Functors and Inheritance

Yes, a derived class can override the operator() of a base functor class. This allows for polymorphic behavior when using functors in an inheritance hierarchy.

Here's an example:

1#include <iostream>
2
3class BaseFunc {
4 public:
5  virtual void operator()() const {
6    std::cout << "BaseFunc operator()\n"; }
7};
8
9class DerivedFunc : public BaseFunc {
10 public:
11  void operator()() const override {  
12    std::cout << "DerivedFunc operator()\n";
13  }
14};
15
16void CallFunc(const BaseFunc& func) {
17  func();  
18}
19
20int main() {
21  BaseFunc base;
22  DerivedFunc derived;
23
24  // Calls BaseFunc::operator()
25  CallFunc(base);
26
27  // Calls DerivedFunc::operator()
28  CallFunc(derived);
29}

1BaseFunc operator()
2DerivedFunc operator()

In this example, BaseFunc is the base functor class, and DerivedFunc is a derived class that overrides the operator().

The CallFunc function accepts a reference to a BaseFunc object and calls its operator(). When a DerivedFunc object is passed to CallFunc, the overridden operator() in DerivedFunc is called instead of the one in BaseFunc.

This demonstrates the polymorphic behavior of functors in an inheritance hierarchy.

Some key points to note:

• The operator() in the base class is declared as virtual to enable polymorphism.
• The operator() in the derived class is marked with override to indicate that it is overriding the base class's operator().
• The CallFunc function accepts a reference to the base functor class, allowing it to work with both the base and derived functor objects.

Functor inheritance can be useful when you need to create a family of related functors with slight variations in their behavior. The base functor class can define the common interface and default behavior, while derived functors can override or extend the base behavior as needed.