Odd behavior when Java converts int to byte?

In Java, an int is 32 bits. A byte is 8 bits .

Most primitive types in Java are signed, and byte, short, int, and long are encoded in two's complement. (The char type is unsigned, and the concept of a sign is not applicable to boolean.)

In this number scheme the most significant bit specifies the sign of the number. If more bits are needed, the most significant bit ("MSB") is simply copied to the new MSB.

So if you have byte 255: 11111111 and you want to represent it as an int (32 bits) you simply copy the 1 to the left 24 times.

Now, one way to read a negative two's complement number is to start with the least significant bit, move left until you find the first 1, then invert every bit afterwards. The resulting number is the positive version of that number

For example: 11111111 goes to 00000001 = -1. This is what Java will display as the value.

What you probably want to do is know the unsigned value of the byte.

You can accomplish this with a bitmask that deletes everything but the least significant 8 bits. (0xff)

So:

byte signedByte = -1;
int unsignedByte = signedByte & (0xff);

System.out.println("Signed: " + signedByte + " Unsigned: " + unsignedByte);

Would print out: "Signed: -1 Unsigned: 255"

What's actually happening here?

We are using bitwise AND to mask all of the extraneous sign bits (the 1's to the left of the least significant 8 bits.) When an int is converted into a byte, Java chops-off the left-most 24 bits

1111111111111111111111111010101
&
0000000000000000000000001111111
=
0000000000000000000000001010101

Since the 32nd bit is now the sign bit instead of the 8th bit (and we set the sign bit to 0 which is positive), the original 8 bits from the byte are read by Java as a positive value.


132 in digits (base 10) is 1000_0100 in bits (base 2) and Java stores int in 32 bits:

0000_0000_0000_0000_0000_0000_1000_0100

Algorithm for int-to-byte is left-truncate; Algorithm for System.out.println is two's-complement (Two's-complement is if leftmost bit is 1, interpret as negative one's-complement (invert bits) minus-one.); Thus System.out.println(int-to-byte( )) is:

  • interpret-as( if-leftmost-bit-is-1[ negative(invert-bits(minus-one(] left-truncate(0000_0000_0000_0000_0000_0000_1000_0100) [)))] )
  • =interpret-as( if-leftmost-bit-is-1[ negative(invert-bits(minus-one(] 1000_0100 [)))] )
  • =interpret-as(negative(invert-bits(minus-one(1000_0100))))
  • =interpret-as(negative(invert-bits(1000_0011)))
  • =interpret-as(negative(0111_1100))
  • =interpret-as(negative(124))
  • =interpret-as(-124)
  • =-124   Tada!!!

byte in Java is signed, so it has a range -2^7 to 2^7-1 - ie, -128 to 127. Since 132 is above 127, you end up wrapping around to 132-256=-124. That is, essentially 256 (2^8) is added or subtracted until it falls into range.

For more information, you may want to read up on two's complement.

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Java

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