Java notes 1 - operators control-flow

Java - Basics
Prepared by
Mohammed Sikander
Technical Lead
Cranes Varsity

Language Basics
Variables
Operators
Control-flow
Arrays

Converting Ideas to Program
mohammed.sikander@cranessoftware.
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C++ Compilation Process
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JAVA Compilation Process
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First Java Program
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 The class can be public [not compulsory]
 main should be public [compulsory]
 main should be static [compulsory]
 return type of main should be void [compulsory]
 Argument of String [] is compulsory.
 only the order of public and static can be
swapped.
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 If class is public, filename should
match class name.
 If class is not public, filename can be
different.
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 javac
FirstProgram.java
 Java
SecondProgram
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Naming Conventions
Name Convention
class Names hould start with uppercase letter and be a
noun
e.g. String, Color, Button, System, Thread etc.
interface Name should start with uppercase letter and be an
adjective e.g. Runnable, Remote, ActionListener etc.
method Name should start with lowercase letter and be a verb
e.g. actionPerformed(), main(), print(), println() etc.
variable Name should start with lowercase letter e.g.
firstName, orderNumber etc.
package Name should be in lowercase letter e.g. java, lang,
sql, util.
Constants Name should be in uppercase letter. e.g. RED,
YELLOW, MAX_PRIORITY etc.
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CamelCase in java naming
conventions
 Java follows camelcase syntax for
naming the class, interface, method
and variable.
 If name is combined with two words,
second word will start with uppercase
letter always
 actionPerformed( ),
 firstName,
 ActionEvent,
 ActionListener etc.
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Variables naming - Rules
 All variable names must begin with a
letter,
an underscore ( _ ), or a dollar sign
($).
 [a-z/_/$] [a-z/_/$/0-9]
 Case sensitive – similar to c/c++
Samples of acceptable
variable names: YES
Samples of unacceptable
variable names: NO
Grade Grade(Test)
GradeOnTest GradeTest#1
Grade_On_Test 3rd_Test_Grade
GradeTest Grade Test (has a space)
int $a = 5;
int a$b = 8;
int _x = 4;
int _ = 9;
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Variable naming – Guidelines
 The convention is to always use a
letter of the alphabet. The dollar sign
and the underscore are discouraged.
 Variable should begin with lowercase,
multiple word variable can use
camelCasing.
 All characters in uppercase are
reserved for final(constants)
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Datatypes
 Byte - 1 byte
 Short - 2 byte
 Int - 4
 Long - 8
 Float - 4 (IEEE 754)
 Double - 8 (IEEE 754)
 Char - 2 (unicode)
 Boolean - 1 (size undefined)
 Note : java does not support unsigned.
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LITERALS
Integer literals
 Decimal [1-9][0-9]
 Octal 0[0-7]
 Hexadecimal 0x[0-9,A-F]
 Binary (JAVA SE7) 0b[0-1]
 Underscore can appear in literals (JAVA SE7)
 long creditCardNum =
1234_5678_9012_3456L;
Floating-point literals
 Double 4.5 or 4.5d 1.23e2
 Float 4.5f 1.23e2f
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Write the Output
class ArithmeticOperation {
public static void main(String [] args )
{
byte a = 5;
byte b = 10;
byte sum = a + b;
System.out.println("Sum = " +
sum);
}
}mohammed.sikander@cranessoftware.
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int sum = a + b;
byte sum = (byte)(a +
b);

Write the Output
Java Program
public static void main(String [] args ) {
float res = 9 % 2.5f;
System.out.println("res = " + res);
}
}
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C/ C++ Program
int main( )
{
float res = 9 % 2.5f;
printf(" %f " ,res);
}

Arithmetic Operators
 int / int => int
 float / int => float
 byte + byte => int
 % operator can be applied to float-
types
 Eg : 5.0 % 2.4 = 0.2
 9 % 2.5 => 1.5; 10 % 2.5 => 0
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Output
public static void main(String [] args ) {
byte a = 5;
byte b = 10;
int sum = a + b;
byte diff = (byte)(a – b);
System.out.println("Sum = " + sum);
System.out.println(“diff = " + diff);
}
}
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Conversion from Small to
Large
public class Conversion {
public static void main(String args[]) {
byte b = 12;
int i = 300;
double d = 323.142;
System.out.println("nConversion of byte to int.");
i = b;
System.out.println("i and b " + i + " " + b);
System.out.println("nConversion of int to double.");
d = i;
System.out.println("d and i " + d + " " + i);
}
}
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Conversion from Large to
Small
public class Conversion {
public static void main(String args[]) {
byte b;
int i = 300;
double d = 323.142;
System.out.println("nConversion of int to byte.");
b = (byte) i;
System.out.println("i and b " + i + " " + b);
System.out.println("nConversion of double to int.");
i = (int) d;
System.out.println("d and i " + d + " " + i);
System.out.println("nConversion of double to byte.");
b = (byte) d;
System.out.println("d and b " + d + " " + b);
}
}
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Write the Output
public class ByteDemo {
public static void main(String [] args)
{
byte x = 127;
System.out.println(x);
x++;
System.out.println(x);
}
}
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Code to Print ASCII / UNICODE
Value of a Character
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public class CharAsciiValue {
public static void main(String[] args)
{
char a = 'A';
System.out.println(a + " " + (int)a);
}
}

Question
 Consider the statement
 double ans = 10.0+2.0/3.0−2.0∗2.0;
 Rewrite this statement, inserting
parentheses to ensure that ans = 11.0
upon evaluation of this statement.
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What should be the datatype of
res variable?
public static void main( String [] args)
{
int a = 5;
int b = 10;
____ res = a > b;
System.out.println(res);
}
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int main( )
{
int a = 5;
if(a)
printf(“True");
else
printf(“False”);
}
public static void main(String []
args)
{
int a = 5;
if(a)
System.out.print("TRUE
");
else
System.out.print(“FALS
E");
}
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Relational Operators
 Result of Relational operators is of type boolean.
 int x = 5 > 2; //In valid – cannot assign boolean to int.
 boolean flag = false;
 if(flag)
 { }
 if(flag = true)
 { }

 int x = 5;
 if(x = 0)
 { }
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Logical Operators
 && , || - Short-circuiting operators, Same as
C
 & , | - evaluates both condition.
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if(++x > 5 & ++ y > 5)
{
}
Both x and y are updated
if(++x > 5 && ++ y > 5)
{
}
Only x is updated.

Write the Output
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Try with following Changes
1. Change the value of x to 8.
2. Replace && with &
3. Try with OR Operators (| | and | )

Bitwise Operators
 Can be applied on integral types only
 & , | , ^ , ~, >> , << , >>>
 >>> (unsigned right shift)
 If left and right are boolean, it is
relational.
 (a > b & a > c)
 If left and right are integral, it is
bitwise.
 (a & c)
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 To print in Binary Form
 Integer.toBinaryString(int)
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Output:
X = 23
10111

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Control Flow - if
 If (boolean)
◦ Condition should be of boolean
if( x ) invalid is not same as if(x != 0 ) (valid)
{ {
} }
boolean x = false;
if( x ) if(x = true )
{ {
} }
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Control Flow - switch
 The switch statement is multiway branch
statement. It provides an easy way to
dispatch execution to different parts of your
code based on the value of an expression.
 It often provides a better alternative than a
large series of if-else-if statements
 Case label can be of
◦ Byte, short, char, int
◦ Enum , String (SE 7 & later)
◦ Character , Byte, Short,Integer(Wrapper class)
◦ Break is not a must after every case (similar to
C).
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Control Flow - switch
public static void main(String [] args)
{
byte b = 1;
switch(b)
{
case 1 : System.out.println("Case 1");
case 2 :System.out.println("Case 2");
}
}
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Iteration Statements
 While
 Do-while
 For
 Enhanced for loop for accessing array
elements
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Enhanced for Loop
 It was mainly designed for iteration
through collections and arrays. It can
be used to make your loops more
compact and easy to read.
 int [ ] numbers = {5 , 8 , 2, 6,1};
 For(int x : numbers)
◦ S.o.println(x);
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Enhanced For with
Multidimensional Arrays
int sum = 0;
int nums[][] = new int[3][5];
for(int i = 0; i < 3; i++)
for(int j=0; j < 5; j++)
nums[i][j] = (i+1)*(j+1);
// use for-each for to display and sum the values
for(int x[ ] : nums) {
for(int y : x) {
System.out.println("Value is: " + y);
sum += y;
}
}
System.out.println("Summation: " + sum);
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 Method 2: No. of columns are same in all rows
 Datatype [ ][ ] matrix;
 Matrix = new datatype[row][col];
 Method 3 : Each row can have diff. no of
elements(cols)
 Datatype [ ][ ] matrix;
 Matrix = new datatype[row][ ];
 For(I = 0 ; I < row ; i++)
◦ Matrix[i] = new datatype[col]
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break
 Unlabelled break & continue – similar to c/c++
 Labelled break
search:
for (i = 0; i < arrayOfInts.length; i++) {
for (j = 0; j < arrayOfInts[i].length; j++) {
if (arrayOfInts[i][j] == searchfor) {
foundIt = true;
break search;
}
}
} mohammed.sikander@cranessoftware.
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Kangaroo
https://www.hackerrank.com/challenges/kangaroo
 There are two kangaroos on an x-axis ready to jump in the
positive direction (i.e, toward positive infinity). The first
kangaroo starts at location x1 and moves at a rate
of v1 meters per jump. The second kangaroo starts at
location x2 and moves at a rate of v2 meters per jump.
Given the starting locations and movement rates for each
kangaroo, can you determine if they'll ever land at the
same location at the same time?
 Input Format
 A single line of four space-separated integers denoting the
respective values of x1, v1, x2, and v2.
 Output Format
 Print No if they cannot land on the same location at the same
time.
 Print YES if they can land on the same location at the same
time; also print the location where they will land at same time.
 Note: The two kangaroos must land at the same location after making the same
number of jumps.
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