How to Convert Int to Unsigned Byte and Back

Question

I need to convert a number into an unsigned byte. The number is always less than or equal to 255, and so it will fit in one byte.

I also need to convert that byte ba

Answer 1

A byte is always signed in Java. You may get its unsigned value by binary-anding it with 0xFF, though:

int i = 234;
byte b = (byte) i;
System.out.println(b); // -22
int i2 = b & 0xFF;
System.out.println(i2); // 234

Answer 2

Even though it's too late, I'd like to give my input on this as it might clarify why the solution given by JB Nizet works. I stumbled upon this little problem working on a byte parser and to string conversion myself. When you copy from a bigger size integral type to a smaller size integral type as this java doc says this happens:

https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.3 A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the sign of the resulting value to differ from the sign of the input value.

You can be sure that a byte is an integral type as this java doc says https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html byte: The byte data type is an 8-bit signed two's complement integer.

So in the case of casting an integer(32 bits) to a byte(8 bits), you just copy the last (least significant 8 bits) of that integer to the given byte variable.

int a = 128;
byte b = (byte)a; // Last 8 bits gets copied
System.out.println(b);  // -128

Second part of the story involves how Java unary and binary operators promote operands. https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2 Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

If either operand is of type double, the other is converted to double.

Otherwise, if either operand is of type float, the other is converted to float.

Otherwise, if either operand is of type long, the other is converted to long.

Otherwise, both operands are converted to type int.

Rest assured, if you are working with integral type int and/or lower it'll be promoted to an int.

// byte b(0x80) gets promoted to int (0xFF80) by the & operator and then
// 0xFF80 & 0xFF (0xFF translates to 0x00FF) bitwise operation yields 
// 0x0080
a = b & 0xFF;
System.out.println(a); // 128

I scratched my head around this too :). There is a good answer for this here by rgettman. Bitwise operators in java only for integer and long?

Answer 3

The Integer.toString(size) call converts into the char representation of your integer, i.e. the char '5'. The ASCII representation of that character is the value 65.

You need to parse the string back to an integer value first, e.g. by using Integer.parseInt, to get back the original int value.

As a bottom line, for a signed/unsigned conversion, it is best to leave String out of the picture and use bit manipulation as @JB suggests.

Answer 4

Except char, every other numerical data type in Java are signed.

As said in a previous answer, you can get the unsigned value by performing an and operation with 0xFF. In this answer, I'm going to explain how it happens.

int i = 234;
byte b = (byte) i;
System.out.println(b);  // -22

int i2 = b & 0xFF;      
// This is like casting b to int and perform and operation with 0xFF

System.out.println(i2); // 234

If your machine is 32-bit, then the int data type needs 32-bits to store values. byte needs only 8-bits.

The int variable i is represented in the memory as follows (as a 32-bit integer).

0{24}11101010

Then the byte variable b is represented as:

11101010

As bytes are unsigned, this value represent -22. (Search for 2's complement to learn more on how to represent negative integers in memory)

Then if you cast is to int it will still be -22 because casting preserves the sign of a number.

1{24}11101010

The the casted 32-bit value of b perform and operation with 0xFF.

 1{24}11101010 & 0{24}11111111
=0{24}11101010

Then you get 234 as the answer.