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Adding two unicode chars in java

I'm stuck in my program that emulates a 8bit processor. I have an array of characters in Unicode(hex) that I call memory. I'm trying to pass my operand of type int a two byte word which will be two unicode values in my array. Lets say myArray[1] has '\u00BF' and myArray[2] has '\u00FF' so I want to pass '\uBFFF' to my operand. I was thinking of parsing the char to its bytes value into a string, concatenating the two, and then passing it to an int. Any help would be GREATLY appreciated!

Can a special character can be taken in a character variable with an alphabet?

I want to take input in a character variable which is A#.
Is it possible to do that?
Example:
char E[]={'E','F#','G#','A','B','C#','D#'};
To solve this I have taken array type as string. Which is giving me problem to get it's ascii to calculate hash value and also with respect to sorting as well.

No, you should use String to store these since there are more than 1 character. A char can only have 1 character. You can use the default hashCode() implementation of String for hash value and the default compareTo() for sorting.

You can write any character, include specials characters... For example 'Ñ'. But you can't write 'F#' in a char.

You can't store two characters in one char variable. Remember that 'F#' isn't a char, it's a String ! You should use String to store them. Then you can use compareTo() method for check if two strings are equal or not. Check This out

Reading one character at a time from a large string

I have a big string having at most 100000 character. Instead of using string.charAt[index] to read a character from the string, I converted that string into char array using string.toCharArray() method and now i am working with charArray[index]. which takes less time than string.charAt[index] method. However i want to know that, is there any other way which is faster than string.toCharArray(); method?

I do not think there is a faster way. But please correct me!
A String instance is backed by a char array. charAt() does some index checks which may be the cause for it being slower than working with the array returned by toCharArray(). toCharArray() simply does a System.arraycopy() of the backing array.

Casting int to char in java- must I use ASCII?

Q: When casting an int to a char in Java, it seems that the default result is the ASCII character corresponding to that int value. My question is, is there some way to specify a different character set to be used when casting?
(Background info: I'm working on a project in which I read in a string of binary characters, convert it into chunks, and convert the chunks into their values in decimal, ints, which I then cast as chars. I then need to be able to "expand" the resulting compressed characters back to binary by reversing the process.
I have been able to do this, but currently I have only been able to compress up to 6 "bits" into a single character, because when I allow for larger amounts, there are some values in the range which do not seem to be handled well by ASCII; they become boxes or question marks and when they are cast back into an int, their original value has not been preserved. If I could use another character set, I imagine I could avoid this problem and compress the binary by 8 bits at a time, which is my goal.)
I hope this was clear, and thanks in advance!

Your problem has nothing to do with ASCII or character sets.
In Java, a char is just a 16-bit integer. When casting ints (which are 32-bit integers) to chars, the only thing you are doing is keeping the 16 least significant bits of the int, and discarding the upper 16 bits. This is called a narrowing conversion.
References:
http://java.sun.com/docs/books/jls/second_edition/html/conversions.doc.html#20232
http://java.sun.com/docs/books/jls/second_edition/html/conversions.doc.html#25363

The conversion between characters and integers uses the Unicode values, of which ASCII is a subset. If you are handling binary data you should avoid characters and strings and instead use an integer array - note that Java doesn't have unsigned 8-bit integers.

What you search for in not a cast, it's a conversion.
There is a String constructor that takes an array of byte and a charset encoding. This should help you.

I'm working on a project in which I
read in a string of binary characters,
convert it into chunks, and convert
the chunks into their values in
decimal, ints, which I then cast as
chars. I then need to be able to
"expand" the resulting compressed
characters back to binary by reversing
the process.
You don't mention why you are doing that, and (to be honest) it's a little hard to follow what you're trying to describe (for one thing, I don't see why the resulting characters would be "compressed" in any way.
If you just want to represent binary data as text, there are plenty of standard ways of accomplishing that. But it sounds like you may be after something else?

how to read and extract an integer from a string in java

Given a string
7 + 45 * 65
How to check whether a given character of this string is an integer and then store the whole integer value in an integer variable?
E.g. for 65, check if 6 is an integer, if yes, then store 65 in another integer variable. You can assume that the string can be converted into a character array.

Given that this looks like homework below are some tips for a simple way to parse and store each integer value.
Check out the API documentation for the Character class. This will contain methods for determining whether a character is a digit.
Consider using a StringBuilder to store the intermediate numerical result as you read in each digit of the number.
Check the Integer class API for methods to help with parsing the String value (stored within your StringBuilder) and turning it into an int.
Finally, consider using a List (e.g. LinkedList) to store the int value.

For a quick and dirty soluiton I would use StringTokenizer and try { Integer.parseInt() } catch (NumberFormatException){}

Check out this post on exactly the same topic:
Java Programming - Evaluate String math expression
It looks like BeanShell has the cleanest method to do what you need. You could also try the JavaScript Engine method (although BeanShell looks much cleaner to me).

Easiest solution would be to use java.util.Scanner. You can set Scanner.useDelimeter("\\D+") which will mean skip any non-digit characters, and then call Scanner.nextInt() to get next Integer from the String.
If you want to work with characters, then use Character.isDigit(char c).