Core java day1

Origin of Java
• World War-2 need of a platform independent
language
• Green by Sun Micro Systems comes up
• Fails due to marketing issues et al
• Analog-to-Digital Transit, Computer H/W
advancement, Concept of World Wide Web and
Internet, Need of security
• Green transformed to Oak
• 1994 Java makes official release
sohamsengupta@yahoo.com1

Three Editions of Java
• J2SE (Java 2 Standard Edititon)
• J2EE (Java 2 Enterprise Edition)
• J2ME (Java 2 Micro Edition)

Java Data Types
Numeric: 1. Whole Numbers
2. Fractions
Whole Numbers: byte 1 byte default value 0
short2 bytes default value 0
int  4 bytes default value 0
long 8 bytes default value 0
Fractions: float 4 bytes default value 0.0f
double 8 bytes default value 0 .0
Symbolic: char  2 bytes (Sun Unicode Character) default
value‘u0000’
Logical : boolean 1 byte default value false
User Defined: class and interface (to be described later on)

Your First Java Program
• In C consider the code snippet…
#include<stdio.h>
void main(){
printf(“Hello From Soham”);
}•
The corresponding Java Code would be…
class A
{
public static void main(String[] args){
System.out.print(“Hello from Soham”);
}}

Before I say more on Java some Do’s
Soham was writing a program as on LHS. After typing out 1000 lines he
felt he needed an extra statement in if branch, but poor Soham! What he
did was
• if(condition)
statement-1;
statement-2;
.
.
.
Statement-1000;
• if(condition)
statement-1;
ssttaatteemmeenntt--11AA;;
statement-2;
.
.
.
Statement-1000

Poor Me! I intended something else
I was supposed to type…
• if(condition){
statement-1;
ssttaatteemmeenntt--11AA;;
}}
statement-2;
.
.
.
statement-1000
• Moral:
1. When you come across some
if-else branch, or for, while,
switch, or any method or
block, at once type out the
braces and then carry on with
your code
2. It’ll not only save you from
my condition, but also it’ll
save you quite a lot of time of
compilation errors saying…
“ } required”

More on Java
Answers to some FAQ about Java 2
• A java file is saved in .java extension
• A java file, if successfully compiled, produces x+y number of .class
files where x and y are the number of classes and interfaces in that
file
• In java, you can’t put anything except comments outside a class or
interface block
• To run java program you need Java Virtual Machine(JVM) which
comes as a part of JDK.
• Successful compilation generates .class files which contain byte
codes that is interpreted by JVM. So, java development generally
uses both compiler and interpreter.
• Java Byte codes are platform independent but JVM is different for
different Operating Systems.

Features of Java 2 (J2SE 1.4)
1. Platform independent…means runs on any HW+OS environment
2. Java was developed using C++ but excludes the features like
pointers.
3. Java is a purely typed language as it does not allow automatic type
demotion (More on this later on)
4. Java is object oriented, secure due to its various security features,
reusable and portable,form-free and case sensitive and supports
general logical statements like C/C++
5. Java is not only platform friendly but also is very much developer
friendly. As support of this statement, Java does not have concept of
garbage value. It will never let the program perform read operation
on a non initialized local variable .Also, it saves you from getting
tampered data due to overflow and/or underflow since it doesn’t
allow automatic type demotion

Java Source File Name & Class Name
1. A java source file may contain any number of classes and interfaces
and there is no such hard-and-first rule that file name and class/
interface name should have relation…. But wait friends, this is
applicable as long as the file contains no public class/interface.
More concisely, a java source file must have the same name as that
of the public class/interface in it. It’s thus implied that a java source
file can contain one and only one public class/interface
2. To compile a Java file, say, A.java you have to give the command
…>javac A.java
3. If successful compilation occurs, you will surely want to run it. And
your command is going to be …> java MyClass where MyClass is
the class inside A.java that has then main method.
4. Remember, MyClass need not always have the main method.

My Second Java Program
#include<stdio.h>
void main()
{
int x=940;
printf(“U scored good marksn”);
printf(“Your marks is %d”,x);
}
class A
{
public static void main(String[] ar)
{
int x=940;
System.out.println(“U scored good
marks”);
System.out.print(“Your marks is ”+x);
}
}

Printing an output on the console
• We generally use the syntax System.out.print() or System.out.println()
to output text on the console. Difference between them is that println()
can be used with no arguments where as print() can’t be used without
an argument. Also, println() automatically appends a trailing new line
character (‘n’)
• Don’t ask me more about System.out because my plan is to climb up
the Java tree step by step and I don’t want to stumble down the stairs.
Yet, FYI, System is class under java.lang package and out is an static
object belonging to System class and of type java.io.PrintStream.
• Note the syntax: System.out.print(“Your marks is ”+x);
• Here + is concatenation operator instead and apart
from being the traditional addition operator.

Dual Nature of + operator
Look at the code snippet below and see the outputs
int x=9; int y=6;
15
System.out.println(x+y);
int x=9; int y=6;
System.out.println(“Sum is ”+x+y);
Sum is 96
int x=9; int y=6;
System.out.println(“Sum is ”+(x+y));
Sum is 15
int x=9; int y=6;
System.out.println(“Product is ”+x*y);
Product is 54
int x=9; int y=6;
System.out.println(x+y+ “ is the sum”);
15 is the sum
int x=9; int y=6;
System.out.println(x-y+ “ is difference”);
3 is difference
int x=9; int y=6;
System.out.println(“Difference is”+x-y);
Compilation error:
Operator – cannot be applied to
java.lang.String,int

From the Experts’ Desk
1. Since you can’t always afford to remember all the precedence rules and
this sort of things, the Java Guru recommends that you should always
use parentheses while using arithmetic expressions in the simplest way
2. As you are mostly accustomed to C/C++, first you’ll ask for a
counterpart of scanf() and cin>>. Yes. You can take input from console
through keyboard. But as I’ve told you, wait till I make you climb to
that level! Java, unlike C/C++ is not meant to be used as a mere
programming language with console as you did with Turbo C++,
generating Pascal triangles or Fibonacci Series et al.
3. Today java is more of a technology than of a language itself. Java can
carry out robust networking, enterprise web development to small
mobile device programming.
4. Note that Java handles all inputs as String, unlike C then tries to
convert to the intended type.

Java As a Typed language: code snippet
byte x=9;
System.out.println(x);
Here output will be 9 It’s OK to assign a value to
a type within range
byte x=129;
Error: Possible loss of
precision: found int ,
required byte
Value beyond range. Range
of byte –128 to +127
byte x=(byte)129;
Output: -127 (data
tampered due to overflow)
Explicit type casting may
cost you to worry later
int y=5; byte x=y;
precision: found int ,
required byte
Though value is in range,
type int can’t be
automatically demoted.
int y=5; byte x=(byte)y;
OK. Output: 5 Explicit type demotion ok
here but can cause OF/UF
long y=8; int x=y;
Guess yourself Automatica type demotion
not allowed so error.
float x=9.8; // Error
Should be float x=9.8f;
Error: possible loss of
precision: found double
required float
9.8 is double and 9.8f/9.8F
is float. This is because java
is memory efficient

Java A Typed Language: Contd.
1. boolean is the only data type that can’t be converted to
any type nor any other type be converted to boolean.
2. Implicit type demotion is not allowed in java. You have
to do it explicitly. But before that make sure that it
causes no overflow and/or underflow or both.
3. Although char is 2 bytes and so is short, yet they are not
compatible. char is automatically converted to int or
higher. For hierarchy consult any standard book.
4. char is unsigned strictly in java and note the following.
char ch=-90; // Error
char ch=90; System.out.println(ch); // output : Z
char ch=90; System.out.println((int)ch);// output 90

Operators: Unary & Binary
1. Rule for Unary Operator:
“If the operand is of a type which is below int in the hierarchy, the
output is converted to int, else the type of the output is the type of
the operand.”
2. Rule for Binary Operator:
“ If the Binary operator takes 2 operators one of type T1 and the
other of type T2, and max(T1,T2) is less than or equal to int, the
output is converted to int itself, else the output is of type
max(T1,T2)”
This rule is not applicable to increment and decrement operators
Hold your breath dears! Lots of surprises await you in the next
slide.

Look at the code snippets below
byte x=9; x=-x;
precision found: int
required: byte
Rule-1: inputbyte
Output int
byte x=9; byte y=x+1;
Same Error Rule:2 i/p byte, int
O/p int
byte x=7; byte y=2;
short z=x+y;
Same Error! Apply Rule-2
long x=89; byte y=8;
int z=x+y;
Error! Apply Rule-2. Output is
of type long
byte x=9; x=x+8; Error Apply Rule-2: Output is
of type int
byte x=9; x+=8; x++; OK Rule-2 Not applicable
for += and ++ operator
char ch=‘%’; int x=100
System.out.println(x+ch
+ “ pure am I”);
Output 137 pure am I Be careful. To avoid
this, use parentheses or
String ch=“%”;

Note the Following
• byte x=‘c’; // is OK
• int x=12;
byte y=x; // Error
• final int x=12; byte y=x; // IS OK
• final int x=134; byte y=x; // Error out of range
• This holds only when source data is less than or
equal to int.
• Adding the keyword “final” before a variable
makes it constant. It can’t be changed and any
code to change this will result in compilation error

Note the following code snippets
public static void main(String[] args){
int x;
x++;
}
Error: variable
x might not
have been
initialized.
In Java there is no concept
of garbage value. Without
initializing a local variable
u can’t perform read
operation on it.
int x;
if(6>4){x=8; }
OK. Output will
be 8
6>4 is evaluated at compile
time. So, since it’s true
always, x must be
initialized.
int x;
if(6<4){ x=8;}
Error:variable
x might not
have been
initialized.
6<4 is evaluated at compile
time. So, since it’s false
always, x must not be
initialized.
Replace numerics by variables say, a &
b. Observe the result, it’s error
if(a>b){ x=8;} System.out.print(x);
But making them final will be ok if a>b
is true
Error:variable
x might not
have been
initialized.
a>b or a<b is not evaluated
during compilation time.
So, it’s not sure if x is
initialized or not. Hence
Error

Core java day1

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