Bitwise shift operators. Signed and unsigned

Question

I'm practising for the SCJP exam using cram notes from the Internet.

According to my notes the >> operator is supposed to be signed right shift, with the sign bit being brought in from the left. While the left shift operator << is supposed to preserve the sign bit.

Playing around however, I'm able to shift the sign with the << operator (f.e. Integer.MAX_VALUE << 1 evaluates to -2, while I'm never able to shift the sign with the >> operator.

I must be misunderstanding something here, but what?

Fitzwater

See what -2 << 31 gives you. – Tracheo

Answer 1

">>" is signed because it keeps the sign. It uses the most left digit in binary representation of a number as a filler. For example:

    | this value is used as a filler 
    11011011 
 >> 11101101  

    01010010
 >> 00101001 

">>>" is unsigned version of this operator. It always use zero as a filler:

    11011011 
>>> 01101101  

    01010010
>>> 00101001

In binary representation the most left digit determines sign of the number. So, if it's '1' then we have negative value and if it's '0' - then our number is positive. That's why using the most left digit as a filler allows to keep sign permanent.

Stanzel

Ah, ok then I understand how the right shift operators are supposed to work. Thanks! But why am I able to change sign with "<<"? – Fitzwater

Because it shifts in another direction and the most left digit depends on the original number. It uses 0 as a filler but appends it to the right side of your number. – Stanzel

The sign of an integer is determined by the value of its leftmost bit. If you shift in a 0 when it is 1 or vice versa, the sign of the result changes. – Tracheo

Ok, so then basically the info in my notes is not correct on how the left shift operator (<<) works. If I get you right it: 1) Fills with 0s from the right. 2) WILL shift the (leftmost) sign bit and replace it with a bit from right. – Fitzwater

Answer 2

The idea behind the shifts is that they can act as multiplying and dividing by powers of 2 ( << 1 is equivalent to *= 2, >> 2 is equivalent to /= 4), which is why the signed version of shifting exists. Unsigned shifting doesn't preserve the sign, necessarily, though. The << operator doesn't actually preserve the sign, as you suggest; it simply happens to in your example. Try doing a left shift on 2,147,483,647; it doesn't stay positive. The reason that they don't bother trying to make a 'signed' left shift is because, if the number shifts from positive to negative (or viceversa), then you went outside the bounds of the variable type anyway.

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