Why does the code snippet \'byte value = 0xfe; int result = value & 0xff;\' result in a value of 254?

Understanding Bitwise Operations: Unraveling value & 0xff in Java

The question arises from a Java code snippet that performs a bitwise operation using the & operator:

<code class="java">byte value = 0xfe;
int result = value & 0xff;</code>

This operation appears to yield a value of 254 when printed. However, understanding how this works requires delving deeper into the intricacies of data types and bitwise operations in Java.

Bitwise Operations and Data Promotion

The & operator in Java performs a bitwise AND operation, where each bit in the result is set to 1 if both corresponding bits in the operands are 1, and 0 otherwise. Normally, this operation results in a value of the same type as its operands.

In the code snippet, value is a byte, which is an 8-bit signed integer. The value 0xfe corresponds to -2 in signed representation and 254 in unsigned representation. The other operand, 0xff, is an int literal representing an unsigned 255.

However, the & operator is defined to operate only on int values. Therefore, value is first promoted to an int (ff ff ff fe) before the operation is performed.

Unveiling the Result

The & operation is then applied between the promoted value and the int literal 0xff (00 00 00 ff), resulting in the value 00 00 00 fe. This corresponds to 254 in unsigned representation, which is printed as the result.

In summary, the expression value & 0xff sets result to the unsigned value resulting from putting the 8 bits of value in the lowest 8 bits of result. This is necessary because byte is a signed type in Java, and direct conversion from byte to int would result in an incorrect signed representation.

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