virtual function that is const in the base class a

问题:

Can anyone explain the output of the following code?

#include <iostream>
#include <string>
class Animal
{
public:
    Animal(const std::string & name) : _name(name) { }
    ~Animal() { }
    virtual void printMessage() const
    {
        std::cout << "Hello, I'm " << _name << std::endl;
    }

private:
    std::string _name;
    // other operators and stuff
};

class Cow : public Animal
{
public:
    Cow(const std::string & name) : Animal(name) { }
    ~Cow() { }
    virtual void printMessage()
    {
        Animal::printMessage();
        std::cout << "and moo " << std::endl;
    }
};

int main() {
    Cow cow("bill");
    Animal * animal = &cow;
    cow.printMessage();
    animal->printMessage();
}

The output is

Hello, I'm bill
and moo
Hello, I'm bill

I don't understand why. The pointer animal points at an object of type Cow. printMessage is a virtual function. Why isn't the implementation of the Cow class the one that is called?

回答1:

Cow isn't overriding the virtual function from Animal because it has a different signature. What's actually happening is that Cow is hiding the function in Animal.

The result of this is that calling printMessage on an Animal will just use the version in Animal, regardless of the one in Cow (it isn't overriding it), but calling it from Cow will use the one in Cow (because it hides the one from Animal).

To fix this, remove the const in Animal, or add the const in Cow.

In C++ 2011, you will be able to use the override keyword to avoid subtle traps like this:

class Cow : public Animal 
{ 
public: 
    Cow(const std::string & name) : Animal(name) { } 
    ~Cow() { } 
    virtual void printMessage() override
    { 
        Animal::printMessage(); 
        std::cout << "and moo " << std::endl; 
    } 
};

Notice the added override after printMessage(). This will cause the compiler to emit an error if printMessage doesn't actually override a base class version. In this case, you would get the error.

回答2:

You have two different versions of printMessage, one which is const and one which isn't. The two are unrelated, even though they have the same name. The new function in Cow hides the one in Animal, so when you call it directly the compiler only considers the Cow version.

回答3:

It took me a while to understand Peter Alexander's hiding answer, but another way to understand it is as follows:

Say that you mispelled the method name in the Cow class, but spelled it correctly in Animal class:

Animal::printMessage()
Cow::mispelledPrintMessage()

then when you have an

Animal *animal;

you can ONLY call

animal->printMessage();

but you CANNOT call

animal->mispelledPrintMessage();

because mispelledPrintMessage() doesn't exist in the Animal class. Its a brand new method in the Cow class, so it cannot be polymorphically called thru a base pointer.

So having the Animal method have const in the signature, but not in Cow method is kinda analogous to a slightly mispelled method name in the derived class.

PS: Another 4th solution (1 making both methods const, 2 making both methods non-const, or 3 using new 2011 override keyword), is to use a cast, to force the Animal pointer into a Cow pointer:

((Cow*)animal)->printMessage();

But this is a very ugly HACK, and I would not recommend it.

PS: I always try to write my toString() methods with const in the signature in the Base class and all derived classes for this very reason. Plus, having const in the toString() signature allows you call toString() either with a const or non-const object. Had you instead left out the const, and tried to pass call toString() with a const object, GCC compiler would complain with the discards qualifiers error message:

const Bad' as `this' argument of `std::string Bad::toString()' discards qualifiers

for this code:

#include <iostream>
#include <string>

class Bad
{
public:
        std::string toString()
        {
                return "";
        }
};

int main()
{
        const Bad       bad;
        std::cout << bad.toString() << "\n";
        return 0;
}

So, in conclusion, since Cow doesn't change any data members, the best solution probably is to add const to printMessage() in derived Cow class, so that both BASE Animal and DERIVED Cow classes have const in their signatures.

-dennis bednar -ahd 310

回答4:

Correction to Dennis's post.

Note you state as a solution you can cast the animal (which is a Animal*) to a Cow*. However, once a Cow* your printMessage from the Animal class will not be available. Therefore ((Cow*)animal)->printMessage() needs to be ((Cow*)animal)->mispelledPrintMessage()

回答5:

Just tried it. Add the const to the Cow class and it will work.

i.e. virtual void printMessage () const for both classes.

回答6:

virtual in subclass-->no need. You just add const to function in subclass to make it same signature