Thread: Virtual & Pure virtual destructors

here it is.....
from Thinking in C++ by Bruce Eckel

While pure virtual destructors are legal in Standard C++, there is an added constraint when using them: you must provide a function body for the pure virtual destructor. This seems counterintuitive; how can a virtual function be "pure" if it needs a function body? But if you keep in mind that constructors and destructors are special operations it makes more sense, especially if you remember that all destructors in a class hierarchy are always called. If you could leave off the definition for a pure virtual destructore, what function body would be called during destruction? Thus, it's absolutely necessary that the compiler and linker enforce the existence of a function body for a pure virtual destructor.

If it's pure, but it has to have a function body, what's the value of it? The only difference you'll see between the pure and non-pure virtual destructor is that the pure virtual destructor does cause the base class to be abstract, so you cannot create an object of the base class (although this would also be true if any other member function of the base class were pure virtual).

great thread. and nice new avatar bmj

read the quote from Eckel that I posted above and then look at this code:

Code:

class test
     {
     public:
          test();
          virtual ~test()=0;
     };

class test1 : public test
     {
     public:
          test1();
          ~test1();
     };

test::test()
     {
     std::cout << "test()" << std::endl;
     }

test1::test1()
     {
     std::cout << "test1()" << std::endl;
     }
test::~test()
     {
     std::cout << "~test()" << std::endl;
     }

test1::~test1()
     {
     std::cout << "~test1()" << std::endl;
     }

This is how its supposed to be done. and yes, pure virtual functions have "= 0" at the end. It's very strange but the code for a pure virtual destructor DOES get called.

copy and paste to your compiler. I already compiled it.

then I guess Thinking in C++ is wrong then right?

virtual ~classname(){};

is not pure. It's virtual but not pure. I don't know what your book says but trust me on this one.

Thinking in C++ second edition... page 668

Originally posted by Troll_King
I've consulted 3 books so far and I have not seen that syntax for a destructor in an abstract class.

An abstract base class doesn't have to have ALL methods pure virtual. Just one. The abstract class examples you are seeing don't have to have pure virtual destructors.