我在用Java创建16位字符时遇到问题

Question

How can I create a variable character that can hold a four byte value?

I am trying to write an program to encrypt messages in java, for fun. I figured out how to use RSA, and managed to write a program that will encrypt a message and save it to a .txt file. For example if "Quiet" is entered the outcome will be "041891090280". I wrote my code so that the number would always have length that is a multiple of six. So I thought that I could convert the numbers into a hash code. The first three letters are "041" so I could convert that into ")".

However I am having trouble created a char with a number greater than 255. I have looked around online and found a few examples, but I can't figure out how to implement them. I created a new method just to test them.

int a = 256;
char b = (char) a;
char c = 0xD836;
char[] cc = Character.toChars(0x1D50A);
System.out.println(b);
System.out.println(c); 
System.out.println(cc);

The program outputs

?

?

?

I am only getting two bytes. I read that Java uses Unicode which should go up to 65535 which is four bytes. I am using eclipse if that makes a difference.

I apologize for the noob question. And thanks in advance.

edit I am sorry, I think I gave too much information and ended up being confusion. What I want to do is store a string of numbers as a string of unicode characters. the only way I know how to do that is to break up the number string small enough to fit it into a character. then add the characters one by one to a new string. But I don't know how to add a variable unicode character to a string.

Answer 1

All chars are 16-bit already. 0 to 65535 only need 16-bit and 2^16 = 65536.

Note: not all characters are valid and in particular, 0xD800 to 0xDFFF are used for encoding code points (characters beyond 65536)

If you want to be able to store all possible 16-bit values I suggest you use short instead. You can store the same values but it may be less confusing to use.