Why does `value & 0xff` return an unsigned integer in Java when `value` is a byte?

Unveiling the Mystery of value & 0xff in Java

In Java, the & operator performs bitwise AND operation on two integers, resulting in a new integer. However, when one operand is of type byte, an intriguing phenomenon occurs.

Consider the following code:

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

When printing result, you astonishingly get 254 (unsigned), defying expectations. To understand this behavior, we must delve into the nuances of Java's type system.

In Java, byte is a signed type, meaning it can represent both positive and negative values. However, when it is promoted to an integer, it is treated as an unsigned value. When the & operator is applied to value (promoted to integer) and the constant 0xff (which is an unsigned integer literal), the result is an unsigned integer containing the lowest 8 bits of the promoted value.

In other words, result will have the same binary representation as value, with all the bits above the 8th bit cleared to 0. This effectively sets result to the unsigned value obtained by discarding the sign bit from value.

Therefore, value & 0xff provides a technique to convert a signed byte to an unsigned integer, maintaining the original 8-bit value. This is often useful when working with data where the sign bit is not relevant or desired.

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