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# Bash Bitwise OR Assignment
The `|=` operator in Bash is the bitwise inclusive OR assignment operator. It performs a bitwise OR operation between a variable (the left-hand side operand) and an evaluated expression (the right-hand side operand), and immediately assigns the resulting integer back to the variable.
## Syntax
Because Bash variables are untyped strings by default, the `|=` operator must be executed within an arithmetic evaluation context.
```bash theme={"dark"}
(( variable |= expression ))
```
Alternatively, it can be used with the `let` builtin:
```bash theme={"dark"}
let "variable |= expression"
```
## Mechanics
1. **Operand Evaluation:** The right-hand side expression is fully evaluated as an integer.
2. **Bitwise OR (`|`):** Bash converts both the current value of the variable and the evaluated right-hand side into their underlying binary representations (typically 64-bit signed integers, depending on the system architecture). It then compares them bit by bit. If a bit is `1` in *either* operand, the corresponding bit in the result is set to `1`. It is set to `0` only if both bits are `0`.
3. **Assignment (`=`):** The final binary result is converted back to a base-10 integer and assigned to the left-hand side variable.
## Equivalence
The compound assignment `|=` is strictly syntactic sugar for a standard bitwise OR operation followed by an assignment:
```bash theme={"dark"}
(( x |= y ))
# is functionally identical to:
(( x = x | y ))
```
## Evaluation Example
When the operator is applied, the operation occurs at the binary level:
```bash theme={"dark"}
x=10 # Binary representation: 1010
(( x |= 6 )) # Binary representation: 0110
# Bitwise OR result: 1110 (Decimal 14)
echo $x # Outputs: 14
```
## Operator Precedence
In Bash arithmetic, the `|=` operator shares the second-lowest precedence level with other assignment operators (such as `=`, `+=`, `&=`, etc.), sitting strictly above the comma operator (`,`), which has the lowest precedence. Assignment operators evaluate right-to-left. Consequently, the entire right-hand side expression is evaluated before the bitwise OR assignment takes place.
```bash theme={"dark"}
x=2
(( x |= 1 << 2 )) # The bitwise shift (1 << 2) evaluates to 4 first.
# Then, x |= 4 is executed.
# x becomes 6.
```
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