{


w = w++;

}



While Loop Example


# include <stdio.h>

main ( )

{

int i=1; int j,k;

while (i<=10)

{

j=i*i; k=j*i;

printf (“ i = %d ”, i ) ;

printf (“ j = %d ”, j ) ;

printf (“ k = %d \n”, k ) ;

i=i+1;

}

return 0; }

The Infinite while loop


while(1)

{

Statement1;

Statement2;

}

Always returns 1, the statements inside thestatement block will be executed over andover- that is, the while loop will continueforever.

You can set certain conditions inside the while loopto break the infinite loop as soon as theconditions are met.

The do-while statement


Initializationdo{…Increment Or Decrement ;

} while ( Test Expression ) ;

do-while Statement Syntax



Initializationdo{…Increment Or Decrement ;

} while ( Test Expression ) ;

do-while Statement Syntax

Only after do-while loop

Don’t forget to write ;



int w = 1;

do

{


w = w++;

} while ( w <= 10 ) ;printf (“ Goodbye \n ”);


do-while Loop Example


# include <stdio.h>

main ( )

{

int i=1; int j,k;

do

{

j=i*i; k=j*i;

printf (“ i = %d ”, i ) ;

printf (“ j = %d ”, j ) ;

printf (“ k = %d \n”, k ) ;

i=i+1;

} while (i<=10);

return 0; }

Examples


Example 1

Write a program to calculate the

summation and Average of 20 values

read from a data statement.

Examples


# include <stdio.h>

main ( )

{

int i,x,sum; float av;

sum=0;

for ( i=1 ; i<=20 ; i++ )

{

printf(“Please enter new no \n”);

scanf(“%d”,&x);

sum=sum+x;

}

av=sum/20;

printf(“Sum=%d \n Av=%f\n”, sum , av);

return 0; }

Examples


Example 2


summation and Average of N numbers


Examples


# include <stdio.h>

main ( )

{

int i,n,x,sum; float av;

sum=0;

printf(“How many Numbers ? \n”);

scanf(“%d”,&n);

for ( i=1 ; i<=n ; i++ )

{


scanf(“%d”,&x);

sum=sum+x;

}

av=sum/n;


return 0; }

Flowcharts Examples


Example 3 Nested Loop

Write a program and Draw its flowchart to compute all

possible products of X and Y

X and Y varying from 1 through 9

(100 products)

1x0=0 2x0=0 … 9x0=0

1x1=0 2x1=1 … 9x1=9

1x2=0 2x2=2 … 9x2=18

1x3=0 2x3=3 … 9x3=27

. .

. .

1x9=0 2x9=9 … 9x9=81

Flowcharts Examples


Example 3 Nested Loop

# include <stdio.h>

int main ( )

{

int x,y;

for ( x=1 ; x <= 9 ; x++)

for ( y=1 ; y <= 9 ; y++)

printf(“%d x %d = %d\n”,x,y,x*y);

return 0;

}

Flowcharts Examples


Example 4

Draw a flowchart and write a Program that

Calculate and Display The Factorial of n

𝒏! = 𝒏 × 𝒏 − 𝟏 × 𝒏 − 𝟐 … × 𝟑 × 𝟐 × 𝟏

Flowcharts Examples


# include <stdio.h>

int main ( )

{

int n,fact; fact=1;

printf(“Enter your N \n”);

scanf(“%d”,&n);

for( int i=1;i<=n;i++)

fact=fact*i;

printf(“ N! =%d \n”,fact);

return 0;

}

Home Work..!!

Rewrite The previous examples using:

a) while Loop

b) do-while Loop


Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

C Programming

Course Contents


Arrays8

C Programming

Examples


Example 2


summation and Average of N numbers


C Programming

Examples


# include <stdio.h>

main ( )

{

int i,n,x,sum; float av;

sum=0;

printf(“How many Numbers ? \n”);

scanf(“%d”,&n);

for ( i=1 ; i<=n ; i++ )

{


scanf(“%d”,&x);

sum=sum+x;

}

av=sum/n;


return 0; }C Programming

Arrays


An array is a collection of variables that are of the samedata type.

Each item in an array is called an element.

All elements in an array are referenced by the name of the array and are stored in a set of consecutive memory slots.

The array takes two forms :

One dimensional array (1-D) ; where the data is arranged horizontally or vertically.

Multi-dimensional array; where the data is arranged in two

dimensional (2-D), three dimensional (3-D),…etc.

C Programming

Arrays


The following is the syntax form to declare an array:

The data-type is the type specifier that indicates what data type the declared array will be.

Array-Name is the name of the declared array. When choosing a name for an array, you should follow the same rules of naming a variable.

Array-Size defines how many elements the array can contain. Note that the brackets are required in declaring an array.

data-type Array-Name [Array-Size];

C Programming

Arrays Example


Where int specifies the data type of the array whose name is arrat_int.

The size of the array is 8, which means that the array can store 8 elements

int array_int [8];

C Programming

Indexing Arrays


You can access each of the elements in the array separately.

The following declaration declares an array called day of integer numbers:

You can access the elements in the array of day one after another: day[0], day[1], …, day[6].

The important thing to remember is that all arrays in C are indexed starting at 0.

The first element in the array of day is day[0].

int day [7];

C Programming

Initializing Arrays


You can initialize each element in an array by two way.

For instance if we have the array :

Initialize the first element in the array: like this:

Here the value 24 is assigned to the first element of the array number, number[0].

int number [5];

number[0] = 24;

C Programming

Initializing Arrays


The second way to initialize an array is to initializeall elements in the array together.

For instance, the following statement initializes allelements of the array, number:

Here the integers inside the braces are assigned tothe corresponding elements of the array number.

That is, 100 is given to the first element(number[0]), 4 to the second element (number[1]),and so on.

int number [5]= {100, 4, 24, 34,5, 16};

C Programming

Initializing Arrays


The second way to initialize an array is to initializeall elements in the array together.

For instance, the following statement initializes allelements of the array, number:

Here the integers inside the braces are assigned tothe corresponding elements of the array number.

That is, 100 is given to the first element(number[0]), 4 to the second element (number[1]),and so on.

int number [5]= {100, 4, 24};

C Programming

Initializing Arrays


# include <stdio.h>

main ( )

{

int i; int A[10];

for ( i=0 ; i<10 ; i++ )

{

A[i]=i+1;

printf(“A[%d]=%d \n”,i,A[i]);

}

return 0; }

C Programming

Example


Example


Summation and Average of 10 numbers read

from a data statement.( Using Array )

C Programming

Example


# include <stdio.h>

main ( )

{

int i; int A[10]; int sum=0; float av;

for ( i=0 ; i<10 ; i++ )

{

printf(“Enter the value of Element [%d] \n”,i);

scanf(“%d”,A[i]);

sum=sum+A[i]

}

av=sum/10;

printf(“Sum=%d \n Av=%f\n”, sum , av);return 0; }

C Programming

Two Dimensional Arrays


int A[3][4] ;

Rows

Columns

C Programming



int A[3][4] = { { 1 , 2 , 3 , 4 },

{ 5 , 6 , 7 , 8 },

{ 9 , 8 , 7 , 6 }} ;

Assigning Static Value

A[3][4] = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 8 , 7 , 6 }} ;

OR

C Programming



Example 1

Write a program to calculate and display

the Summation of two Matrices A and B.

A = 𝟏 𝟐 𝟑𝟒 𝟓 𝟔𝟕 𝟖 𝟗

, B = 𝟗 𝟖 𝟕𝟔 𝟓 𝟒𝟑 𝟐 𝟏

C Programming

Example


# include <stdio.h> main ( ){

int A[3][3] = { { 1 , 2 , 3 } ,

{ 4 , 5 , 6 } ,

{ 7 , 8, 9 } } ;

int B[3][3] = { { 9 , 8 , 7 } ,

{ 6 , 5 , 4 } ,

{ 3 , 2, 1 } } ;

return 0; }

C Programming

Example



int A[3][3] = { { 1 , 2 , 3 } , { 4 , 5 , 6 } , { 7 , 8, 9 } } ;

int B[3][3] = { { 9 , 8 , 7 } , { 6 , 5 , 4 } , { 3 , 2, 1 } } ;

return 0; }

C Programming

Example



int A[3][3] = { { 1 , 2 , 3 } , { 4 , 5 , 6 } , { 7 , 8, 9 } } ;

int B[3][3] = { { 9 , 8 , 7 } , { 6 , 5 , 4 } , { 3 , 2, 1 } } ;

int S[3][3] ; int i,j;

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

printf(“S[%d][%d] = %d \n ”, i , j , S [ i ][ j ] ) ; }

return 0; }

C Programming

Example


for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;


return 0; }

S3,3 = A3,3 + B3,3

= 𝟏 𝟐 𝟑𝟒 𝟓 𝟔𝟕 𝟖 𝟗

+𝟗 𝟖 𝟕𝟔 𝟓 𝟒𝟑 𝟐 𝟏

=𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎

at i = 0 j = 0

C Programming

Example


for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;


return 0; }

S3,3 = A3,3 + B3,3

= 𝟏 𝟐 𝟑𝟒 𝟓 𝟔𝟕 𝟖 𝟗

+𝟗 𝟖 𝟕𝟔 𝟓 𝟒𝟑 𝟐 𝟏

=𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎

at i = 0 j = 1

C Programming

Example


for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;


return 0; }

S3,3 = A3,3 + B3,3

= 𝟏 𝟐 𝟑𝟒 𝟓 𝟔𝟕 𝟖 𝟗

+𝟗 𝟖 𝟕𝟔 𝟓 𝟒𝟑 𝟐 𝟏

=𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎

at i = 1 j = 1

C Programming

Example


for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;


return 0; }

S3,3 = A3,3 + B3,3

= 𝟏 𝟐 𝟑𝟒 𝟓 𝟔𝟕 𝟖 𝟗

+𝟗 𝟖 𝟕𝟔 𝟓 𝟒𝟑 𝟐 𝟏

=𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎

at i = 2 j = 0

C Programming

Example


for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;


return 0; }

S3,3 = A3,3 + B3,3

= 𝟏 𝟐 𝟑𝟒 𝟓 𝟔𝟕 𝟖 𝟗

+𝟗 𝟖 𝟕𝟔 𝟓 𝟒𝟑 𝟐 𝟏

=𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎𝟏𝟎 𝟏𝟎 𝟏𝟎

at i = 2 j = 2

C Programming

Example



int A[3][3] = { { 1 , 2 , 3 } , { 4 , 5 , 6 } , { 7 , 8, 9 } } ;

int B[3][3] = { { 9 , 8 , 7 } , { 6 , 5 , 4 } , { 3 , 2, 1 } } ;

int S[3][3] ; int i,j;

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;


return 0; }

C Programming



Example 2


the Summation of any two 3x3 Matrices.

C Programming

Example 2


# include <stdio.h> main ( ){ int A[3][3] , B[3][3] , S[3][3]; int i,j ;

.

.

.

.

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;



Example 2



for ( i=0 ; i<3 ; i++ ) /* Rows of Metrics A*/

for ( j=0 ; j<3 ; j++ ) /* Columns of Metrics A */

{ printf(“Please Enter A[%d][%d] = \n ”, i , j ) ;

scanf(“%d”,A[i][j]); }

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;



Example 2







for ( i=0 ; i<3 ; i++ ) /* Rows of Metrics B*/

for ( j=0 ; j<3 ; j++ ) /* Columns of Metrics B */

{ printf(“Please Enter B[%d][%d] = \n ”, i , j ) ;

scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

{ S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;





Example 3


the Summation of any two 3x3 Matrices.

Then from Summation Matrix Find :

a) The Largest and Smallest element.

b) The average

c) Matrix Transpose

C Programming

Example 3










scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

return 0; }

C Programming

Example 3










scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

int L=S[0][0], M=S[0][0];

for ( i=0 ; i<3 ; i++ ) /* Rows*/

for ( j=0 ; j<3 ; j++ ) /* Columns*/

if (L<S[i][j])

L=S[i][j];

return 0; }

C Programming

Example 3 (a)



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )


scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

int L=S[0][0], M=S[0][0];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (L<S[i][j])

L=S[i][j];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (M>S[i][j])

M=S[i][j];

printf(“Largest Element = %d \n ”, L ) ;

printf(“Smallest Element = %d \n ”, M ) ;

return 0; }

C Programming

Example 3 (a) (b)



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )


scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

int L=S[0][0], M=S[0][0];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (L<S[i][j])

L=S[i][j];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (M>S[i][j])

M=S[i][j];



int sum = 0 ;

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

sum=sum + S[i][j];

float average =sum/9 ;

printf(“Average = %d \n ”, average ) ;

return 0; }

C Programming

Example 3 (a) (b) (c)



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )


scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

int L=S[0][0], M=S[0][0];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (L<S[i][j])

L=S[i][j];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (M>S[i][j])

M=S[i][j];



int sum = 0 ;

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

sum=sum + S[i][j];



int ST[3][3];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

ST[i][j]=S[j][i];

return 0; }

C Programming




for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )


scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

int L=S[0][0], M=S[0][0];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (L<S[i][j])

L=S[i][j];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (M>S[i][j])

M=S[i][j];



int sum = 0 ;

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

sum=sum + S[i][j];



int ST[3][3];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

ST[i][j]=S[j][i];

return 0; }

C Programming

Do Not Forget to

add the display

statements for ST




for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )



for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )


scanf(“%d”, B[i][j]); }

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

S [ i ] [ j ] = A [ i ] [ j ] + B [ i ] [ j ] ;

int L=S[0][0], M=S[0][0];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (L<S[i][j])

L=S[i][j];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

if (M>S[i][j])

M=S[i][j];



int sum = 0 ;

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

sum=sum + S[i][j];



int ST[3][3];

for ( i=0 ; i<3 ; i++ )

for ( j=0 ; j<3 ; j++ )

{ printf (“\n”); ST[i][j]=S[j][i];

printf (“%d”, ST[i][j]); }

return 0; }

C Programming

Matrix Multiplication


Example 4


the Multiplication of any two 3x3 Matrices.

C Programming



Remember

C Programming

=

New Matrix

Dimensions

Example 4


# include <stdio.h> main ( ){ int A[4][3]={ 2 , 3 , 4 ,

1 , 2 , 1 ,2 , 1 , 0 ,1 , 0 , 2 };

int B[3][4]={ 10 , 0 , 20 , 10 ,0 , 20 , 0 , 20,

20 , 10 , 10 , 0 };int M[4][4];int ra = 4 , ca = 3 , rb = 3, cb = 4 ;int i , j , k ;for (i=0;i<ra;i++)for (j=0;j<cb;j++){M[i][j]=0;for(k=0;k<ca;k++)M[i][j]= M[i][j]+ A[i][k]*B[k][j];}

printf("\n The M is \n");

for(i=0;i<ra;i++)

{

printf("\n");

for (j=0;j<cb;j++)

printf("%4d",M[i][j]);

}

return 0 ; }

C Programming


Home Work ..!!

Write a program to calculate thedeterment of 3x3 Matrix.

Course Contents

Introduction


1





Decision6

Iteration7

Arrays8

C Functions9

C Programming

Course Contents


C Functions9

C Programming

Functions


Why Function

o To make large programs Manageable,

Programmers modularize them into

subprograms.

o These subprograms are called

Functions

C Programming

Functions


C Standard FunctionsOR

Built in Function

C Standard Functions


The C language has a standard library which is acollection of predefined functions and other programelements which are accessed through header files.

One of the most important header file <math.h>contains useful mathematical functions.

C Programming

Note

Every mathematical function return doubledata type

Some Mathematical in <math.h>


Function Descriptioncos(x) Cosine of x in Radian

exp(x) Exponential of x (base e)

log(x) Natural Logarithm of x (base e)

log10(x) Common Logarithm of x (base e)

pow(x,p) x to the power p = xP

sin(x) Sine of x in Radian

sqrt(x) Square root of x = 𝒙

tan(x) Tangent of x in Radian

Example 1


Write a program to calculate the following:

a) cos (x)b) e x

c) x 3/2

d) 𝒙Test your program with

x = 8, 16, 32

Example 1


# include <stdio.h>

# include <math.h>

main ( )

{

int x=8; double a,b,c,d;

do

{

a = cos(x*180/(22/7));

b=exp(x);

c=pow(x, 3/2);

d=sqrt(x);

printf(“At x = %d \n A = %f \n B= %f \n C= %f \n D=%f \n ”,x,a,b,c,d);x=x*2;

} while ( x<=32 );

return 0;

}

Example 1


# include <stdio.h>

# include <math.h>

main ( )

{

int x=8; double b ;

b=exp(x);

return 0;

}

x

b

8

Function 1

Function 2

…

…

exponential Function

…

…

math

main

Example 1


# include <stdio.h>

# include <math.h>

main ( )

{

int x=8; double b ;

b=exp(x);

return 0;

}

x

b

8

Function 1

Function 2

…

…


…

…

math

main

8

Example 1


# include <stdio.h>

# include <math.h>

main ( )

{

int x=8; double b ;

b=exp(x);

return 0;

}

x

b

8

Function 1

Function 2

…

…


…

…

math

main

8

8886110.520508

Some Header Files in C Library


Header File

Description

<ctype.h> Defines function to test characters

<float.h> Defines constant to float

<limits.h>Defines the integer limits on your local computer

<math.h> Defines mathematical function

<stdio.h> Defines functions for standard input and output

<stdlib.h> Defines utility functions

<string.h> Defines functions for processing string

<time.h> Defines time and date functions

User Defined Function


You Have to Know

o Function Type

o Function Name

o Arguments to the function

Typevoid

returnable

Name

Func_1

n1

Arguments

with

without

Build Your Function


Func_Type Func_name( type1 Arg1, type2 Arg2 )

C Programming

{

statment1 ;

statment2 ;

.

.

.

return variable;

}

Function’s

Body

Where I can put my Function ??


# include <stdio.h>

func_type Func_1( Argument1, Argument2 )

{

statment1 ;

statment2 ;

…

…

return ;

}

main ( )

{

…

Func1 Call ;

…

return 0;

}

C Programming

Main Function

Function’s Body



# include <stdio.h>

func_type Func_1( typeArgument1, typeArgument2 )main ( )

{

…

Func1 Call ;

…

return 0;

}


{

statment1 ;

statment2 ;

…

…

return ;

}

C Programming

Function’s Body

Main Function

;Function DeclarationOR



# include <stdio.h>

func_type Func_1( typeArgument1, typeArgument2 )main ( )

{

…

Func1 Call ;

…

return 0;

}


{

statment1 ;

statment2 ;

…

…

return ;

}

C Programming

Function’s Body

Main Function

;Function Declaration

Please Do not

forget ; after


Simple Example


# include <stdio.h>

int sum ( int x, int y)

{

z = x + y;

return z ;

}

main ( )

{

int a, b, c;

a = 3;

b = 5;

c = sum( a , b );

printf("C=%d \n", c);

return 0;

}

C Programming

Simple Example


# include <stdio.h>


{

z = x + y;

return z ;

}

main ( )

{

int a, b, c;

a = 3;

b = 5;

c = sum( a , b );


return 0;

}

C Programming

Simple Example


# include <stdio.h>


{

z = x + y;

return z ;

}

main ( )

{

int a, b, c;

a = 3;

b = 5;

c = sum( a , b );


return 0;

}

C Programming

• Why we use x, y,z in Function’s Body

While use a,b,c in main Function …?

• Effect of use the same variable name.

Simple Example by Method 2


# include <stdio.h>

int sum ( int, int ) ;

main ( )

{

int a, b, c;

a = 3;

b = 5;

c = sum( a , b );


return 0;

}


{

z = x + y;

return z ; }

C Programming

Simple Example by Method 2


# include <stdio.h>

int sum ( int, int ) ;

main ( )

{

int a, b, c;

a = 3;

b = 5;

c = sum( a , b );


return 0;

}


{

z = x + y;

return z ; }

C Programming

Again Do not

forget ; after


Why Functions


ExampleWrite a program to calculate the

factorial of any integer number. Then use

your program to calculate the value of Y.

𝒀 =𝒏! − 𝒌!

𝒏 − 𝒌 !

C Programming

Example


# include <stdio.h>

main ( )

{

int n ;

printf(“Please Enter The Value of n \n");

scanf(“%d”,&n);int factorial =1;

for ( int i=n ; i>=1 ; i-- )

factorial = factorial * i ;

printf(“Factorial =%d \n", factorial);

return 0;

}

C Programming

Example


# include <stdio.h>

int fact ( int ) ;

main ( )

{

int n, k, y;

printf(“Please Enter The Value of n and k \n");

scanf(“%d %d”,&n,&k);y = ( fact(n) * fact(k) ) / fact( n – k ) ;

printf(“Y=%d \n", y);

return 0;

}

int fact ( int fc )

{

int factorial =1;

for ( int i=fc ; i>=1 ; i-- )

factorial = factorial * i ;

return factorial ;

}

C Programming

LOGO

Eng. Ibrahim Elewah

Higher Technological Institute 10th of Ramadan City6th of October Branch

Electrical and Computer Engineering Department

284

HTI Student Book

Main Reference “C For Dummies”

by Dan Gookin 2nd Edition

computer programming all chapters

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