解决所有Java字节都已签名这一事实的最佳方法是什么？

Question

In Java, there is no such thing as an unsigned byte.

Working with some low level code, occasionally you need to work with bytes that have unsigned values greater than 128, which causes Java to interpret them as a negative number due to the MSB being used for sign.

What's a good way to work around this? (Saying don't use Java is not an option)

Answer 1

It is actually possible to get rid of the if statement and the addition if you do it like this.

byte[] foobar = ..;
int value = (foobar[10] & 0xff);

This way Java doesn't interpret the byte as a negative number and flip the sign bit on the integer also.

Answer 2

When reading any single value from the array copy it into something like a short or an int and manually convert the negative number into the positive value it should be.

byte[] foobar = ..;
int value = foobar[10];
if (value < 0) value += 256 // Patch up the 'falsely' negative value

You can do a similar conversion when writing into the array.

Answer 3

使用int通常比使用short更好，因为java无论如何都在内部使用32位值（即使是字节，除非在数组中），因此使用int将避免不必要的转换为字节码中的短值。

Answer 4

Probably your best bet is to use an integer rather than a byte. It has the room to allow for numbers greater than 128 without the overhead of having to create a special object to replace byte.

This is also suggested by people smarter than me (everybody)

• http://www.darksleep.com/player/JavaAndUnsignedTypes.html
• http://www.jguru.com/faq/view.jsp?EID=13647

Answer 5

The best way to do bit manipulation/unsigned bytes is through using int s. Even though they are signed they have plenty of spare bits (32 total) to treat as an unsigned byte. Also, all of the mathematical operators will convert smaller fixed precision numbers to int . Example:

short a = 1s;
short b = 2s;
int c = a + b; // the result is up-converted
short small = (short)c; // must cast to get it back to short

Because of this it is best to just stick with integer and mask it to get the bits that you are interested in. Example:

int a = 32;
int b = 128;
int foo = (a + b) | 255;

Here is some more info on Java primitive types http://mindprod.com/jgloss/primitive.html

One last trivial note, there is one unsigned fixed precision number in Java. That is the char primitive.

Answer 6

I know this is a very late response, but I came across this thread when trying to do the exact same thing. The issue is simply trying to determine if a Java byte is >127.

The simple solution is:

if((val & (byte)0x80) != 0) { ... }

If the real issue is >128 instead, just adding another condition to that if-statement will do the trick.

Answer 7

I guess you could just use a short to store them. Not very efficient, but really the only option besides some herculean effort that I have seen.