Bitwise right shift operator in Java

Question

In Java, -4 >> 2 gives -1 but -5 >> 2 gives -2. Can anybody explain why? Here is a sample code:

byte r=-5;
r>>=2;
System.out.println(r);

Also in this scenario >> and >>> operators give the same answer. Can anyone explain that as well?

Answer 1

You can take a look at the bits. Using two's complement notation, the bits for -4 and -5 are, showing only the last 8 bits for brevity:

-4: 1111 1100      
-5: 1111 1011

Bit shifting to the right 2 positions, with sign extension:

-4 >> 2: 1111 1111 (-1)
-5 >> 2: 1111 1110 (-2)

Normally, you think of >>> not using sign extension, and this is true, but in this case:

r >>>= 2;

... the value r is promoted to int for the bit-shift operation using binary numeric promotion, but the compound assignment operator casts the returned value back to byte, and the shifted-in zero "disappears".

byte r = -5;     // 1111 1100
r >>>= -2;       // promoted to int:   11111111 11111111 11111111 11111010
                 // bit shift:         00111111 11111111 11111111 11111110
                 // cast back to byte: 11111110 (-2)

The JLS, Section 15.26.2, talks about the casting operation done in compound assignment operators:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

That is, in this case, the result of the bit-shift is casted back to byte.

The same cast-back-to-byte operation occurs when the value of r is -4.

Note that if the assignment part wasn't done, then you wouldn't see the same answer, because it wouldn't cast the result back to byte:

System.out.println(r >>> 2);

Then you would see:

1073741822