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&^= operator is the bitwise clear (AND NOT) assignment operator in Go. It clears the bits of the left-hand operand based on the set bits (the 1s) of the right-hand operand, and assigns the resulting value back to the left-hand operand.
x = x &^ (y). The parentheses around y are critical because &^ has a higher precedence than lower-precedence operators like + or -. For example, an assignment like x &^= a + b evaluates as x = x &^ (a + b), not (x &^ a) + b.
Additionally, the Go specification guarantees that the expression x is evaluated only once. This is a vital distinction from manual expansion, ensuring safe and efficient execution even when x is a complex expression like a map index (m[key] &^= y), which explicitly lacks an addressable memory location in Go.
Mechanics
The operator evaluates the binary representation of both operands bit by bit.- If a bit in the right operand
yis1, the corresponding bit in the left operandxis forced to0. - If a bit in the right operand
yis0, the corresponding bit in the left operandxremains unchanged.
x &^ (y):
Bit in x | Bit in y | Result in x |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Code Visualization
To observe the bitwise manipulation, it is best to view the operands as binary literals:Type Constraints
Because&^= performs bitwise arithmetic, the Go compiler requires both operands to be of the same integer type (e.g., int, uint8, int32) or untyped constants representable by values of that type. It cannot be applied to floating-point numbers, complex numbers, or non-numeric types.
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