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    <p>All shifts are done mod 32 for ints and mod 64 for longs.</p> <p>From <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.19" rel="noreferrer">section 15.19 of the spec</a>:</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 &amp; (§15.22.1) with the mask value 0x1f. The shift distance actually used is therefore always in the range 0 to 31, inclusive.</p> <p>If the promoted type of the left-hand operand is <code>long</code>, then only the six 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 &amp; (§15.22.1) with the mask value 0x3f. The shift distance actually used is therefore always in the range 0 to 63, inclusive.</p> </blockquote> <p>As for <em>why</em> the language was designed that way - I don't know, but C# has the same design decision. Here's what the annotated ECMA C# spec says:</p> <blockquote> <p>C# deliberately keeps implementation-defined behaviors to a miinimum. They are accepted only when the performance impact of forcing uniform behavior would be excessive (such as for some floating-point precision issues). Hence, the size of each integral type is precisely specified, and the character set is fixed to Unicode.</p> <p>For shift operations, too, uniform behavior is specified. It can be achieved using a single extra instruction (&amp; 0x1F or &amp; 0x3F) that incurs only a tiny cost on modern processors, especially since it does not reference memory. Unlike for floating-point operations, the difference in shift behavior would be dramatic if left to the whim of the processors; rather than a small difference in precision, completely different integral results would be produced.</p> <p>In making this decision the committe studied reference materials for a number of different processor architectures. There is little consistency in the behavior for shift counts outside the range -32..+32 for 32-bit operands, and respectively -64..+64 for 64-bit operands.</p> </blockquote> <p>(There's then a list of some examples.)</p> <p>This seems an entirely reasonable explanation to me. Consistency is definitely important, and if it would be impossible to implement <em>different</em> consistent behaviour in a performant way on some systems, I think this is a reasonable solution.</p>
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    1. POIn Java, when using bitshifts, why does 1 << 32 != 1 << 31 << 1?
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    1. POIn Java, when using bitshifts, why does 1 << 32 != 1 << 31 << 1?
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    1. COI suppose it is because it is quite pointless to shift an int of 32 bits: the result will be always 0, no?
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    2. COWell, the situation would arise when the amount you want to shift by is say, an user defined variable. So yes, it should just zero the number, but I don't want to insert a check just for the 32 bit case.
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    3. COUsing "b > 32 ? 0 : a << b" isn't *so* bad...
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