Why Do C Copy Assignment Operators Return References Instead of Copies?

The Necessity of Reference/Const Reference Return in Copy Assignment Operator

The copy assignment operator in C raises questions regarding its return type. Why does it return a reference or a const reference instead of a copy of the new object? To clarify this concept, consider the following scenario:

A a1(param);
A a2 = a1;
A a3;

a3 = a2; // The problematic line

Assuming that the copy assignment operator is defined as follows:

A A::operator=(const A& a)
{
    if (this == &a)
    {
        return *this;
    }
    param = a.param;
    return *this;
}

Returning a reference from the copy assignment operator has significant advantages over returning a copy. By returning a reference, it allows for minimal work, as it only copies values from one object to another.

However, returning by value creates additional overhead. Each time the assignment operator is called, it calls a constructor and destructor, resulting in unnecessary resource consumption. For instance:

A& operator=(const A& rhs) { /* ... */ };

a = b = c; // Calls assignment operator twice. Efficient.

In contrast:

A operator=(const A& rhs) { /* ... */ };

a = b = c; // Calls assignment operator twice, calls copy constructor twice, calls destructor twice for temporary values. Inefficient.

Therefore, returning a reference or a const reference from the copy assignment operator optimizes performance by avoiding unnecessary object creation and destruction, enhancing efficiency and code maintainability.

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