CODEWITHSUNDEEP
javaoperatorsbytenegative-number
zaozaoer
zaozaoer

Reputation: 387

>> and >>> operator with negative byte value in java

I have a sample code snippet like this:

    byte a = -0b00001111;
    byte b = a;
    byte c = a;

    System.out.println("a=" + a );
    System.out.println("b=" + (b >> 1) );
    System.out.println("c=" + (c >>> 1) );

and, it prints:

a=-15

b=-8

c=2147483640

I don't quite understand how b and c became those 2 values respectively, could someone demonstrate me in steps how those 2 values were calculated please?

Upvotes: 1

Views: 44

Answers (1)

rgettman
rgettman

Reputation: 178343

For byte a, you have the literal 0b00001111, which is binary for 15, so a is -15. The bit representation of -15 for a byte is:

11110001

In Java, unary numeric promotion occurs on the operands of bit-shift operators <<, >>, and >>>.

Unary numeric promotion (§5.6.1) is performed on each operand separately.

This means that the value is promoted to int before shifting.

The code (b >> 1) promotes b to an int, then shifts the value with sign extension. This means that if the value was already negative, then a 1 bit is shifted to ensure it's still negative.

11110001

is promoted to

11111111 11111111 11111111 11110001

which is -15 as an int. After shifting to the right one bit, with sign extension:

11111111 11111111 11111111 11111000

which is -8.

However, for the code (c >>> 1), the >>> unsigned right shift operator does not perform sign extension, even if the promotion to int does maintain the sign and the value.

11110001

is promoted to

11111111 11111111 11111111 11110001

which is -15 as an int as before. After unsigned shifting to the right one bit:

01111111 11111111 11111111 11111000

The first bit is now 0. Without the most significant bit set, the value is now 231 - 8, or 2147483640.

Upvotes: 1

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