如何在Java中存储无符号字节的存储（数组）以对其进行位操作？

Question

I need to store (an array of) unsigned bytes in Java to do bit manipulations on them. I want to do standard bit manipulations. shift ((sVal>>8) & 0xff); , or |, and &, and comparisons. and things like w |= ~0xff; How do I store them?

byte b = 0x96; //the value is 150

I get a compile error: Type mismatch: cannot convert from int to byte

so I cast it

byte b = (cast) 0x96;

Now I see the value has changed to -106.

If I store as a char, that will promote it to 2 bytes.

How should I store the value as an unsigned byte in Java so that I can do bit manipulations on them?

Will there be issues using int ?

Answer 1

There is no unsigned primitive data type in Java -- bytes are signed. You'll need to figure out some other way to represent your data, depending on what you want to do.

EDIT:

People have given you a number of suggestions and explanations, and you haven't explained much about the requirements of your situation ('a lot of them and fast' isn't helpful). But I noticed something else about your original post.

You say

byte b = (cast) 0x96;

Now I see the value has changed to -106.

I'm wondering if you understand that the value of the byte didn't change at all. The byte is still hexadecimal 96, and you could see that if you printed it out in hexadecimal. The -106 is merely the decimal number represented by hex 96.

If ALL you want to do is bit manipulation, then you could STILL store it all as bytes. Only if you also need arithmetic operations -- add, subtract, multiple, divide, greater than, less than -- would you need to worry about the signed nature of the bytes.

Answer 2

There is no problem using bit operations (apart from unsigned-shift-right) if you are not doing sign involving arithmetics.

Yes, converting to char or int extends the sign, but you can do:

byte b = (byte)0x96;
int n = 0xFF & b; // Again 0..255

(Maybe one should warn, trying to store bytes in a String with chars, is a kind of abuse involving encoding conversion.)

By the way, the classes Integer, Long and such have many interesting bit operations. The class BitSet is also an interesting playground.

Answer 3

You can do bit manipulations on signed bytes--they produce the same results you'd expect. The one thing you have to watch for is >> ; this will fill in the leftmost bits with 1 if the signed byte is negative, but if you use >>> (three > signs), it will always fill the leftmost bits with 0, which in essence treats it like an unsigned byte even though it's signed.

If you want the integer value of a byte and you want a value from 0 to 255:

int value = ((int)yourByte) & 0xFF;

Or, in Java 8, you can use Byte.toUnsignedInt .

[Actually, the cast isn't necessary, because yourByte and 0xFF will automatically get promoted to int before the & is applied. However, I never remember that rule without looking it up, so the explicit cast is helpful to me.]

If you want to do an ordered comparison of two bytes (or one byte and an integer literal) and treat them as unsigned, you can convert them both to int using one of the above methods and compare the int s.

Answer 4

You can use binary operations on bytes with no restrictions, just need to be careful with the right shift, depending on what you are willing to achieve. With Integer.toBinaryString(int i) you can see what is happening right there, although it is not the best representation:

public static void main(String[] args) {
        byte b1 = (byte) 0x96;
        int i1 = b1 & 0xFF;
        System.out.println(Integer.toBinaryString(i1));
        byte b2 = (byte) (b1 << 1);
        int i2 = b2 & 0xFF;
        System.out.println(Integer.toBinaryString(i2));
            System.out.println(Integer.toBinaryString(i1 & i2));
    }