Fundamental of Programming (C) Group...

Lecturer: Vahid Khodabakhshi

Sharif University of TechnologyDepartment of Computer Engineering

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Fundamental of Programming (C) Group 4

Lecture 5

Structured Program Development

CE 40153 - Fall 98

Structured Program Development – Lecture 5

Sharif University of TechnologyDepartment of Computer Engineering 2/65

How to develop a program?

RequirementsProblem Analysis

DesignDesigning algorithm

Pseudo code Flow chart

ImplementationImplementing algorithm in

c/c++

Validation Test

Maintenance

Software Development

Life-Cycle

Refinement



Requirements Discovery• Problem Analysis

– Problem Identification

– Abstraction

• Inputs and outputs determination

• Defining their relation

Write a program to compute an employee'sweekly pay?

How many hours did you work?

How much do you get paid per hour?

employee's weekly pay equal to multiplication of Hours by Pay Rate



Problem vs. Solution• A problem is something that causes trouble

• The solution is how to solve or fixe the problem



Algorithm• A procedure for solving a problem in terms of actions

and the order in which these actions are to beexecuted in a computer program (program control)

• Action

• Order



Execution order• Sequential execution, normally, statements in a

program are executed one after the other in theorder in which they’re written

• Various C statements enable you to specify that thenext statement to be executed may be other thanthe next one in sequence. This is called transfer ofcontrol



Structured Programming• goto statement that allows programmers to specify a transfer of

control to one of many possible destinations in a program

• Research had demonstrated that programs could be writtenwithout any goto statements

• Programs produced with structured techniques were clearer, easierto debug and modify and more likely to be bug free in the firstplace.

• Research had demonstrated that all programs could be written interms of only three control structures, namely the sequencestructure, the selection structure and the repetition structure



Pseudo code• An artificial and informal language that helps

you develop algorithms

– similar to everyday English

– are not executed on computers

– consists only of action statements

pay-calculation program:

1. Read Hours and Pay Rate2. weekly pay = Read Hours * Pay Rate



Flowchart• A flowchart is a diagram that

depicts the flow of an algorithm

– pay-calculation program

• Each symbol represents a different type of operation

START

Display message “How many hours did you

work?”

Read Hours

Display message “How much do you get paid

per hour?”

Read Pay Rate

Multiply Hours by Pay Rate. Store result

in Gross Pay.

Display Gross Pay

END



Terminals• represented by rounded rectangles

• indicate a starting or ending point

START

END

START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

ENDTerminal

Terminal

Flow line



Input/Output Operations• indicate an input or output operation

Display message “How many

hours did you work?”

Read Hours

START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Input/Output Operation



Processes• indicates a process such as a

mathematical computation or variable assignment

START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Multiply Hours by Pay Rate.

Store result in Gross Pay.

Process



Execution START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Output Operation

Variable Contents:Hours: ?Pay Rate: ?Gross Pay: ?




Execution START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Variable Contents:Hours: 40Pay Rate: ?Gross Pay: ?


40

Input Operation

(User types 40)



Execution START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Output Operation

Variable Contents:Hours: 40Pay Rate: ?Gross Pay: ?

How much do you get paid per hour?



Execution START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Variable Contents:Hours: 40Pay Rate: 20Gross Pay: ?


20

Input Operation

(User types 20)



Execution START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Variable Contents:Hours: 40Pay Rate: 20Gross Pay: 800


20

Process

(Gross Pay = Hours * Pay Rate)



Execution START


work?”

Read Hours


per hour?”

Read Pay Rate


in Gross Pay.

Display Gross Pay

END

Variable Contents:Hours: 40Pay Rate: 20Gross Pay: 800

Your gross pay is 800

Output Operation



Connectors• Sometimes a flowchart will not fit on one page

– A connector (represented by a small circle) allows you to connect two flowchart segments

A

ASTART

END



Modules• A program module (such as a function in C) is

represented by a special symbol

• The position of the module symbol indicates the point the module is executed

• A separate flowchart can be constructed for the module

START

END

Read Input

Call calc_payfunction

Display results

module



Control Structures• Sequence• Decision selection statement

– The if statement is called a single-selection statement because it selects or ignores a single action.

– The if…else statement is called a double-selection statement because it selects between two different actions.

– The switch statement is called a multiple-selection statement because it selects among many different actions

• Repetition– while– do…while– for



Sequence Structure• a series of actions are performed in sequence

– The pay-calculating example



Compound Statements• A statement is a specification of an action to

be taken by the computer as the programexecutes

• Compound Statements is a list of statements enclosed in braces, { }



Decision Structure• One of two possible actions is taken,

depending on a condition

• Selection structures are used to chooseamong alternative courses of action

YESNO YESNOx < y?

Process BProcess A



Decision Structure• The flowchart segment below shows how a

decision structure is expressed in C as anif/else statement

YESNO

x < y?

Calculate a as x times 2.

Calculate a as x plus y.

if (x < y)

a = x * 2;

else

a = x + y;

Flowchart C programming language



Example• Compound statement is way to group many

statements together so they are treated as one

if (grade >= 60)

printf( "Passed.\n" ); // { printf( "Passed.\n" ); }

else

{

printf( "Failed.\n" );

printf( "You must take this course again.\n" );

}



Decision Structure• The flowchart segment below shows a

decision structure with only one action toperform

if (x < y)

a = x * 2;


YESNOx < y?

Calculate a as x times 2.



Combining Structures

Display “x is within limits.”

Display “x is outside the limits.”

YESNOx > min?

x < max?

YESNO

Display “x is outside the limits.”

if (x > min){

if (x < max)printf("x is within the limits");

elseprintf("x is outside the limits");

}else

printf("x is outside the limits");



Exampleint k = 1, m = 4;

if (k < 2 || m == 3){

m = 2 + k, printf("%d", m);}else{

k = 1, printf("%d", k);}

int k = 1, m = 4;

if (k < 2 || m == 3){

m = 2 + k;printf("%d", m);

}else{

k = 1;printf("%d", k);

}

Which is more readable?



Exampleif(x)

if(y)printf("Yes");

else printf("No");

if(x){

if(y)printf("Yes");

else printf("No");

}

if(x){

if(y)printf("Yes");

}else

printf("No");

if (x < 0) sign = -1;

else if (x == 0) sign = 0;

else sign = 1;

if (x < 0.25) count1++;

else if (x >= 0.25 && x < 0.5) count2++;

else if (x >= 0.5 && x < 0.75) count3++;

else count4++;



Case Structure• One of several possible actions is taken,

depending on the contents of a variable



Case Structure• indicates actions to perform depending on the

value in years_employed

CASEyears_employed

1 2 3 Other

bonus = 100 bonus = 200 bonus = 400 bonus = 800

If years_employed = 1, bonus is set to 100



If years_employed is any other value, bonus is set to 800



switch• A switch statement allows a single variable (integer

or char) to be compared with several possibleconstants

– A constant can not appear more than once, and there can only be one default expression



switchswitch (variable)

{

case const:

statements...;

default:

statements...;

}

switch (c = toupper(getch())){

case ‘R’: printf("Red");break;

case ‘G’:printf("Green");break;

default:printf("other");

}



Exampleswitch(betty) {

case 1: printf("betty = 1\n");

case 2: printf("betty=2\n"); break;

case 3: printf("betty=3\n"); break;

default: printf("Not sure\n");

}

CASEbetty?

1 2 3 Other

betty = 1 betty = 2 betty = 3 Not sure



Repetition Structure• A loop tests a condition, and if the condition

exists, it performs an action. Then it tests thecondition again. If the condition still exists, theaction is repeated. This continues until thecondition no longer exists

x < y? Process AYES



Repetition Structure• The flowchart segment below shows a

repetition structure expressed in C as a whileloop

while (x < y)

x++;


x < y? Add 1 to xYES



Whilewhile (loop_repetition_condition)

statement;

OR

//Compound statementwhile (loop_repetition_condition) {

statement1;statement2; // …

}



Controlling a Repetition Structure• The action performed by a repetition structure

must eventually cause the loop to terminate.Otherwise, an infinite loop is created

• In this flowchart segment, x is never changed. Once the loop starts, it will never end

• How can this flowchart be modified so it is no longer an infinite loop?

x < y? Display x

YES



Controlling a Repetition Structure

• Adding an action within the repetition thatchanges the value of x

x < y? Display x Add 1 to x

YES



A Pre-Test Repetition Structure• This type of structure is known as a pre-test

repetition structure. The condition is tested BEFORE any actions are performed

– if the condition does not exist, the loop will never begin

x < y? Display x Add 1 to x

YES



Example

int counter = 0;while (counter < 9) {

printf("%d\n", counter ++);}

int counter = 0;while (counter < 1000) ;

while (1);

int counter = 0;while (counter < 9) {

printf("%d\n", counter);counter++;

}

int counter = 0;while (counter < 9)

printf("%d\n", counter ++);

int counter = 9;while (counter > 0)

printf("%d\n", counter --);



A Post-Test Repetition Structure• The condition is tested AFTER the actions are

performed

– A post-test repetition structure always performs its actions at least once

Display x

Add 1 to x

YESx < y?

do

{

printf(…);

x++;

} while (x < y);

C programming language



do-whiledo

statement;while (loop_repetition_condition)

OR

do //Compound statement {

statement1;statement2; // …

}while (loop_repetition_condition)



Example

Input : 5 integern1, n2, n3, n4, n5

Output: The summation of n1, n2, .., n5

Input example: 2 3 4 5 6

Output example: 20

Can you

identify the

input and

output?

Draw a flowchart for the

following problem:

Read 5 integer and display the

value of their summation.



Input n1

Input n2

Input n3

input n4

input n5

output sum

sum ← n1+n2+n3+n4+n5

start

2n1

Assume input example:2 3 4 5 6

3n2

4n3

5n4

6n5

20sum

end

This flowchart does

not use loop, hence

we need to use 6

different variables



counter ← 1, sum ← 0

counter < 6

sum ← sum + n

false

true

counter++

output sum

input n

1counter

sum 0

1 < 6 true

2n

0 + 22

2

2 < 6 true

3

2 + 35

3

3 < 6 true

4

5 + 49

4

4 < 6 true

5

9 + 514

5

5 < 6 true

6

14 + 620

6

6 < 6 false

Assume input example:2 3 4 5 6

This loop is

counter-controlledThe counter

Increases by 1

Uses only

3 variables



C Programming Language

Decreasing Counter-Controlled Loop


counter > 0

sum←sum+ x

false

true

counter--

output sum

input x



For

for (initial_value ; condition; update_counter)

statement;

OR

// Compound statement

for (initial_value ; condition; update_counter) {

statement;

statement; // …

}



int x, sum, i;

sum = 0;

for (i = 0; i < 5; i++) {

scanf(“%d”,&x);

sum = sum + x;

}printf(“%d”,sum);


counter < 6

sum ← sum + n

false

true

counter++

output sum

input n



Example:

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

printf(“%d\t”, num);

printf(“have come to exit\n”);

num

???

_



Example:

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



num

1

_



num

1

_

Example:

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





num

1

1 _

Example:

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





Example:

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



num

2

1 _



Example:

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



num

2

1 2 _



Example:

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



num

3

1 2 _



Example:

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



num

3

1 2 3 _



Example:

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



num

4

1 2 3 _



Example:

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



num

4

1 2 3 have come to exit_



Example*********************

#include <stdio.h>

int main(){

int iCounter = 0;int jCounter = 0;

while (iCounter < 7){

jCounter = 0;while (jCounter < iCounter){

printf("*");jCounter++;

}printf("\n");iCounter++;

}

getch();return 0;

}

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