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    <p>The Java specification explains the <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.19" rel="noreferrer">shift operators</a> as follows:</p> <blockquote> <p>If the promoted type of the left-hand operand is <code>int</code>, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator <code>&amp;</code> (<a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.22.1" rel="noreferrer">§15.22.1</a>) with the mask value <code>0x1f</code>. The shift distance actually used is therefore always in the range 0 to 31, inclusive.</p> </blockquote> <p>The value of <code>32 &amp; 0x1f</code> is zero.</p> <p>If the left operand is <code>long</code>, then you get an extra bit for the right operand, expanding the upper limit to 63 instead of 31.</p> <hr> <p>In order to have any specific expected value from shifting -1 to the right, you need to specify the underlying binary representation of integers (e.g., two's complement) as well as the number of bits (e.g., 32). Each programming language can define those differently, but in the interest of keeping things simpler for the implementation, they'll usually specify that shifting by more than the number of available bits is not allowed. That's often because the underlying CPU hardware doesn't support it, either. After all, if you want to shift that many bits, the left operand no longer matters since the result will always be the same.</p>
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    1. POWhy is (-1 >>> 32) = -1?
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    1. USRob Kennedy
    1. USRob Kennedy
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    1. POWhy is (-1 >>> 32) = -1?
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    1. COBTW it means you can get the top bit of `x` using `x >> 63` or `x >> -1` whether x is int or long. ;)
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      1. USPeter Lawrey
    2. COShifting may also depend on 32/64 bit system.
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      1. USTheCottonSilk
    3. CO@TheCottonSilk Not in Java; shifting works the same way, doesn't matter if you're using a 32-bit or 64-bit operating system or JVM.
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      1. USJesper
 

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