Why must virtual function be implemented?

[Absurd]

When I try to compile this:

class Base {
public:
    virtual void functionA() {cout<<"Base::functionA"<<endl;}
    virtual void functionB(){cout<<"Base::functionB"<<endl;}
    virtual ~Base();
};

class Derived1: public Base {
public:
    virtual void functionA() {cout<<"Derived1::functionA"<<endl;}
};

It does not compile, saying:

undefined reference to `vtable for Base'
undefined reference to `Base::~Base()'

However, when I change the above to this:

class Base {
public:
    virtual void functionA() {cout<<"Base::functionA"<<endl;}
    virtual void functionB(){cout<<"Base::functionB"<<endl;}
    virtual ~Base() {} // notice empty implementation added
};

class Derived1: public Base {
public:
    virtual void functionA() {cout<<"Derived1::functionA"<<endl;}
};

It compiles just fine.
Why must a virtual function be implemented, but non virtual function can stay unimplemented as long as you don't call them, of course...

[phantomotap]

Why must virtual function be implemented?

O_o

The fact is that virtual functions in general don't have to be implemented.

You can search for "pure virtual" to find the information.

The destructor is kind of unique. Simply put, a destructor for a class is always called at the end of an object's lifetime.

A `Derived1` object is also object `Base` object so the lifetime of a `Derived1` is also the lifetime of a `Base` object.

The `Derived1` class destructor is calling the `Base` class destructor even if you can't see the code.

Soma

[laserlight]

Why must a virtual function be implemented, but non virtual function can stay unimplemented as long as you don't call them, of course...

The exact requirement of the standard is actually:

A virtual function declared in a class shall be defined, or declared pure in that class, or both; but no diagnostic is required.

I believe the "no diagnostic is required" part means that you're pretty much in the realm of undefined behaviour when you do declare a non-pure virtual function and yet fail to define it, and in this case your compiler/linker has chosen to issue a diagnostic.

As for why this rule exists: it is probably a consideration for expected implementations, but you'll need someone more familiar with these details to explain exactly why.