Why Return a Reference, Not a Value, from a C Copy Assignment Operator?

Returning by Reference in Copy Assignment Operator: A Convoluted Concept Unveiled

The concept of returning a reference from the copy assignment operator in C can be perplexing. Why not simply return a copy of the new object? To unravel this enigma, let us delve into an example involving a class A and its assignment operator:

class A {
public:
    A(int param);
    A& operator=(const A& a);

private:
    int param;
};

int main() {
    A a1(10);
    A a2 = a1;
    A a3;

    a3 = a2; // The problematic line
}

A::A(int param) : param(param) {}

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

Initially, returning a value from the assignment operator seems like a viable option. However, returning by value in this context poses several drawbacks. With each call to the assignment operator, both a constructor and destructor are invoked. Consider the following assignment chain:

a = b = c;

When you return a value, this chain triggers two calls to the assignment operator, two calls to the copy constructor, and two calls to the destructor. This wasteful process consumes unnecessary resources and offers no tangible benefits.

In contrast, returning by reference minimizes overhead significantly. Values are copied from one object to another without the need for additional constructor or destructor calls. This optimization is particularly crucial in complex scenarios where multiple assignments occur.

In summary, returning by value in the copy assignment operator offers no advantages but incurs significant performance penalties. By contrast, returning by reference ensures efficient resource utilization while maintaining the functionality of the operator.

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