procedural programming & fundamentals of programming · lab exercises • distributed at the...

25.04.2018

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Procedural Programming &Fundamentals of ProgrammingExercise 1 (SS 2018) ￭ 24.04.2018

Procedural Programming & Fundamentals of Programming

Principles of the exercises

ti.uni‐due.de Joachim Zumbrägel

• We have 4 exercise dates

• We discuss topics of the lecture

• We discuss exercises and their solutions

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Principles of the practical trainings


• A lab takes 90 minutes

• Each of the eight lab consists of

– homework tasks in the handout, checked during the lab

– test before the lab (PP only)

– the practical training itself


Schedule


Date Lecture(3h)

Exercise(3h)

Practical Training for PP: Tu., We. (2h)Practical Training for FOP: Mo., Tu., We. (2h)

10.04.18L1

17.04.18L2

24.04.18 E1

08.05.18L3 P1 Variables, Data Types, Statements, Hello World

15.05.18 E2 P2 Boolean expressions, Conditional Executions,

29.05.18 E3 P3 Finite and Infinite Loops, Break and Continue

05.06.18L4 P4 Functions: Parameters and Arguments, Void

12.06.18 E4 P5 Arrays

19.06.18 P6 Addressing: Direct and Indirect, Pointers

26.06.18 P7 Strings and Files

03.07.18 P8 Math Functions, Linked Lists, and Binary Trees

10.07.18 NT Repetition Date for P1‐P8, used for missed labs

17.07.18

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Practical training time and location


Procedural Programming• Tuesday 12:00-13:30, 14:00-15:30

• Wednesday 15:45-17:15

Fundamentals of Programming • Monday 08:15-09:45, 10:00-11:30, 17:30-19:00

• Tuesday 08:00-12:00

• Wednesday 17:30-19:00

Rooms: BA 028, BB 219


Lab programming environment

Restrictions:• Simple as possible (syntax high lighting)• Available under Linux and Microsoft

Result:• Compiler

gcc

• Editor Jedit


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Programming environment for you

Under linuxGcc: sudo apt-get install gccJedit: sudo apt-get install jedit

Under WindowsGcc: Download and install mingw (http://mingw.org)

Extend the Environment variable „PATH“ with the path where gcc is installed

Jedit: Download and install jedit (http://jedit.org)

That‘s all



Lab tests

• Duration 5 minutes• Multiple Choice• 5 Questions (about the current lab respectively

about the corresponding handout)• Passed with 2 right answers• Example: What are the right symbols to

enclose a block in C? a) (); b) [ ]; c) {}; d) “ “;


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Lab exercises

• Distributed at the beginning of the practical part of the lab

• Some simple tasks• Example (first lab):

Write a program asking the user to input two integer numbers and then calculates the sum of those two numbers.



What will I learn in the 1. Seminar/Exercise


• Exercises– Flowchart, – Pseudocode– C-Code

• C-Programming– Introduction from source to executable file– C-Basics (main und return)– Operators– Basic data types– Inputs/Outputs– If and switch

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Flow chart symbols



Welcome


Start Excercise

Say „hello and quiet please“

If to loud

Continue with excercise ….

Show KISS

Hear Response

Stop Excercise

Check loudness

no

yes(unfortunately)

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Kiss principle


• Keep it simple [and] stupid• Keep it simple and short• Keep it simple and smart• Keep it simple and straightforward• Keep it simple and small• Keep it simple and safe


Solving Strategies

• Restate the problem• Divide the problem• Start with what you know• Reduce the problem• Look for analogies• Experiment• Don’t get frustrated


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Excercise 1 (Task)

The following Ripple-Carry-Adder is constructed using a cascade of multiple full adders (FA). The Ripple-Carry-Adder should be operated on 20 MHZinput frequency, in which all input signals are changing at the same time. Each full adder has a propagation delay of 2ns.

Considering the limitation of input frequency and propagation delay, how many full adders in the cascade can produce the correct outputs? (Hint: Correct output means that the sum of the input vectors (a,b) should at least be delivered correctly at the output before the input changes next time).



Exercise 1 (Questions/Task)

1. What do I know and what do I have to know to solve the problem?

2. Calculate the max. number of full adders (n) for the given circuit

3. Create now a program which generalizes the approach so that we can calculate n for a given input frequency and a given propagation delay– Build up the formula to solve this problem – Think about INPUT and OUTPUT

(measurement unit, dimension) – Draw a flow chart– Write the program


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Program for exercise 1

/* PROG: NrOfAdders

CREATED:24.04.2018 by JOACHIM ZUMBRÄGEL */

#include <stdio.h>

int main()

{

float frequency, propagationDelay, nrOfSubCircuits;

printf("Frequency(in MHZ):\n");

scanf("%f", &frequency);

printf("Propagation delay (in ns):\n");

scanf("%f", &propagationDelay);

nrOfSubCircuits = 1000/frequency/propagationDelay;

printf("\nNr of Sub-Circuits <= %.0f",nrOfSubCircuits);

return 0;

}



From program to executable


source file<filename>.c

Other object file(libraries)

object file(machine code)

Header files<headerfile>.h

executable file<filename>.exe

editor

compiler

linker

preprocessed file(text replacement)

preprocessor

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Preprocessor

# indicates that the so called preprocessor comes into a play.

#include<stdlib.h> //includes the stdlib.h

#define MAX_ADDERS 10 //Definition of

after preprocessing ⇒the term „#include<stdlib.h>“ is replaced with

the contents of the file <stdlib.h> and ⇒the term MAX_ADDERS is replaced with 10.



Main function and its return value

Every C-Programm has its main function:int main(){::return <integervalue>;

}

The returned integer value is catched by the operating system and can be used for further processing. 0 stands for success.


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Comments

/* */ to enclose more than one line/* AUTHOR: J. Zumbrägel

CREATED: 01.05.2017

LAST UPDATED: 02.05.2017*/

// if you want to comment one line// Use comments to explain



Simple data types

• Four basic arithmetic type identifiers in C: char, int (Integer types)float, double (floating point types)

• There are optional specifiers:signed, unsigned, short, long

• The actual size of integer types varies by implementation, but rules are existing like:char<=short<=int<=long<=long long

• Size can be evaluated by sizeof(<data type>)


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Variables

• Before variables can be used in a program, all variables must be defined:- int x;- unsigned short int // or unsigned short- float y;

• Variables can and should be initialized:- unsigned long int ulCounter=0;

• You can define “constant” variables which will be remain unchanged after initialization.- const int maxNrOfItems = 10;



Input and Output

• Inputscanf(format string, address of variable(s))int age;scanf(„%d“,&age) //reads in age„%d“ is the format string&age is the address of the variable age

• Outputprintf (format string,variable(s))int age;printf(„You are %d“,age) //prints age=> You are 20


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Format string

Format strings allow the usage of data type placeholders in strings. The placeholder will be later on replaced with values of variables:

- “%d” // Value of type integer- “%f” // Value of type float- “%c” // Value of type Character

Hint: A % in a string always points to the start of a format specifier. If you want to output the sign % you have to write %% in the format string.



Escape sequences

Escape sequences, otherwise known as escape codes, are special type of characters that are used inside the string literals. Examples:\n => newline\t => tab\” = double quote (“)• printf(“Hello, \“interested students\“!“);=> Hello, “interested students“!

• printf(“Hello,\nare you fine?“)=> Hello,

are you fine?


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Operators (not complete)

C supports a lot of operators. They are symbols used within an expression to specify the manipulations to be performed while evaluating that expression. C operators are:• arithmetic: +, -, *, /, %• assignment: =• augmented assignment: +=, -=, *=, /=,%=

a+=value; //equivalent to a=a+value;

To be continued:



Programming style

• Use spaces in source code (easier to read)• Use BlockCodes• Descriptive names for variables and use

CamelCase. Example: (int nrOfAdders)• Initialize your variables• Use constant or defines for fixed values• Use comments • Avoid long lines and complex statements


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Exercise 2

• Write a program which converts a decimal number (0-9) into a nibble (half byte) .The output should be printed in the form:nibble = 0101Don‘t use if statements for the solution.Hint: make use of the operators “/” and “%”

Example : int n; n = 9;n/2 => 4n%2 => 1



Relational Operators

Operator Meaning

> Greater than

< Less than

>= Greater than or equal to

<= Less than or equal to

== Equal to

!= Not equal to


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Relational Expression

• Consists of operands and a relational operator between them.

• The result is a boolean value, represented as either true (1) or false (0).

• The result of a relational expression can beassigned to a variable.

Example: logResult=(x < y)

//logResult is set to 1(true) if x smaller y



Logical Operators

• The logical operators allow logical combination of relational expressions

• The && Operator realizes Logical AND• The || Operator realizes Logical OR• The ! Operator realizes Logical NOTExample: logResult=((x && y) || (!z))

// logResult is set to 1 (true) // if x AND y are non-zero OR if z is equal 0


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if - else statement

The if statement executes a code only when the condition is true. The else part is optional. If-else can be nested!if (condition){

statement(s)_1;//executed when condition true (1)

}

else

{statement(s)_2;//executed when condition false (0)

}



if vs. if-else


statement(s)

condition?

true

false

if

statement(s)_1

condition?

true

false

statement(s)_2

if-else

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The conditional expression

A conditional expression is a „short“ if else!(expression)?(execute if true):(execute if false)

Example:conResult=(x>y)? x : y;

//is equal to

if (x>y)conResult=x;

elseconResult=y;



Switch (1)

The switch statement allows the selection of statements from several available choices.switch (expression of type integer){

case (choice 1): // executed if expression == choice 1

case (choice 2): // executed if expression == choice 2

:case (choice n):

// executed if expression == choice n default:

// default execution }


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Switch (2)

Take care: When a case is true, all following statements are executed regardless of the case choices.

=> Use break to leave the switch.

switch (expression of type integer){

case (choice 1): statement(s); break; //leave the switch block

case (choice 2): :

}ti.uni‐due.de Joachim Zumbrägel


Exercise 3

Write a program which simulates the behaviourof an AND Gate and an OR Gate. The gates have two inputs. The program works as follows:

By entering a number the user can choose which gate he wants to simulate. 1 for the AND Gate and 2 if he wants two simulate an OR Gate. Then the user has to enter the two input values. After that the program shows the result of the simulation.


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Material

BOOK:• C - Einführung und professionelle Anwendung.

IT-Studienausgabe- Ulla Kirch Prinz, Peter Prinz- ISBN: 978-3826617669

WWW• http://de.wikibooks.org/wiki/C-Programmierung


Exercise 1 (Flow Chart)

Start

Input frequency (ín MHZ) Input propDelay (in ns)

n = 1000/f/pd

Output n

End

Exercise 1 (program)

/* PROG: NrOfAdders

CREATED:24.04.2018 by JOACHIM ZUMBRÄGEL */

#include <stdio.h>

int main()

{

float frequency; float propDelay, float nrOfSubCircuits; printf("Frequency(in MHZ):\n"); scanf("%f", &frequency); printf("Propagation delay (in ns):\n"); scanf("%f",&propagationDelay); nrOfSubCircuits = 1000/frequency/propDelay; printf("\nNr of Sub-Circuits <= %.0f", nrOfSubCircuits); return 0;

}

Exercise 2 /* Program converts a decimal value * from 1-9 into a nibble * AUTHOR: Joachim Zumbrägel * DATE CREATED:2018-04-24 */ #include <stdio.h> #include <stdlib.h> int main() { unsigned int n=0, b0=0, b1=0, b2=0, b3=0; printf("Convert n(0-9) into a nibble:\n”); printf(“Input n please:"); scanf("%u",&n); b0 = n%2; n = n/2; b1 = n%2; n = n/2; b2 = n%2; n = n/2; b3 = n%2; printf("Nibble is:%d%d%d%d\n", b3,b2,b1,b0); return EXIT_SUCCESS;

}

Exercise 3 #include <stdio.h> #define SIM_AND 1 #define SIM_OR 2 int main() { int select,a,b; printf("Select (1=AND), (2=OR)"); scanf("%d",&select); switch (select) { case SIM_AND: printf("Enter a (0 or 1):"); fflush(stdin); scanf("%d",&a); printf("Enter b (0 or 1):"); fflush(stdin); scanf("%d",&b); printf("result (AND):%d\n", (a && b)); break; case SIM_OR: printf("Enter a (0 or 1):"); fflush(stdin); scanf("%d",&a); printf("Enter b (0 or 1):"); fflush(stdin); scanf("%d",&b); printf("result (OR):%d\n", (a || b)); break; default: printf("Wrong input\n"); } return 0;

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