# Bitwise operators

Bitwise operators treat their operands as a sequence of 32 bits (zeroes and ones), rather than as decimal, hexadecimal, or octal `numbers`. For example, the decimal number nine has a binary representation of `1001`. Bitwise operators perform their operations on such binary representations, but they return standard JavaScript numerical values.

The following table summarizes JavaScript's bitwise operators:

Operator Usage Description
Bitwise AND `a & b` Returns a `1` in each bit position for which the corresponding bits of both operands are `1`s.
Bitwise OR `a | b` Returns a `1` in each bit position for which the corresponding bits of either or both operands are `1`s.
Bitwise XOR `a ^ b` Returns a `1` in each bit position for which the corresponding bits of either but not both operands are `1`s.
Bitwise NOT `~ a` Inverts the bits of its operand.
Left shift `a << b` Shifts `a` in binary representation `b` (< 32) bits to the left, shifting in `0`s from the right.
Sign-propagating right shift `a >> b` Shifts `a` in binary representation `b` (< 32) bits to the right, discarding bits shifted off.
Zero-fill right shift `a >>> b`   Shifts `a` in binary representation `b` (< 32) bits to the right, discarding bits shifted off, and shifting in `0`s from the left.

## Signed 32-bit integers

The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format, except for zero-fill right shift which results in an unsigned 32-bit integer.

Two's complement format means that a number's negative counterpart (e.g. `5` vs. `-5`) is all the number's bits inverted (bitwise NOT of the number, or ones' complement of the number) plus one.

For example, the following encodes the integer `314`:

```00000000000000000000000100111010
```

The following encodes `~314` (the ones' complement of `314`):

```11111111111111111111111011000101
```

Finally, the following encodes `-314`, (the two's complement of `314`):

```11111111111111111111111011000110
```

The two's complement guarantees that the left-most bit is `0` when the number is positive and `1` when the number is negative. Thus, it is called the sign bit.

The number `0` is the integer that is composed completely of 0 bits.

```0 (base 10) = 00000000000000000000000000000000 (base 2)
```

The number `-1` is the integer that is composed completely of 1 bits.

```-1 (base 10) = 11111111111111111111111111111111 (base 2)
```

The number `-2147483648` (hexadecimal representation: `-0x80000000`) is the integer that is composed completely of 0 bits, except the first (left-most) one.

```-2147483648 (base 10) = 10000000000000000000000000000000 (base 2)
```

The number `2147483647` (hexadecimal representation: `0x7fffffff`) is the integer that is composed completely of 1 bits, except the first (left-most) one.

```2147483647 (base 10) = 01111111111111111111111111111111 (base 2)
```

The numbers `-2147483648` and `2147483647` are the minimum and the maximum integers representable through a 32-bit signed number.

## Bitwise logical operators

Conceptually, the bitwise logical operators work as follows:

• The operands are converted to 32-bit integers and expressed by a series of bits (zeroes and ones). Numbers with more than 32 bits get their 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```
• Each bit in the first operand is paired with the corresponding bit in the second operand: first bit to first bit, second bit to second bit, and so on.
• The operator is applied to each pair of bits, and the result is constructed bitwise.

### & (Bitwise AND)

Performs the AND operation on each pair of bits. `a` AND `b` yields 1 only if both `a` and `b` are `1`. The truth table for the AND operation is:

a b a AND b
0 0 0
0 1 0
1 0 0
1 1 1
```.    9 (base 10) = 00000000000000000000000000001001 (base 2)
14 (base 10) = 00000000000000000000000000001110 (base 2)
--------------------------------
14 & 9 (base 10) = 00000000000000000000000000001000 (base 2) = 8 (base 10)
```

Bitwise ANDing any number `x` with `0` yields `0`.

### | (Bitwise OR)

Performs the OR operation on each pair of bits. `a` OR `b` yields 1 if either `a` or `b` is `1`. The truth table for the `OR` operation is:

a b a OR b
0 0 0
0 1 1
1 0 1
1 1 1
```.    9 (base 10) = 00000000000000000000000000001001 (base 2)
14 (base 10) = 00000000000000000000000000001110 (base 2)
--------------------------------
14 | 9 (base 10) = 00000000000000000000000000001111 (base 2) = 15 (base 10)
```

Bitwise ORing any number `x` with `0` yields `x`.

### ^ (Bitwise XOR)

Performs the XOR operation on each pair of bits. `a` XOR `b` yields 1 if `a` and `b` are different. The truth table for the `XOR` operation is:

a b a XOR b
0 0 0
0 1 1
1 0 1
1 1 0
```.    9 (base 10) = 00000000000000000000000000001001 (base 2)
14 (base 10) = 00000000000000000000000000001110 (base 2)
--------------------------------
14 ^ 9 (base 10) = 00000000000000000000000000000111 (base 2) = 7 (base 10)
```

Bitwise XORing any number `x` with `0` yields `x`.

### ~ (Bitwise NOT)

Performs the NOT operator on each bit. NOT `a` yields the inverted value (or one's complement) of `a`. The truth table for the `NOT` operation is:

a NOT a
0 1
1 0
``` 9 (base 10) = 00000000000000000000000000001001 (base 2)
--------------------------------
~9 (base 10) = 11111111111111111111111111110110 (base 2) = -10 (base 10)
```

Bitwise NOTing any number `x` yields `-(x + 1)`. For example, `~-5` yields `4`.

Note that due to using 32-bit representation for numbers both `~-1` and `~4294967295` (232-1) results in `0`.

## Bitwise shift operators

The bitwise shift operators take two operands: the first is a quantity to be shifted, and the second specifies the number of bit positions by which the first operand is to be shifted. The direction of the shift operation is controlled by the operator used.

Shift operators convert their operands to 32-bit integers in big-endian order, and return a result of the same type as the left operand. Only the low five bits of the right operand will be used.

### << (Left shift)

This operator shifts the first operand the specified number of bits to the left. Excess bits shifted off to the left are discarded. Zero bits are shifted in from the right.

For example, `9 << 2` yields 36:

```.    9 (base 10): 00000000000000000000000000001001 (base 2)
--------------------------------
9 << 2 (base 10): 00000000000000000000000000100100 (base 2) = 36 (base 10)
```

Bitwise shifting any number `x` to the left by `y` bits yields `x * 2 ** y`.
So e.g.: `9 << 3` translates to: `9 * (2 ** 3) = 9 * (8) =`` 72`.

### >> (Sign-propagating right shift)

This operator shifts the first operand the specified number of bits to the right. Excess bits shifted off to the right are discarded. Copies of the leftmost bit are shifted in from the left. 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".

For example, `9 >> 2` yields 2:

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

Likewise, `-9 >> 2` yields `-3`, because the sign is preserved:

```.    -9 (base 10): 11111111111111111111111111110111 (base 2)
--------------------------------
-9 >> 2 (base 10): 11111111111111111111111111111101 (base 2) = -3 (base 10)
```

### >>> (Zero-fill right shift)

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.

For non-negative numbers, zero-fill right shift and sign-propagating right shift yield the same result. For example, `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)
```

However, this is not the case for negative numbers. For example, `-9 >>> 2` yields 1073741821, which is different than `-9 >> 2` (which yields `-3`):

```.     -9 (base 10): 11111111111111111111111111110111 (base 2)
--------------------------------
-9 >>> 2 (base 10): 00111111111111111111111111111101 (base 2) = 1073741821 (base 10)
```

## Examples

The bitwise logical operators are often used to create, manipulate, and read sequences of flags, which are like binary variables. Variables could be used instead of these sequences, but binary flags take much less memory (by a factor of 32).

Suppose there are 4 flags:

flag A
we have an ant problem
flag B
we own a bat
flag C
we own a cat
flag D
we own a duck

These flags are represented by a sequence of bits: DCBA. When a flag is set, it has a value of `1`. When a flag is cleared, it has a value of `0`.

Suppose a variable `flags` has the binary value `0101`:

```let flags = 5;   // binary 0101
```

This value indicates:

• flag A is `true` (we have an ant problem)
• flag B is `false` (we don't own a bat)
• flag C is `true` (we own a cat)
• flag D is `false` (we don't own a duck)

Note: Since bitwise operators are 32-bit, `0101` is actually `00000000000000000000000000000101`, but the preceding zeroes can be neglected since they contain no meaningful information.

A bitmask is a sequence of bits that can manipulate and/or read flags. Typically, a "primitive" bitmask for each flag is defined:

```let FLAG_A = 1; // 0001
let FLAG_B = 2; // 0010
let FLAG_C = 4; // 0100
let FLAG_D = 8; // 1000
```

New bitmasks can be created by using the bitwise logical operators on these primitive bitmasks. For example, the bitmask `1011` can be created by ORing `FLAG_A`, `FLAG_B`, and `FLAG_D`:

```let mask = FLAG_A | FLAG_B | FLAG_D; // 0001 | 0010 | 1000 => 1011
```

Individual flag values can be extracted by ANDing them with a bitmask, where each bit with the value of one will "extract" the corresponding flag. The bitmask masks out the non-relevant flags by ANDing with zeroes (hence the term "bitmask"). For example, the bitmask `0100` can be used to see if `FLAG_C` is set:

```// if we own a cat
if (flags & FLAG_C) { // 0101 & 0100 => 0100 => true
// do stuff
}
```

A bitmask with multiple set flags acts like an "either/or". For example, the following two are equivalent:

```// if we own a bat or we own a cat
// (0101 & 0010) || (0101 & 0100) => 0000 || 0100 => true
if ((flags & FLAG_B) || (flags & FLAG_C)) {
// do stuff
}
```
```// if we own a bat or cat
let mask = FLAG_B | FLAG_C; // 0010 | 0100 => 0110
if (flags & mask) {         // 0101 & 0110 => 0100 => true
// do stuff
}
```

Flags can be set by ORing them with a bitmask, where each bit with the value one will set the corresponding flag, if that flag isn't already set. For example, the bitmask `1100` can be used to set `FLAG_C` and `FLAG_D`:

```// yes, we own a cat and a duck
let mask = FLAG_C | FLAG_D; // 0100 | 1000 => 1100
flags |= mask;              // 0101 | 1100 => 1101
```

Flags can be cleared by ANDing them with a bitmask, where each bit with the value zero will clear the corresponding flag, if it isn't already cleared. This bitmask can be created by NOTing primitive bitmasks. For example, the bitmask `1010` can be used to clear `FLAG_A` and `FLAG_C`:

```// no, we don't have an ant problem or own a cat
let mask = ~(FLAG_A | FLAG_C); // ~0101 => 1010
flags &= mask;                 // 1101 & 1010 => 1000
```

The mask could also have been created with `~FLAG_A & ~FLAG_C` (De Morgan's law):

```// no, we don't have an ant problem, and we don't own a cat
let mask = ~FLAG_A & ~FLAG_C;
flags &= mask;                 // 1101 & 1010 => 1000
```

Flags can be toggled by XORing them with a bitmask, where each bit with the value one will toggle the corresponding flag. For example, the bitmask `0110` can be used to toggle `FLAG_B` and `FLAG_C`:

```// if we didn't have a bat, we have one now,
// and if we did have one, bye-bye bat
// same thing for cats
let mask = FLAG_B | FLAG_C;
flags = flags ^ mask;          // 1100 ^ 0110 => 1010
```

Finally, the flags can all be flipped with the NOT operator:

```// entering parallel universe...
flags = ~flags;               // ~1010 => 0101
```

### Conversion snippets

Convert a binary `String` to a decimal `Number`:

```let sBinString = '1011';
let nMyNumber = parseInt(sBinString, 2);
alert(nMyNumber); // prints 11, i.e. 1011
```

Convert a decimal `Number` to a binary `String`:

```let nMyNumber = 11;
let sBinString = nMyNumber.toString(2);
alert(sBinString); // prints 1011, i.e. 11
```

You can create multiple masks from a set of `Boolean` values, like this:

```function createMask() {
let nMask = 0, nFlag = 0, nLen = arguments.length > 32 ? 32 : arguments.length;
for (nFlag; nFlag < nLen; nMask |= arguments[nFlag] << nFlag++);
}
// etc.

```

### Reverse algorithm: an array of booleans from a mask

If you want to create an `Array` of `Booleans` from a mask you can use this code:

```function arrayFromMask(nMask) {
// nMask must be between -2147483648 and 2147483647
throw new TypeError('arrayFromMask - out of range');
}
for (let nShifted = nMask; nShifted;
aFromMask.push(Boolean(nShifted & 1)), nShifted >>>= 1);
}

alert('[' + array1.join(', ') + ']');
// prints "[true, true, false, true]", i.e.: 11, i.e.: 1011
```

You can test both algorithms at the same time…

```let nTest = 19; // our custom mask

```

For the didactic purpose only (since there is the `Number.toString(2)` method), we show how it is possible to modify the `arrayFromMask` algorithm in order to create a `String` containing the binary representation of a `Number`, rather than an `Array` of `Booleans`:

```function createBinaryString(nMask) {
// nMask must be between -2147483648 and 2147483647
for (let nFlag = 0, nShifted = nMask; nFlag < 32;
nFlag++, sMask += String(nShifted >>> 31), nShifted <<= 1);
}

let string1 = createBinaryString(11);
let string2 = createBinaryString(4);
let string3 = createBinaryString(1);