Binary in java - guide me

[Manisekar Chinnasami]

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);
}
}

[Peter van de Riet]

It makes perfectly sence:

decimal 1111111 is binary 100001111010001000111
decimal 100000 is binary 11000011010100000

100001111010001000111
000011000011010100000 &
---------------------
000001000010000000000

binary 1000010000000000 = decimal 33792

Maybe you want to fill your "int a" and "int b" a different way:

class Test2 {
public static void main(String[] args) {
int a, b = Integer.parseInt("1111111", 2), c = Integer.parseInt("100000", 2), d;
a = b & c;
d = b | c;
System.out.println("& = " + Integer.toBinaryString(a) + " | = " + Integer.toBinaryString(d));
}
}

The output is: & = 100000 | = 1111111

[Jesper de Jong]

Originally posted by Manisekar Chinnasami:
int a, b=1111111,c=100000,d;

You are not assigning binary numbers to the variables b and c here, but decimal numbers. Java doesn't have a way to express integer literals as binary numbers. You could try this instead:

int a, b = Integer.parseInt("1111111", 2), c = Integer.parseInt("100000", 2), d;

[Rob Spoor]

Use Integer.toBinaryString() to get a binary representation back.

[Manisekar Chinnasami]

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 ???

[Henry Wong]

100 & 111 = 100

This part is probably done in decimal too -- and worked by dumb luck.

ie....

 
100 (decimal) = 0110 0100 (binary)
111           = 0110 1111
AND           = 0110 0100 ---> 100 decimal

Henry

[Henry Wong]

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

how can i convert int c to String ???

To repeat what everyone is saying... The value of c is assigned the decimal value of 1011111. Not the binary value of 1011111. Even if you convert it to string, using Integer.toString(), you are converting the wrong value.

Henry

[Manisekar Chinnasami]

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 ...

[Ulf Dittmer]

For all our eyes' sake, please KeepItDown. Writing in all uppercase is considered rude all over the net.

[Henry Wong]

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

I know this seems to be rubbing salt on the wound, but you don't seem to understand that this is more your issue, than Java's. There is no such thing as "INITIALISE THE NUMBERS AS BINARY" -- integers are in the integer format.

Decimal, Octal, Hexidecimal, are literal formats used to help the programmer specify that integer value. In your case, you want to specify a literal value -- and it is that, that you want in binary.

If there was a huge problem with mentally converting a binary (literal) value to hexidecimal or octal -- then programmers would have pushed to have binary literals added a long time ago.

Quite frankly, it takes seconds to mentally convert a binary literal value to its octal or hexidecimal counterpart. And you don't need a calculator.

110110101 --> 110 110 101 --> 0665

or

110110101 --> 1 1011 0101 --> 0x1B5


But... To answer your question. There is currently no easy way to specify a integer literal as a binary in Java. So... if you absolutely need to have it done as a binary integer literal, then you can't do it. Sorry.

Henry

[Manisekar Chinnasami]

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))));
.....
.....

[Stefan Wagner]

Originally posted by Manisekar Chinnasami:


finally i have got the solution for my problem ...
bitWiseCheck = Integer.parseInt(Integer.toBinaryString((Integer.parseInt(Integer.toString(binVal[j]),2)) & (Integer.parseInt(Integer.toString(vsVal),2))));

I don't think so.
Try your code for -3 and -2 for instance, or 16 and 15.

I guess all you need is

int bitWiseCheck = binVal[j] & vsVal;

That is doing a bitwise AND.

Integer.toBinaryString (bitWiseCheck); 

Will show you the binary representation.