How does Attribute Inheritance in C++ Work?

问题

Suppose you wanted to reproduce the following Python snippet:

class Base:
    name = "base"

    def announce(self):
        print(self.name)

class Derived(Base):
    name = "derived"

Base().announce()
Derived().announce()

... which would output:

"base"
"derived"

Initially, you may be inclined to write something like the following:

#include <iostream>
#include <string>

struct Base {
    std::string name = "base";
    void announce() {
        std::cout << name << std::endl;
    }
};

struct Derived : public Base {
    std::string name = "movie";
};

int main() {
    Base().announce();
    Derived().announce();
    return 0;
}

But here Derived.name simply shadows Base.name. Calling Derived.announce() references Base.name and prints "base".

Is there any way to implement this behaviour? Ideally without class templates if possible.

回答1:

C++ doesn't work like Python (which isn't surprising, they are two very different languages after all), and member variables defined in an inherited class really defines a new variable that is unrelated to the variables of the parent class.

One possible solution is to create a second (possibly protected) Base constructor which takes the name as an argument, and then the Derived class can use it to initialize the member:

struct Base {
    Base() = default;  // A defaulted default constructor

    std::string name = "base";
    void announce() {
        std::cout << name << std::endl;
    }

protected:
    explicit Base(std::string name)
        : name{ std::move(name) }  // Initialize the member
    {
    }
};

struct Derived : public Base {
    Derived()
        : Base("movie")  // Initialize the base class using the special constructor
    {
    }
};

The default Base default constructor is needed, because if you declare another constructor the compiler won't automatically generate a default constructor for you.

来源：https://stackoverflow.com/questions/61314511/how-does-attribute-inheritance-in-c-work

标签

c++

inheritance