why is 1>>32 == 1?

Question

I\'m wondering if perhaps this is a JVM bug?

java version \"1.6.0_0\" OpenJDK Runtime Environment (IcedTea6 1.4.1) (6b14-1.4.1-0ubuntu13) OpenJDK 64-Bit Server VM (b

Answer 1

http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.22.1

15.19 Shift Operators

If the promoted type of the left-hand operand is int, 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 & (§15.22.1) with the mask value 0x1f. The shift distance actually used is therefore always in the range 0 to 31, inclusive.

If the promoted type of the left-hand operand is long, 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 & (§15.22.1) with the mask value 0x3f. The shift distance actually used is therefore always in the range 0 to 63, inclusive.

(emphasis mine)

Answer 2

It's not a bug. In n >> m, it only looks at the last five bits of m - so any number greater than 31 will be reduced to that number mod 32. So, (256 >> 37) == 8 is true.

Edit: This is true if you're working with ints. If it's longs, then it looks at the last six bits of m, or mods by 64.