Manisekar Chinnasami

yup i know the exception is thrown at that point ... how can i solve that ... but when i use 'AF' as input i am getting the result ... in this case, the arraylength is 2 only ...

i dont find any of the array is overloaded .... but i am getting arrayoutofbound exception ...

the concept of the program is ... if ypu enter the input as 'ABDF' , the output should be 110101 ... for A.B,D,F which is present 1 is replaced .. for the missing c and e , 0 is added ...

the length is assigned as 6 .. so if we enter 'A' , then the output should be 100000 ... but i am getting an arrayoutofbound exception ... can you help me to find the error...

here is the program ..............

class TCheck
{
public static void main(String [] args)
{
int blen = 6;
String aString = "ABC";
int binArray[] = new int[blen];
int c,k=0,asciiVal=65,x=0;
char chArray[] = aString.toCharArray();
for(int i=0;i<chArray.length;i++)
System.out.println(chArray[i]);
for(int i =0;i<blen;i++)
{
if(chArray[x] == asciiVal)
{
binArray[k] = 1;
x++;
}
else
binArray[k] = 0;

k++;
asciiVal++;

}
for(int i=0;i<k;i++)
System.out.println(binArray[i]);
int l =0,tenPower,binaryValue=0,n=k-1;
for(int i=0;i<k;i++)
{
tenPower = (int)Math.pow(10,n);
binaryValue = binaryValue +( binArray[i] * tenPower);
n--;
}
System.out.println("binaryValue --------- "+binaryValue);
}
}

how can i convert an long value into binary ?? and i have to store it as long ...

(is there any way to store as binary !!! )

binVal = Integer.parseInt(Integer.toBinaryString(25));//for int it is working

thanks for your reply ...

finally i have got the solution for my problem ...

i have converted it to String and performed the & operation and back to binary number and stored as integer type ...

it working ....

.....
.....
bitWiseCheck = Integer.parseInt(Integer.toBinaryString((Integer.parseInt(Integer.toString(binVal[j]),2)) & (Integer.parseInt(Integer.toString(vsVal),2))));
.....
.....

THEN HOW COULD I DO IT WITH THE BINARY NUMBERS ???

THE ACTUAL PROBLEMS IS,

CONSIDER A BINARY NUMBER 110110101, HAVE TO CHECK BITWISE DIGIT TO ZERO OR ONE, IF IT IS ONE, THE NUMBER LEFT UNCHANGED,ELSE IF IT IS 0, THEN IT SHOULD BE CHANGED TO ONE.

IF I USE SHIFT OPERATORS WILL IT WORK ???

IS THERE ANY WAY TO INITIALISE THE NUMBERS AS BINARY ??(i think it is not there in Java)

GIVE ME A RITE SOLUTION FOR THIS ...

actually i m getting the values as integer ...

b = Integer.parseInt("1011111", 2)

here the value "1011111" is in String format ... but i m getting those values in the interger format ...

int c = 1011111;
int b = Integer.parseInt(c,2);

how can i convert int c to String ???

i need to do logical AND mulitiplication for two binary numbers ....

100 & 111 = 100

when i do that it is working for this 3 bit .... but when i go further , it is not working ...

1111111 & 100000 = 100000 ( but the output is 33792)

here is the simple program ... find where i have done the mistake ...

class Test1
{
public static void main(String [] args)
{
int a, b=1111111,c=100000,d;
a = b & c;
d = b | c;
System.out.println("& = "+a+" | = "+d);
}
}

yup ... you are correct, Kelvin. i didnt notice that ... thanks

yup .. you are correct ...

when i tried to compile the following program, i got an incompatible type error ... (for returning an array) ....

import java.lang.reflect.Array;

class ResizeArray
{
static int z;
public int resizeArr(int [] b,int n)
{

z = n;
b = new int[10];
b = (int[])ArrayUtils.expand(b);
return b;

}

public static Object expand(Object a)
{
Class cl = a.getClass();
if (!cl.isArray()) return null;
int length = Array.getLength(a);
int newLength = z;
Class componentType = a.getClass().getComponentType();
Object newArray = Array.newInstance(componentType, newLength);
System.arraycopy(a, 0, newArray, 0, length);
return newArray;
}

public static void main(String args[])
{
ResizeArray ra = new ResizeArray();
int a[] = {1,2,3,4,5};
System.out.println("main length "+a.length);
int d=3;
ra.resizeArr(a,d);
}
}

watz the error in it ???

class Test
{
public int methode()
{
int a = {1,2,3,4,5};
return a; // how to return it ??? return a[] or anyother way?
}
}

give me a solution ...

i can resize the array ...

COPYING ANY ARRAY IS THE OLD BUSINESS ...

here is a sample program for resizeing an array ...

import java.lang.reflect.Array;

public class ArrayResize {

public static void main (String arg[]) {

int i[] = new int[5];
for(int j=0;j<5;j++)
{
System.out.println("The i array length is " + i.length);
i = (int[])ArrayUtils.expand(i);
}
}
public static Object expand(Object a) {
Class cl = a.getClass();
if (!cl.isArray()) return null;
int length = Array.getLength(a);
int newLength = length +1; // the size of the array is incremented by 1
Class componentType = a.getClass().getComponentType();
Object newArray = Array.newInstance(componentType, newLength);
System.arraycopy(a, 0, newArray, 0, length);
return newArray;
}
}

but for resizing two arrays with different values , i feel struck ...

hello friends ,

In my program, i am about to resize tow arrays' size.

i could do only for one using, java.lang.reflect.Array;

i dont know how to resize two arrays with different values ie. one array should be incremented by one(that i ve done). the other should to decremented by the calculated values (might be 2 or 1).

someone guide me ...

HELLO FRIENDS,

FINALLY I GOT THE SOLUTION FOR THIS QUERY ...

THANKS FOR YOUR HELP ...

HERE IS THE PROGRAM ...

class Arr
{
public static void main(String[] args)
{
Arr ao = new Arr();
int a[] = {1,2,3,2,1};
int k=0;


for(int i=0;i<a.length;i++)
{
for(int j=0;j<a.length;j++)
if(i!=j)
if(a[i] == a[j])
{
k++;
break;
}
}

int b[] = new int[k];
k=0;

for(int i=0;i<a.length;i++)
{
for(int j=0;j<a.length;j++)
if(i!=j)
if(a[i] == a[j])
{
b[k] = a[i];
k++;
break;
}
}

int g,m=0;
for(int l=0;l<a.length;l++)
{
g = 0;
for(int j=0;j<b.length;j++)
{

if(a[l] == b[j])
{
break;

} g++;
if(g == b.length)
{ a[m] = a[l];
m++;
}

}
}

for(int i=0;i<m;i++)
System.out.println(a[i]);
}
}