As I was using bit-shifting on byte
, I notice I was getting weird results when using unsigned right shift (>>>
). With int
, both right shift (signed:>>
and unsigned:>>>
) behave as expected:
int min1 = Integer.MIN_VALUE>>31; //min1 = -1
int min2 = Integer.MIN_VALUE>>>31; //min2 = 1
But when I do the same with byte
, strange things happen with unsigned right shift:
byte b1 = Byte.MIN_VALUE; //b1 = -128
b1 >>= 7; //b1 = -1
byte b2 = Byte.MIN_VALUE; //b2 = -128
b2 >>>= 7; //b2 = -1; NOT 1!
b2 >>>= 8; //b2 = -1; NOT 0!
I figured that it could be that the compiler is converting the byte
to int
internally, but does not seem quite sufficient to explain that behaviour.
Why is bit-shifting behaving that way with byte in Java?
This happens exactly because byte
is promoted to int
prior performing bitwise operations. int -128
is presented as:
11111111 11111111 11111111 10000000
Thus, shifting right to 7 or 8 bits still leaves 7-th bit 1, so result is narrowed to negative byte
value.
Compare:
System.out.println((byte) (b >>> 7)); // -1
System.out.println((byte) ((b & 0xFF) >>> 7)); // 1
By b & 0xFF
, all highest bits are cleared prior shift, so result is produced as expected.
Shift operators for byte
, short
and char
are always done on int
.
Therefore, the value really being shifted is the int
value -128
, which looks like this
int b = 0b11111111_11111111_11111111_10000000;
When you do b2 >>= 7;
what you are really doing is shifting the above value 7 places to the right, then casting back to a byte
by only considering the last 8 bits.
After shifting 7 places to the right we get
0b11111111_11111111_11111111_11111111;
When we convert this back to a byte, we get just 11111111
, which is -1
, because the byte
type is signed.
If you want to get the answer 1 you could shift 31 places without sign extension.
byte b2 = Byte.MIN_VALUE; //b2 = -128
b2 >>>= 31;
System.out.println(b2); // 1
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