Copy Assignment Operator: Returning Reference vs. Value
In C , copy assignment operators do not usually return a copy of the new object but instead return a reference or a const reference. Why is this the preferred approach?
Rationale for Returning by Reference
First, returning by reference minimizes the computational overhead associated with assignment. It involves simply copying values from one object to another, avoiding the creation and destruction of temporary objects. This is especially beneficial when dealing with complex or large objects where constructing and deleting copies can be resource-intensive.
Consequences of Returning by Value
On the other hand, if a copy assignment operator returns by value, it triggers the following sequence for each assignment:
This sequence repeats for each assignment in a chain, resulting in excessive constructor and destructor calls. Consider the following scenario:
A a1(param); A a2 = a1; A a3; a3 = a2; // Line in question
If operator= returns by value, it would require two constructor and destructor calls for assignments to a2 and a3, whereas returning by reference avoids this overhead.
Additional Considerations
Returning by reference also allows for the assignment operator to return an lvalue, enabling further modifications of the assigned object. In contrast, returning by value results in an rvalue, which cannot be modified directly.
Conclusion
While returning by reference is the standard practice for copy assignment operators in C , it is important to weigh the specific context and requirements when deciding on a suitable return type. However, for most scenarios involving efficiency and object management, returning a reference remains the preferred choice.
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