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# Bash Bitwise Operators

Last modified: July 3, 2022

## 1. Overview

Bitwise operators allow simple, fast, and basic operations on the individual bits of the variables involved. They may seem a heritage of the assembly language era. Yet, we still need them in cryptography, computer graphics, hash functions, compression algorithms, encoding, network protocols, and other types of applications.

In this tutorial, we’ll explore their usage in Bash after clarifying the correspondence between numbers and the bits used to represent them.

## 2. Binary Representation of Numbers – A Quick Refresher

We can represent the same numerical quantities with different bases, using methods to convert from one base to another. In particular, we need to be familiar with binary integer numbers.

Bash performs all its arithmetic on *intmax_t* numbers with no checking for overflow. As detailed in *stdint.h*, *intmax_t* designates a signed integer type capable of representing any value between -2^63 and 2^63-1 (on 64-bit computers). **So, Bash’s bitwise operators work only on signed integers, whose binary representation is in two’s complement. **This means:

- all numbers are represented with 32 bits or 64 bits, depending on the processor
- the first bit on the left indicates the sign, which is 0 for positive numbers and 1 for negative numbers
- if the number is positive, its value in base 10 results from a summation of powers of 2
- if the number is negative, its absolute value corresponds to its opposite, with all bits inverted and to which we have added 1.

Let’s see an example 64-bit number:

```
+17 = 0000000000000000000000000000000000000000000000000000000000010001
= 2**0 + 2**4 = 1 + 16 = 17
```

Its opposite -17 requires inverting all bits and adding 1:

```
+17 = 0000000000000000000000000000000000000000000000000000000000010001
----------------------------------------------------------------
1111111111111111111111111111111111111111111111111111111111101110 +
1 =
----------------------------------------------------------------
-17 = 1111111111111111111111111111111111111111111111111111111111101111
```

**This basic understanding of binary numbers is a prerequisite for using bitwise operators.**

### 2.1. Bash’s Tools for Base Conversion

Bash’s arithmetic expansion allows us to write numbers in any base by prefixing the numeric digits of the base followed by the hash symbol. Thus, we can specify the binary number 1001 and print its value in base 10:

```
$ echo $((2#1001))
9
```

*bc* allows us to convert effortlessly between bases, using the parameters *ibase* (for the input base) and *obase* (for the output base). So, let’s convert the decimal number 9 to base 2:

```
$ echo "ibase=10;obase=2;9" | bc
1001
```

**But these simple examples apply only to unsigned integers**, while bitwise operators require the two’s complement notation for both positive and negative integers. **In addition, the internal representation of numbers has a fixed number of bits.**

### 2.2. Bash’s Representation of Positive and Negative Integers

Let’s define a function that shows us the individual bits that make up a number, positive or negative. **Our goal is to print the individual bits used internally by Bash so that we know what we are applying the bitwise operators on**:

```
# Given a decimal number, prints its two's complement with the number of bits used by Bash
twos() {
n=$(getconf LONG_BIT) # detect the machine architecture, 32bit or 64bit
printf 'obase=2; 2^%d+%d\n' "$n" "$1" | bc | sed -E "s/.*(.{$n})$/\1/"
}
```

We’ll gloss over this function’s algorithmic and mathematical details. Let’s just copy and paste the function into the terminal to test it:

```
$ twos 100
0000000000000000000000000000000000000000000000000000000001100100
$ twos -100
1111111111111111111111111111111111111111111111111111111110011100
$ twos 9223372036854775807
0111111111111111111111111111111111111111111111111111111111111111
$ twos -9223372036854775808
1000000000000000000000000000000000000000000000000000000000000000
```

**The last two numbers are the maximum and minimum integers on a 64-bit machine**.

Let’s now define the inverse function:

```
# Given a two's complement representation of a signed number, prints its decimal notation
dec() {
printf 'n=%d; ibase=2; v=%s; v-2^n*(v/2^(n-1))\n' "$(getconf LONG_BIT)" "$1"| bc
}
```

Let’s test it:

```
$ dec 1111111111111111111111111111111111111111111111111111111110011100
-100
$ dec 0000000000000000000000000000000000000000000000000000000001100100
100
```

**We will use twos() and dec() in the following examples to better illustrate the operations.**

## 3. Bitwise Operators

**Bash allows us to use bitwise operators via expr, let, declare, or arithmetic expansion.** All the following examples of bitwise

*AND*are therefore valid. But, we should be aware that many operators need to be escaped or quoted:

```
$ expr 1 \& 0
0
$ let A=1\&0
$ echo $A
0
$ declare -i B=1\&0
$ echo $B
0
$ echo $[ 1 & 0 ] # deprecated notation, but still used and working
0
$ echo $((1 & 0))
0
```

Below, we’ll use only the last type of *$((…))*, which is the most straightforward to read and doesn’t require escaping or quoting.

**Let’s not forget to take into account the operator precedence**. For instance, these two expressions are equivalent, but the second one’s less confusing since it makes the order of precedence explicit:

```
$ echo $(( 8 | ~ 4 >> 1 ^ 2 & 6 ))
-1
$ echo $(( 8 | (((~ 4) >> 1) ^ (2 & 6)) ))
-1
```

Finally, **some familiarity with the concept of truth tables would be helpful**, as the tables below of *NOT*, *AND*, *OR*, and *XOR* are the simplest and most direct ways to illustrate their operation.

### 3.1. Bitwise *NOT* Operator (*~*)

**The bitwise NOT operator inverts all bits**. Its truth table is the simplest:

```
A | NOT A
---|-------
0 | 1
1 | 0
```

The arithmetic expansion allows us to test the *NOT* operator:

```
$ echo $(( ~ -3))
2
```

Let’s check what happens to the individual bits:

```
$ twos -3
1111111111111111111111111111111111111111111111111111111111111101
```

The *NOT* operator handles bits in the following way:

```
1111111111111111111111111111111111111111111111111111111111111101
---------------------------------------------------------------- NOT
0000000000000000000000000000000000000000000000000000000000000010
```

Here is the final result:

```
$ dec 0000000000000000000000000000000000000000000000000000000000000010
2
```

As a general rule, we can observe that **applying the NOT operator is always equivalent to subtracting 1 to the opposite of the input number**. This rule is because Bash uses two’s complement, which implements the sign change by inverting all bits and adding 1.

### 3.2. Bitwise *AND* Operator (*&*, *&=*)

**The bitwise AND operator returns 1 only if both bits are equal to 1**:

```
A | B | A AND B
---|---|---------
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
```

The arithmetic expansion allows us to test the *AND* operator:

```
$ echo $(( -1000 & 20 ))
16
```

Let’s check what happens to the individual bits:

```
$ twos -1000
1111111111111111111111111111111111111111111111111111110000011000
$ twos 20
0000000000000000000000000000000000000000000000000000000000010100
```

The *AND* operator handles bits in the following way:

```
1111111111111111111111111111111111111111111111111111110000011000
0000000000000000000000000000000000000000000000000000000000010100
---------------------------------------------------------------- AND
0000000000000000000000000000000000000000000000000000000000010000
```

Here is the final result:

```
$ dec 0000000000000000000000000000000000000000000000000000000000010000
16
```

The *&=* operator performs bitwise *AND* with a variable and stores the result in that variable:

```
$ A=-1000;
$ (( A &= 20 ))
$ echo $A
16
```

Looking at the truth table and knowing that the sign is the first bit on the left, it follows that the output will be a negative number only if the two input numbers are both negative.

### 3.3. Bitwise *OR* Operator (*|*, *|=*)

**The bitwise OR operator returns 1 if at least one of the two bits is equal to 1**:

```
A | B | A OR B
---|---|--------
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 1
```

Let’s use the arithmetic expansion to test the *OR* operator:

```
echo $(( 123 | -321 ))
-257
```

Next, let’s see what happens to the individual bits:

```
$ twos 123
0000000000000000000000000000000000000000000000000000000001111011
$ twos -321
1111111111111111111111111111111111111111111111111111111010111111
```

The *OR* operator handles bits in the following way:

```
0000000000000000000000000000000000000000000000000000000001111011
1111111111111111111111111111111111111111111111111111111010111111
---------------------------------------------------------------- OR
1111111111111111111111111111111111111111111111111111111011111111
```

Here is the final result:

```
$ dec 1111111111111111111111111111111111111111111111111111111011111111
-257
```

The *|=* operator performs bitwise *OR* with a variable and stores the result in that variable:

```
$ A=123;
$ (( A |= -321 ))
$ echo $A
-257
```

In this case, the output will be a positive number only if the two input numbers are both positive.

### 3.4. Bitwise *XOR* Operator (*^*, *^=*)

**The bitwise XOR operator returns 1 if only one of the two bits is equal to 1**:

```
A | B | A XOR B
---|---|--------
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
```

The arithmetic expansion allows us to test the *XOR* operator:

```
echo $(( -314 ^ 537 ))
-801
```

Let’s look at the individual bits:

```
$ twos -314
1111111111111111111111111111111111111111111111111111111011000110
$ twos 537
0000000000000000000000000000000000000000000000000000001000011001
```

The *XOR* operator handles bits in the following way:

```
<code class="language-bash">1111111111111111111111111111111111111111111111111111111011000110
0000000000000000000000000000000000000000000000000000001000011001
---------------------------------------------------------------- XOR
1111111111111111111111111111111111111111111111111111110011011111
```

Here is the final result:

```
$ dec 1111111111111111111111111111111111111111111111111111110011011111
-801
```

The *^=* operator performs bitwise *XOR* with a variable and stores the result in that variable:

```
$ A=-314;
$ (( A ^= 537 ))
$ echo $A
-801
```

In this case, the output will be a positive number only if the two input numbers are both positive or negative. As a general rule, **writing C=$((A ^ B)) is identical to C=$((A & ~B | ~A & B ))**. This is proved by starting from the truth table and applying the Karnaugh map simplification technique.

### 3.5. Left Bitwise Shift (*<<*, *<<=*)

In general, in various programming languages, there are two types of bitwise shifts: logical shifts and arithmetic shifts. Both shift bits left or right. The difference is that **arithmetic shifts preserve the sign in two’s complement notation (except in overflow cases)**, while logical shifts don’t. Some languages offer both types of shifts, while **Bash offers arithmetic shifts exclusively**.

**Bash’s shift is “arithmetic” because it is an arithmetic operation – a multiplication in the left shift case**. The operator takes the number of bits to shift as the second argument. If *n* is the number of bits to be shifted and *x* an integer, then** $((x<<n)) and $((x*2**n)), that is x multiplied by the nth power of 2,**

**are the same operation**:

```
$ n=3
$ x=-1000
$ echo $(( x << n ))
-8000
$ echo $(( x * 2**n ))
-8000
$ x=-8070450532247928857
$ echo $(( x << n ))
9223372036854775608
$ echo $(( x * 2**n ))
9223372036854775608
```

Let’s note that the sign is only kept in the first case because *x*2**n* doesn’t generate overflow. In the second case, by contrast, the multiplication result would be a number smaller than the minimum possible integer, that is, *-2**63* on a 64-bit CPU.

Let’s check what happens to the individual bits in the first case:

```
$ twos -1000
1111111111111111111111111111111111111111111111111111110000011000
```

The left shift operator handles bits in the following way:

- starting from the left, the first bit remains as it is because it’s the sign
- starting from the left, the second, third, and fourth bits leave because the 3-bit shift results in their elimination
- three new bits equal to 0 come to the right (the left shift never adds bits equal to 1)

Here’s the final result:

`1111111111111111111111111111111111111111111111111110000011000000`

We can see that it’s the integer we expect:

```
$ dec 1111111111111111111111111111111111111111111111111110000011000000
-8000
```

The *<<=* operator performs a left shift with a variable and stores the result in that variable:

```
$ A=-1000
$ (( A <<= 3 ))
$ echo $A
-8000
```

As a final note, shifts can result in an implementation-defined or undefined behavior, so we must take care when using them. What we’ve covered here takes into account the operation of Bash 5.

### 3.6. Right Bitwise Shift (*>>*, *>>=*)

The same considerations already expressed for the left bitwise shift apply, i.e., **it’s an arithmetic shift with sign preservation** (in this case, there is no possibility of overflow).

The left shift corresponds to a division. ** $((x>>n)) and $((x/2**n)) are the same operation if x is positive, or if x is negative and a multiple of -2**n. Instead, if x is negative and not a multiple of -2**n, then $((x>>n)) equals $((x/2**n-1))**.

Let’s see two examples:

```
$ x=1030
$ n=3
$ echo $(( x >> n ))
128
$ echo $(( x / 2**n))
128
$ x=-1030
$ echo $(( x >> n ))
-129
$ echo $(( x / 2**n))
-128
```

Let’s check what happens to the individual bits in the second case:

```
$ twos -1030
1111111111111111111111111111111111111111111111111111101111111010
```

The right shift operator handles bits in the following way:

- starting from the right, the first, second, and third bits leave because the 3-bit shift results in their elimination
- starting from the left, the first bit remains as is because it’s the sign
- three new bits equal to 1 come to the left after the first bit (the right shift adds 0s for positive integers and 1s for negative integers)

Here is the final result:

`1111111111111111111111111111111111111111111111111111111101111111`

It’s the expected integer:

```
$ dec 1111111111111111111111111111111111111111111111111111111101111111
-129
```

The *>>=* operator performs a left shift with a variable and stores the result in that variable:

```
$ A=-1030
$ (( A >>= 3 ))
$ echo $A
-129
```

As with the left shift, we must be careful using this operator since it can result in an implementation-defined or undefined behavior. What we’ve seen takes into account the operation of Bash 5.

## 4. Conclusion

In this article, we’ve seen how Bash represents numbers by bits and what basic operations we can do on these bits.

**Knowledge of the basic rules of Boolean algebra is not strictly necessary, but it will undoubtedly help us to understand bitwise operators better**.