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Consider the following snip of java code
byte b=(byte) 0xf1; byte c=(byte)(b>>4); byte d=(byte) (b>>>4); output:
c=0xff d=0xff expected output:
c=0x0f how? as b in binary 1111 0001 after unsigned right shift 0000 1111 hence 0x0f but why is it 0xff how?
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The unsigned right shift operator ( >>> ) (zero-fill right shift) evaluates the left-hand operand as an unsigned number, and shifts the binary representation of that number by the number of bits, modulo 32, specified by the right-hand operand.
That's all about difference between right shift and unsigned right shift operators in Java. Right shift ">>" keeps the sign extension while shifting bit patterns, but right shift without sign doesn't keep the original sign bit intact, it fills with zero.
The right shift operator ( >> ) returns the signed number represented by the result of performing a sign-extending shift of the binary representation of the first operand (evaluated as a two's complement bit string) to the right by the number of bits, modulo 32, specified in the second operand.
Bitwise Zero Fill Right Shift Operator (>>>) Bitwise Zero Fill Right Shift Operator shifts the bits of the number towards the right a specified n number of positions. The sign bit filled with 0's. The symbol >>> represents the Bitwise Zero Fill Right Shift Operator.
The problem is that all arguments are first promoted to int before the shift operation takes place:
byte b = (byte) 0xf1; b is signed, so its value is -15.
byte c = (byte) (b >> 4); b is first sign-extended to the integer -15 = 0xfffffff1, then shifted right to 0xffffffff and truncated to 0xff by the cast to byte.
byte d = (byte) (b >>> 4); b is first sign-extended to the integer -15 = 0xfffffff1, then shifted right to 0x0fffffff and truncated to 0xff by the cast to byte.
You can do (b & 0xff) >>> 4 to get the desired effect.
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