Short circuit on |= and &= assignment operators in C#

Question

I know that || and && are defined as short-circuit operators in C#, and such behaviour is guaranteed by the language specification, but do |= and &= short-circuit too?

For example:

private bool IsEven(int n) {     return n % 2 == 0; }  private void Main() {     var numbers = new int[] { 2, 4, 6, 8, 9, 10, 14, 16, 17, 18, 20 };      bool allEven = true;     bool anyOdd = false;      for (int i = 0; i < numbers.Length; i++)     {         allEven &= IsEven(numbers[i]);         anyOdd |= !IsEven(numbers[i]);     } } 

When the 9 entry is hit, allEven becomes false, meaning that all subsequent entries are irrelevant - the value of allEven is guaranteed to be false for all future calls to that expression. The same goes for anyOdd, which is set to true when it sees 9, and will remain true for all subsequent calls to that expression.

So, do &= and |= shortcut, or is IsEven guaranteed to be called on every iteration? Is there any defined behaviour in the language specification for this case? Are there any corner-cases where such short circuiting would be problematic?

Answer 1

The C# specification guarantees that both sides are evaluated exactly once from left-to-right and that no short-circuiting occurs.

5.3.3.21 General rules for expressions with embedded expressions

The following rules apply to these kinds of expressions: parenthesized expressions (§7.6.3), element access expressions (§7.6.6), base access expressions with indexing (§7.6.8), increment and decrement expressions (§7.6.9, §7.7.5), cast expressions (§7.7.6), unary +, -, ~, * expressions, binary +, -, *, /, %, <<, >>, <, <=, >, >=, ==, !=, is, as, &, |, ^ expressions (§7.8, §7.9, §7.10, §7.11), compound assignment expressions (§7.17.2), checked and unchecked expressions (§7.6.12), plus array and delegate creation expressions (§7.6.10).

Each of these expressions has one or more sub-expressions that are unconditionally evaluated in a fixed order.

The C# specification for compound operators says:

7.17.2 Compound assignment

...

An operation of the form x op= y is processed by applying binary operator overload resolution (§7.3.4) as if the operation was written x op y. Then,

• If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x op y, except that x is evaluated only once.

• Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x or the operator is a shift operator, then the operation is evaluated as x = (T)(x op y), where T is the type of x, except that x is evaluated only once.

...

In your case op is & or |. The short circuiting behavior mirrors that of &/| and not &&/||.

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Note that this only refers to behavior visible in a single threaded scenario. So if the right hand side has no side-effects that are observable in such a scenario, the compiler or JITter is still free to omit the evaluation.

In your example the compiler is free to terminate the loop once it knows the result, since there are no such side-effects. But it's not required to do so.

In particular timing does not count as such a side-effect, and you thus can't rely on your code having constant runtime. This can be problematic in a security context, since it can introduce a timing side-channel.