Does Overloading Operators in C Eliminate Undefined Behavior in Expressions like `i = i`?

Undefined Behavior and Sequence Points Revisited

In this sequel to the topic "Undefined Behavior and Sequence Points," we delve into the behavior of expressions involving user-defined types.

User-Defined Types and Undefined Behavior

Consider the following expression involving a user-defined type Index:

i += ++i;

The behavior of this expression with built-in types is undefined. However, does it still invoke undefined behavior if i is of type Index?

No, it does not. This is because the expression becomes equivalent to:

i.operator+=(i.operator++());

Since overloaded operators are functions, the normal sequencing rules apply. A sequence point exists after the evaluation of i.operator (), so the subsequent modification of i in i.operator =() does not violate any undefined behavior rules.

Similarly, the expressions i.add(i.inc()); and i are well-defined. The first expression is equivalent to:

i.operator+=(i.operator++());

And the second expression is equivalent to:

(i.operator++()).operator++()).operator++();

Each of these expressions has a sequence point after the evaluation of the operator () expression, ensuring that the object i is not modified twice between consecutive sequence points.

Subscript Operator Overload

The expression:

a[++i] = i;

where a is a user-defined type that overloads the subscript operator, is also well-defined. The increment operator returns an Index object, which is then used to index the a array. The assignment operator = is equivalent to the operator[]() method, which is a function call. Therefore, the sequencing rules apply, and a sequence point exists after the evaluation of i. Consequently, the expression is well-defined.

Additional Points

• The number of sequence points associated with an expression does depend on the types of operands involved, as the case of i = i demonstrates.
• In C 03, the expression i is well-defined.

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