Two-week ISTE workshop on Effective teaching/learning of computer programming Dr Deepak B Phatak Subrao Nilekani Chair Professor Department of CSE, Kanwal

Two-week ISTE workshop onEffective teaching/learning of computer

programming

Dr Deepak B Phatak

Subrao Nilekani Chair Professor

Department of CSE, Kanwal Rekhi Building

IIT Bombay

Lecture 2, Fundamental programming concepts

(Simple programs)

Tuesday 29 June 2010

OVERVIEW Revisit ‘Dumbo’ model for a program Structure of a C program

• Explanation of some nuances Sample program

• Notion of variables and data types• Expression evaluation and assignment

Problems needing non-sequential execution of instructions Conditional execution of instructions

• If statement and its variants

The Dumbo Model of computing We saw the abilities of a computer

• a model represented by Dumbo. A ‘program’ contains instructions executed sequentially. An

instruction may

• Accept values as input, store these in designated memory locations

• A value may be computed by evaluating an expression, which is assigned to a location

• One or more values can be copied from memory locations and given to us as output

Computing …

Memory

• Comprises of ‘locations’, each of which can contain a single value

• Each location has a tag or a name. Such name is called a ‘variable’ as the value in that location can change or vary based on operations

• A location is never empty, it always contains some value, may be from some previous computing activity

More about a program (student’s thoughts)

When we discussed our sample programs (prog1.c and prog2.c), we noticed presence of several lines which we did not understand clearly. We will just identify these and take a note of the nuances we need to face as of now

When we have done some more programming and have become more familiar with the computers, we will learn the full meaning of these lines

We look at the problem of evaluating formula v = n1/n2 + x , for x=3, n1 = 2,

and n2 = 5 (I expect v to be 3.4)

Program to evaluate a formula

#include<stdio.h>#include<stdlib.h>// prog3.c// arithmetic operationsint main(){ int n1,n2; float x,v; n1 = 2; n2 = 5; x = 3; v = x + n1/n2; printf("value is %f\n", v); return 0;}

Meaning of different lines

#include<stdio.h> // prog3.c// arithmetic operationsint main(){ int n1,n2; float x,v; n1 = 2; n2 = 5; x = 3; v = x + n1/n2; printf("value is %f\n", v); return 0;}

Does not make any sense, perhaps asking the computer to include some pre-written instructions

Meaning of different lines …

#include<stdio.h> // prog3.c// arithmetic operationsint main(){ int n1,n2; float x,v; n1 = 2; n2 = 5; x = 3; v = x + n1/n2; printf("value is %f\n", v); return 0;}

These appear to be commentsmeant for human readers


#include<stdio.h>// prog3.c// arithmetic operationsint main(){ int n1,n2; float x,v; n1 = 2; n2 = 5; x = 3; v = x + n1/n2; printf("value is %f\n", v); return 0;}

Seems to be something importantbut cannot understand the meaning.Perhaps the line indicates thebeginning of our main program



Appears to define variables for the formula.I don’t understand the words int and float;int perhaps means integer, in which case,float should mean a real number



These are clearly the instructions written to solve the problem. However, the way the formula is written, the computer may first add n1 to x, and the result will be 1.0 which will be wrong



Looks like an output instructionsimilar to cout instruction. I am guessing this because of the word “print”, but why is it is written as ‘printf? Also, what do thefunny symbols inside parenthesis mean?



This is very confusing! Perhaps the line indicates returning Value 0 to some one. But to whom?And why a zero?

In a nutshell

When I execute the programs in the lab from files prog1.c, and prog2.c, these work correctly, collecting input values from me and returning the correct computed value as output• So all those funny lines must have some purpose which

I do not understand now However, I must understand the way my variables are

defined, and the way computations are carried out, so that I can write instructions to correctly implement the desired calculations

Our Computing Environment

The computer has evolved significantly. It can be instructed in a human like language,

A tool called ‘compiler’ can translate these instructions in its machine language

It can exchange with us, values in a form familiar to us (e.g. decimal numbers, fractions) and convert these to and from the internal binary representation

A special program called ‘operating system’ controls its operations and its interaction with us

Our Computing Environment ...

We typically interact with a computer through a keyboard, using which we type our instructions , and data values and through a monitor, which displays to us, any value output by our program

We interact with the operating system (OS), which is a set of prewritten programs to control the computers.

This interaction is either through text interface where the OS shows us a ‘command prompt’, or using a graphical window interface,

• We use Ubuntu, a variant of Linux OS

A session with the computing environment[dbp@localhost ~]$ pwd

/home/dbp

[dbp@localhost ~]$ cd cpp

[dbp@localhost ~]$ cc prog3.c

[dbp@localhost cpp]$ ls

prog1.c prog2.c prog3.c a.out

[dbp@localhost cpp]$ ./a.out

value is 3.000000

Handling Numerical values

We define variable names used in our programs. Such definition allocates a memory location to the chosen name, and also defines the ‘type’ of value which can be stored in that location

int N; : reserves one location in memory for N to store integer values.

Max positive number: + 231-1 Min negative number: - 231-1 How to store larger numbers? We use Long integers: long M; Max magnitude 263-1,

Handling Integer Numerical values … The computer actually uses ‘Binary’, or ‘Hexadecimal’

representation (Base 2 or 16) • Thus value limits are expressed in powers of 2 or of 16

For the time being, we will continue to believe that computer uses Decimal values

For integer values the base is immaterial Our Computer does arithmetic with constraints because of the limited capacity of a memory location What happens if two values are individually within the permissible range, but an arithmetic operation, such as multiplication, gives a larger result value?

Floating Point Representation

Real numbers (fractions) are stored differently, using a “floating point”

Two components of the number are stored in the same location in two separate compartments. One is called mantissa (m), and the other is called exponent (e)

The Number value is taken as m * B e

Computer uses B as 2 (Binary) or 16• We will take it to be our usual 10

In the program we write such values as 2.53, or -7.2E15 “float y;” reserves 1 location to store a number in this format.

Floating point …

No decimal point is actually stored! It is assumed to be at the beginning of mantissa

Can write as -.4127 E 11, or as -41.27 E 9 Mantissa can store about 7 decimal digits. Exponent can be between -100 and +100 double z; mantissa about 14 decimal digits

-4127

+19462 -21

11 -.4127 * 10 11

.19462 * 10 -21

Problems in Numerical Computations

Since computers offer limited precision, the computations carried out on given values often lead to inaccuracies, specially with floating point numbers

Special care needs to be taken to handle such ‘round off’ errors

There are many text books which show how these problems can be handled by using special techniques and algorithms in our programs• Numerical recipes in C (C++, Java)

Example of problems with floating Point arithmetic float w, y=1.5, avogadro = 6.023 E 23 ;

w = avogadro + y; What is the value of w? Exact value of w and avogadro will differ in the 23rd digit. “float” stores 7 digits of mantissa. Digits beyond are ignored. (roundoff error). So the value of w is same as avogadro (6.023 E 23)

Summary of our understanding so far

“God made natural numbers, the rest is the work of man.” -- Leopold Kronecker

“Humans made fractions and decimal numbers, the rest is the work of Dumbos”

-- Deepak Phatak

Mixed Arithmetic

computer converts from float to int and vice versa as needed.

In a program, a number written without decimal point is taken as integer, otherwise as float

int N=20; float y=15.8;

N=y; : 15.8 is rounded down. N=15.

360/N : integer result is calculated.

360.0/N : float result is calculated.

Remarks on Arithmetic

Arithmetic operators: *, /, %, +, -,

*, / have higher precedence than +, - *, / have equal precedence, evaluated

in left to right order. Similarly +, - x = m % n; % means modulo remainder for m=10, and n=4, x will have value 2. % same precedence as *, / () override default precedence.

Use of Comments

Programs are read not only by compilers, but also by programmers and other humans

You should include extra description (“comments”) to help other programmers figure out the logic:

Style 1: “// ......” to end of line Style 2: “/* .... */”

An important notion: Reassignment

int m

m =5;

…

m = 3*m + 1; // Reassignment

This is not an equation, as it would be Mathematically absurd, because it would simplify to: 0 = 2*m+1

Computer interpretation: Calculate the value of the expression on rhs, then store it in the variable on lhs.

More on assignment ...

• “increment” is a special form of reassignment count = count + 1; or count += 1;or, more simply, ++count; or count++;

• Input and Output instructions in C are not “natural”• Handled through special “functions”

Conditional execution

• Program instructions are normally executed in the given sequence

• We can examine conditions, and based on the result, can execute instructions out of sequence

if (condition)

{ statement group 1};

else

{ statement group 2};

Problem requiring conditional execution

• Based on the age of a passenger, the ticket cost for bus travel is different. Let us say, it is Rs 25.50 for an adult, and Rs 12.75 for a child (< 12 Years)

• Suppose we write our program as

int age, float ticketcost;

ticketcost = 12.75;

ticketcost = 25.50;

cout << ticketcost;

We will always get cost as 25.50

If we exchange the two assignments, then we will always get 12.75

Use of if statement

We need to execute either the first or the second instruction based on the value of age, but never both

Use of the if – else statements will permit us to correctly evaluate the ticket cost

... if (age > 12) {ticketcost = 25.50}; else {ticketcost = 12.75}; … What if elders are charged Rs 20?

If – else if ladder

…if (age > 60){ ticketcost = 20.00;}else if (age > 12){ ticketcost = 25.50;}else { ticketcost = 12.75;} …

Example to find Maximum of given numbers// program 4 (prog4.cpp), find maximum of 3 given numbers#include <iostream>using namespace std; int main() { int a, b, c, max; cout << “Give three numbers” << “\n”; cin >> a, b, c; max = a; if (b > max) {max = b}; if (c > max) {max = c}; cout <<“Maximum is”<< max<<"\n"; return 0;

}

Finding Maximum …

If there are more numbers, say 5, we could write: int a, b, c, d, e, max; cin >>a >>b >>c >>d >>e ; max = a; if (b > max) {max = b}; if (c > max) {max = c}; if (d > max) {max = d}; if (e > max) {max = e}; cout<<“Max is”<< max << "\n"; What do we write to find maximum of 200 numbers, or when

number of numbers itself varies every time a user runs the program?

Specifying comparison Conditions

“ a op b” where op is <, >, <=, >=, ==, !=

== : is equal to

!= : is not equal to

Conditions can be combined: • (a > 0) && (a <= 9) : true if a is a positive digit.

&& : conjunction. “and”

|| : disjunction. “or”

In the programming language C, evaluation of any condition

results in a value 1 if the condition is true, and 0 if it is false

Questions?

Question From NIT Kurukshetra

why do we use printf insted of print





Question From JEC_Kukas

How will convert any floating point number in to integer

number

Documents

Two-week ISTE workshop on Effective teaching/learning of computer programming Dr Deepak B Phatak Subrao Nilekani Chair Professor Department of CSE, Kanwal