[JAVA] XOR function

[donchen]

Hi guys,

I have 2 binaries 10010101 and 00101001.

I will like to perform XOR on these 2 binaries. How can I do that ?

Can i just do this

int result = 10010101 ^ 00101001;

The above gave me this 9977844.

[Smenus]

I haven't checked this because I don't think I have a Java SDK installed, but a quick check on the net says that prefixing the numbers by 0b might help.

[donchen]

it doesn't work

[Smenus]

  donchen said:

it doesn't work

OK, sorry. Like I said, it was untested.

[rpgfan]

You need to convert them to integers first. You could do it the hard way and code a function to do that yourself, or you can use the built-in Integer class:

int bin1 = Integer.valueOf("10010101", 2), bin2 = Integer.valueOf("00101001", 2);
int result = bin1 ^ bin2;

Details on the valueOf() method that I used can be at http://java.sun.com/javase/6/docs/api/java....String,%20int)

Note that it returns an instance of the Integer class. However, this can be coerced to a native int type, so you shouldn't have problems with the above code. If you do (my Java is a bit rusty), just call the intValue() method on the result like this:

int bin1 = Integer.valueOf("10010101", 2).intValue(), bin2 = Integer.valueOf("00101001", 2).intValue();
int result = bin1 ^ bin2;

That's probably what happens internally anyway, but for code clarity, I omitted it because it shouldn't be necessary.

[kjordan2001]

Easiest way would probably be:

int result = new BigInteger("10010101",2).xor(new BigInteger("00101001",2)).intValue();

[donchen]

perfect! Its awesome! Thanks! Works perfect! Thank you all!

[donchen]

Sorry... it doesn't work....

Why does it give me a value of 188 ?

Should it be 10111100 when you XOR 10010101 with 00101001 ?

[ViZioN]

  donchen said:

Sorry... it doesn't work....

Why does it give me a value of 188 ?

Should it be 10111100 when you XOR 10010101 with 00101001 ?

188 is the integer representation of 10111100. I think you will need to convert it back to binary.

[rpgfan]

String result = Integer.toBinaryString(Integer.parseInt("10010101", 2) ^ Integer.parseInt("00101001", 2));

The parseInt() method already returns an int rather than an Integer, so I chose to use that. The toBinaryString() method should be easy enough to understand. :p