Binary math in java?

Dear everyone,
A silly question here: How do I convert an integer to its binary form and do some operations such as OR, AND on it?
The only clue I have is Integer.toBinaryString(). Could you please clue me in?
Thanks a lot!
George

Integers are already in binary form. To do boolean arithmetic use &=AND |=OR ^=NOT and ~=XOR:
int a= 0xff;
int b;
int c;
b=a & 0x01;
c=a&b|31; etc.
By the way, I like 0xdeaddeaf. Ive seen several of these, like 0xcafebabe and 0xfeedc0de. Maybe we shouls compile a list!

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How to store binary data in Java

I need to show news in my web aplication.So I am retrieving news from a Feed Server in which the field GEN_UNICODE_MESSAGE_1 gives the news data.News consists of both English/Arabic data.So how will I store the binary content in JAVA and how can I display the news in JSP?
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BufferedInputStream bis=new BufferedInputStream(new ByteArrayInputStream((byte[])msg.getField("GEN_UNICODE_MESSAGE_1").value().toString().getBytes()));
BufferedReader br = new BufferedReader(new InputStreamReader(bis));What on earth is all this?
1. What is the type of msg.getField("GEN_UNICODE_MESSAGE_1").value()?
2. Whatever that type is, you are then converting it to a String.
3. You are then converting that to bytes.
4. You are then casting that to byte[], which it already is,
5. You are then wrapping a ByteArrayInputStream around that.
6. ... and an InputStreamReader round that ...
7. ... and a BufferedReader around that ...
8. ... and reading lines
9. ... and appending them to a StringBuilder
10. ... and converting that to a String.
Which you already at at step 2.
String incoming = msg.getField("GEN_UNICODE_MESSAGE_1").value().toString();Not sure whether that's correct given the charset issues, but if not the problem starts at step 2. In any case steps 3-10 add precisely nothing.
This is all pointless. The key question is (1). Answer that first.
2. The type of String.getButy
>
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String line = null;
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Binary operations in Java

Hi,
are there any binary operations in Java like '&' in C? I didn't found any information in the API description.
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Binary Tree in Java - ******URGENT********

HI,
i want to represent a binary tree in java. is there any way of doing that.
thanx
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HI,
i want to represent a binary tree in java. is there
e any way of doing that.
thanx
sraphsonFirst, what is a binary tree? Do you know how the binary tree looks like on puesdo code? How about a representation in terms of numbers? What is a tree? What is binary? The reason I ask is, what do you know about programming?
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Lulu wrote:
Hi there
I have several questions about binary trees
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treeMap.put("second","Orange");
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If this is a map how do I define if the data should go to the left or to the right of certain node?That is all done for you. In a TreeMap (using the no-args constructor), you can only store objects that are comparable to each other (so, they must implement the Comparable interface!). So the dirty work of deciding if the entry should be stored, or traversed, into the left or right subtree of a node, is all done behind the scenes.
Also note that TreeMap is not backed up by a binary tree, or a binary search tree, but by a red-black tree. A red-black tree is a self balancing binary tree structure.
Should I have dynamical keys so that they increase automatically when adding new data?
According to a webpage I should use Comparator(), is that to find the data in the tree and retrieve the key?
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This resurrected thread was first posted November 25, 1960 at 8:25AM
-------------------------------------------------------------------------

This resurrected thread was first posted November 25,
1960 at 8:25AMIn what time zone?

Binary search in java

Hi,
I keep getting a java.lang.ClassCastException with these two classes when I try to perform a binary search. Any tips?
Phonebook class:
=============
import java.util.*;
public class Phonebook
private static long comparisons = 0;
private static long exchanges = 0;
     public static void main (String[] args)
     // Create an array of Phonebook records
     PhoneRecord[] records = new PhoneRecord[10];
     records[0] = new PhoneRecord("Smith","Bob", 1234367);
     records[1] = new PhoneRecord("Jones","Will", 1234548);
     records[2] = new PhoneRecord("Johnson","Frank", 1234569);
     records[3] = new PhoneRecord("Mc John","Pete", 1234560);
     records[4] = new PhoneRecord("OBrien","Frank", 1234571);
     records[5] = new PhoneRecord("OConnor","Joe", 1234572);
     records[6] = new PhoneRecord("Bloggs","Ricky", 1233570);
     records[7] = new PhoneRecord("empty","empty", 8888888);
     records[8] = new PhoneRecord("empty","empty", 9999999);
     records[9] = new PhoneRecord("Van Vliet","Margreet", 1244570);
     // call menu
     Menu(records);
     } // end main
//================================================
// menu
//================================================
static void Menu(PhoneRecord[] records)
     int option;
     // menu options
     System.out.println("===========Menu==============================");
     System.out.println("=============================================");
     System.out.println(" ");
     System.out.println("1. Find record (Advanced Search) ");
     System.out.println("2. Quick Search ");
     System.out.println("3. Add a record ");
     System.out.println("4. Show database ");
     System.out.println("5. Sort database ");
     System.out.println("6. Exit     ");
     System.out.println(" ");
     System.out.println("=============================================");
     System.out.println("=============================================");
     System.out.println(" ");
     System.out.println("Choose a number ");
     option = Console.readInt();
     // every menu option has its own method
     if (option == 1)
          FindRecord(records);
     if (option == 2)
          QuickSearch(records);
     else if (option == 3)
          AddRecord(records);
     else if (option == 4)
          ShowDatabase(records);
     else if (option == 5)
          quickSort(records, 0, 9);
          ShowDatabase(records);
     // if 6 then terminate the program
     else
               System.out.println("Goodbye!");
} // end of menu
//=================================================
// menu option 1: Find a record - using linear search
//=================================================
static void FindRecord(PhoneRecord[] records)
int option;
do
// the user can search based on first name or last name
System.out.println("Do you want to search for:");
System.out.println("1. First Name");
System.out.println("2. Last Name");
System.out.println("3. End search");
option = Console.readInt();
     // option 1 is search based on first name
     if (option == 1)
          System.out.println("Enter First Name");
          String first = Console.readString();
          int notthere = -1;
          for (int i=0; i < 10; i++)
               if (first.equals(records.first_Name))
                    System.out.println("----------------------------------");
                    System.out.println(records[i].last_Name + ", " + records[i].first_Name);
                    System.out.println(records[i].phonenumber);
                    System.out.println("----------------------------------\n\n");
               // if a record is found, the variable notthere will be > -1
               notthere = i;
          } // end search array
               // if notthere is -1, then there is no record available
               if (notthere < 0)
               System.out.println("------------------------------");
               System.out.println("No record available");
               System.out.println("------------------------------\n\n");
     } // end option 1 First Name
     // option 2 allows the user to search based on last name
     else if (option == 2)
          System.out.println("Enter Last Name");
          String last = Console.readString();
          int notthere = -1;
          for (int i=0; i < 10; i++)
               if (last.equals(records[i].last_Name))
                    System.out.println("----------------------------------");
                    System.out.println(records[i].last_Name + ", " + records[i].first_Name);
                    System.out.println(records[i].phonenumber);
                    System.out.println("----------------------------------\n\n");
               notthere = i;
               // if notthere is -1 then there is no record available
               // if notthere is > -1 then there is a record available
               } // end search array
               if (notthere < 0)
               System.out.println("------------------------------");
               System.out.println("No record available");
               System.out.println("------------------------------\n\n");
     } // end option 2 Last Name
     else
          // if the user types in a wrong number, he or she returns to the menu
          Menu(records);
while (option != 3);
} // end FindRecord
//=================================================
// menu option 2: Quick Search - using binary search
//=================================================
static void QuickSearch(PhoneRecord[] records)
     // Sort array - Using Quicksort
     quickSort(records, 0, 9);
     // allow user to enter the last name
     System.out.println("Enter Last Name");
     String last = Console.readString();
     // use binary search to find the target
     int index = binarySearch(records, last);
     // -1 means that there are no records
     if (index == -1)
     System.out.println("------------------------------");
     System.out.println("No record available");
     System.out.println("------------------------------\n\n");
     // print out the record
     System.out.println("----------------------------------");
     System.out.println(records[index].last_Name + ", " + records[index].first_Name);
     System.out.println(records[index].phonenumber);
     System.out.println("----------------------------------\n\n");
     // return to menu
     Menu(records);
} // end QuickSearch
public static int binarySearch( Comparable [ ] a, Comparable x )
int low = 0;
int high = 9;
int mid;
while( low <= high )
mid = ( low + high ) / 2;
if( a[ mid ].compareTo( x ) < 0 )
low = mid + 1;
else if( a[ mid ].compareTo( x ) > 0 )
high = mid - 1;
else
return mid;
return -1; // not found
//=================================================
// menu option 3: Add a record
//=================================================
static void AddRecord(PhoneRecord[] records)
int option;
int index = 0;
// enter details
do
     // to say that the array is not full yet, I use the variable filled
     int filled = 0;
System.out.println("Enter the First Name");
String frst = Console.readString();
System.out.println("Enter the Last Name");
String lst = Console.readString();
System.out.println("Enter the phone number");
int phn = Console.readInt();
// search the array for the empty slot
     for (int i=0; i < 10; i++)
          if (records[i].first_Name.equals("empty") && filled == 0)
          records[i].first_Name = frst;
          records[i].last_Name = lst;
          records[i].phonenumber = phn;
          filled = 1;
     // Sort array - Using Quicksort
     quickSort(records, 0, 9);
     // Print out sorted values
     for(int i = 0; i < records.length; i++)
          System.out.println("----------------------------------");
          System.out.println(records[i].last_Name + ", " + records[i].first_Name);
          System.out.println(records[i].phonenumber);
          System.out.println("----------------------------------\n\n");
     System.out.println("Do you want to add more records?");
     System.out.println("1. Yes");
     System.out.println("2. No");
     option = Console.readInt();
     if (option == 2)
          Menu(records);
     // sets the database to full
     int empty = 0;
     for (int i=0; i < 10; i++)
          // empty = 1 means that there is an empty slot
          if (records[i].first_Name.equals("empty"))
               empty = 1;
     // if the system didn't find an empty slot, the database must be full
     if (empty == 0)
     System.out.println("Database is full");
     option = 2;
     Menu(records);
while (option != 2);
} // end AddRecord
//=================================================
// menu option 4: Show database
//=================================================
static void ShowDatabase(PhoneRecord[] records)
          // shows the entire database
          for (int i=0; i < 10; i++)
                    System.out.println("----------------------------------");
                    System.out.println(records[i].last_Name + ", " + records[i].first_Name);
                    System.out.println(records[i].phonenumber);
                    System.out.println("----------------------------------");
     Menu(records);
//===============================================
// Sort array
//=============================================
public static void quickSort (Comparable[] a, int left, int right)
     // Sort a[left?right] into ascending order.
     if (left < right) {
     int p = partition(a, left, right);
     quickSort(a, left, p-1);
     quickSort(a, p+1, right);
static int partition (Comparable[] a, int left, int right)
     // Partition a[left?right] such that
     // a[left?p-1] are all less than or equal to a[p], and
     // a[p+1?right] are all greater than or equal to a[p].
     // Return p.
     Comparable pivot = a[left];
     int p = left;
     for (int r = left+1; r <= right; r++) {
     int comp = a[r].compareTo(pivot);
     if (comp < 0) {
     a[p] = a[r]; a[r] = a[p+1];
a[p+1] = pivot; p++;          }
     return p;
} // end class PhoneBook
PhoneRecord class:
================
public class PhoneRecord implements Comparable
public int phonenumber;
public String last_Name;
public String first_Name;
public PhoneRecord(String last_Name, String first_Name, int phonenumber)
this.last_Name = last_Name;
this.phonenumber = phonenumber;
this.first_Name = first_Name;
/* Overload compareTo method */
public int compareTo(Object obj)
PhoneRecord tmp = (PhoneRecord)obj;
// sorting based on last name
String string1 = this.last_Name;
String string2 = tmp.last_Name;
int result = string1.compareTo(string2);
if(result < 0)
/* instance lt received */
return -1;
else if(result > 0)
/* instance gt received */
return 1;
/* instance == received */
return 0;

JosAH wrote:
prometheuzz wrote:
bats wrote:
Hi,
I keep getting a java.lang.ClassCastException with these two classes when I try to perform a binary search. Any tips?
...Looking at your binary search method:
public static int binarySearch( Comparable [ ] a, Comparable x)I see it expects to be fed Comparable objects. So, whatever you're feeding it, it's not a Comparable (ie: it doesn't implement Comparable), hence the CCE.It's even worse: if an A is a B it doesn't make an A[] a B[].
kind regards,
JosMy post didn't make much sense: if there were no Comparables provided as an argument, it would have thrown a compile time error. The problem lies in the compareTo(...) method.

What to use to store binary data in Java

Hello,
I have to read data stored as varbinary in the SQL database. What kind of class, data type I can use to store binary data from database?
I need to store binary data finally in ByteBuffer.
Please, help.
Thank you!

Unfortunately, when I am doing following I am getting an exception:
ByteBuffer lUTF8BIN=ByteBuffer.allocate(6)
byte [] lTempByteArray lTempByteArray = Recordset.getBytes(VarBinColumn);
if (lUTF8BIN.remaining()>=lTempByteArray.length)
lUTF8BIN.put(lTempByteArray);
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Adding binary numbers in java please help!!

Hello all, im a total newbie to java and i need your help urgently, i have two variables
that are integers, that store binary numbers i.e.
int tmpIntOne, tmpIntTwo;
tmpIntOne = 1010;
tmpIntTwo = 1110;
i want to add these numbers together and return a binary result so far when you add them together you get:
2120
however i would like a binary result so that
result = tmpIntOne + tmpIntTwo;
= 11000
Thank you.

Hello all, im a total newbie to java and i need your
help urgently, i have two variables
that are integers, that store binary numbers i.e.
int tmpIntOne, tmpIntTwo;
tmpIntOne = 1010;
tmpIntTwo = 1110;These are not binary numbers. These are decimal numbers whose digits consist of ones and zeros. If you want to interpret the characters "010" as a binary number, then do
int tmp1 = Integer.parseInt("1010", 2);
etc.
i want to add these numbers together
int result = tmp1 + tmp;
and return a
binary result Adding two ints will give an int. Int's are always binary. If you want to display it as a binary String, as opposed to the usual decimal String, then, as suggested, use toBinaryString.

Binary math in java?

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