This operator shifts the first operand the specified number of bits to the right. Excess bits shifted off to the right are discarded. Zero bits are shifted in from the left. The sign bit becomes `0`

, so the result is always non-negative. Unlike the other bitwise operators, zero-fill right shift returns an unsigned 32-bit integer.

Consider the 32-bit binary representations of the decimal (base 10) numbers `9`

and `-9`

:

9 (base 10): 00000000000000000000000000001001 (base 2)
-9 (base 10): 11111111111111111111111111110111 (base 2)

Notice that the binary representation of the negative decimal (base 10) number `-9`

is the two's complement of the binary representation of the positive decimal (base 10) number `9`

. That is, it's calculated by inverting all the bits of `00000000000000000000000000001001`

and adding `1`

.

In both cases, the sign of the binary number is given by its leftmost bit: for the positive decimal number `9`

, the leftmost bit of the binary representation is `0`

, and for the negative decimal number `-9`

, the leftmost bit of the binary representation is `1`

.

Given those binary representations of the decimal (base 10) numbers `9`

, and `-9`

:

For the positive number `9`

, zero-fill right shift and sign-propagating right shift yield the same result: `9 >>> 2`

yields `2`

, the same as `9 >> 2`

:

9 (base 10): 00000000000000000000000000001001 (base 2)
--------------------------------
9 >> 2 (base 10): 00000000000000000000000000000010 (base 2) = 2 (base 10)
9 >>> 2 (base 10): 00000000000000000000000000000010 (base 2) = 2 (base 10)

Notice how two rightmost bits, `01`

, have been shifted off, and two zeroes have been shifted in from the left.

However, notice what happens for `-9`

: `-9 >> 2`

(sign-propagating right shift) yields `-3`

, but `-9 >>> 2`

(zero-fill right shift) yields 1073741821:

-9 (base 10): 11111111111111111111111111110111 (base 2)
--------------------------------
-9 >> 2 (base 10): 11111111111111111111111111111101 (base 2) = -3 (base 10)
-9 >>> 2 (base 10): 00111111111111111111111111111101 (base 2) = 1073741821 (base 10)

Notice how two rightmost bits, `11`

, have been shifted off. For `-9 >> 2`

(sign-propagating right shift), two copies of the leftmost `1`

bit have been shifted in from the left, which preserves the negative sign. On the other hand, for `-9 >>> 2`

(zero-fill right shift), zeroes have instead been shifted in from the left, so the negative sign of the number is not preserved, and the result is instead a (large) positive number.

The left operand will be converted to an unsigned 32-bit integer, which means floating point numbers will be truncated, and number not within the 32-bit bounds will over-/underflow.

The right operand will be converted to an unsigned 32-bit integer and then taken modulo 32, so the actual shift offset will always be a positive integer between 0 and 31, inclusive. For example, `100 >>> 32`

is the same as `100 >>> 0`

(and produces `100`

) because 32 modulo 32 is 0.