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javaintbitoperation

java Bit operations >>> shift

Why if 

int x = -1 // binary: 11111111111111111111111111111111
x = x >>> 31;

we have 00000000000000000000000000000001

but if 

int x = -1
x = x >>> 32;

we have 11111111111111111111111111111111 (again -1)

but not 00000000000000000000000000000000 ?

Solution

  • From Section 15.19 of JLS:

    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 (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

    Emphasis mine. So:

    x >>> n

    is equivalent to: 

    x >>> n & 0x1f  // or x >>> n % 32

    So, x >>> 32 is equivalent to x >>> 32 & 0x1f <==> x >>> 0 == x.

    So the Rule of Thumb is, whenever you shift a number by a multiple of 32(int is 32 bits), you get back the same value.