Why does `value & 0xff` produce an unsigned value in Java, even though `value` is a signed byte?

Bitwise Manipulation with Value & 0xff in Java

In this Java code snippet:

<code class="java">byte value = 0xfe; // corresponds to -2 (signed) and 254 (unsigned)
int result = value & 0xff;</code>

the question arises: why does the bitwise AND operation (&) between value and 0xff produce the result 254 (unsigned) instead of -2 (signed) as expected?

Unlike C, byte is a signed type in Java. Assigning value to int without bitwise manipulation would result in the signed value -2. However, using & with 0xff achieves the intended unsigned value.

The key here is that & operates on int values. When value (a byte) is used with &, it is first promoted to an int. Similarly, 0xff is an int literal. The & operation then yields the 8-bit binary value from value placed in the least significant 8 bits of result.

In this example:

• value is promoted to -2 (11111110 in binary)
• 0xff represents 255 (11111111 in binary)
• value & 0xff thus results in 254 (11111110 in binary)

This bitwise manipulation is commonly used to extract specific bits or maintain the unsigned nature of a value, particularly in low-level programming or data manipulation scenarios.

The above is the detailed content of Why does `value & 0xff` produce an unsigned value in Java, even though `value` is a signed byte?. For more information, please follow other related articles on the PHP Chinese website!