The right shift (>>) operator returns a number or BigInt whose binary representation is the first operand shifted by the specified number of bits to the right. Excess bits shifted off to the right are discarded, and copies of the leftmost bit are shifted in from the left. This operation is also called "sign-propagating right shift" or "arithmetic right shift", because the sign of the resulting number is the same as the sign of the first operand.
Syntax
x >> y
Description
The >> operator is overloaded for two types of operands: number and BigInt. For numbers, the operator returns a 32-bit integer. For BigInts, the operator returns a BigInt. It first coerces both operands to numeric values and tests the types of them. It performs BigInt right shift if both operands become BigInts; otherwise, it converts both operands to 32-bit integers and performs number right shift. A TypeError is thrown if one operand becomes a BigInt but the other becomes a number.
Since the new leftmost bit has the same value as the previous leftmost bit, the sign bit (the leftmost bit) does not change. Hence the name "sign-propagating".
The operator operates on the left operand's bit representation in two's complement. 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)
The binary representation under two's complement of the negative decimal (base 10) number -9 is formed by inverting all the bits of its opposite number, which is 9 and 00000000000000000000000000001001 in binary, 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:
9 >> 2 yields 2:
9 (base 10): 00000000000000000000000000001001 (base 2)
--------------------------------
9 >> 2 (base 10): 00000000000000000000000000000010 (base 2) = 2 (base 10)
Notice how two rightmost bits, 01, have been shifted off, and two copies of the leftmost bit, 0 have been shifted in from the left.
-9 >> 2 yields -3:
-9 (base 10): 11111111111111111111111111110111 (base 2)
--------------------------------
-9 >> 2 (base 10): 11111111111111111111111111111101 (base 2) = -3 (base 10)
Notice how two rightmost bits, 11, have been shifted off. But as far as the leftmost bits: in this case, the leftmost bit is 1. So two copies of that leftmost 1 bit have been shifted in from the left — which preserves the negative sign.
The binary representation 11111111111111111111111111111101 is equal to the negative decimal (base 10) number -3, because all negative integers are stored as two's complements, and this one can be calculated by inverting all the bits of the binary representation of the positive decimal (base 10) number 3, which is 00000000000000000000000000000011, and then adding one.
If the left operand is a number with more than 32 bits, it will get the most significant bits discarded. For example, the following integer with more than 32 bits will be converted to a 32-bit integer:
Before: 11100110111110100000000000000110000000000001
After: 10100000000000000110000000000001
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.
For BigInts, there's no truncation. Conceptually, understand positive BigInts as having an infinite number of leading 0 bits, and negative BigInts having an infinite number of leading 1 bits.
Right shifting any number x by 0 returns x converted to a 32-bit integer. Do not use >> 0 to truncate numbers to integers; use Math.trunc() instead.
Examples
Using right shift
9 >> 2; // 2
-9 >> 2; // -3
9n >> 2n; // 2n