问题
I have
struct IMyInterface
{
virtual method1() = 0;
virtual method2() = 0;
};
GCC insists that I have
struct IMyInterface
{
virtual method1() = 0;
virtual method2() = 0;
virtual ~IMyInterface(){};
};
I dont see why. A pure interface is all about the interface (duh). The destructor is part of the internal implementation details of a concrete implementer of the interface; it does not form part of the interface. I understand the whole slicing issue (or at least I think I do)
So my question is - is GCC right to insist on it and if so why?
回答1:
According to the C++ spec, yes.
You need to declare the destructor virtual because otherwise, later
IMyInterface * ptr = getARealOne();
delete ptr;
won't call the destructor on the derived class (because the destructor isn't in the VTable)
It needs to be non-pure because base class destructors are always called by the sub-class destructor.
To further explain, C++ doesn't have a concept of an interface in the same way that Java or C# do. It's just a convention to use only pure-virtual methods, and think of that as an interface. The other rules about C++ destructors make it need to be non-pure, which breaks the similarity to interfaces in other languages, but those languages didn't exist at the time these rules were made.
回答2:
If you don't declare the virtual d'tor in the base class, deleting objects of derived classes through a pointer to the base class leads to the wrong destructor being called, and thus to undefined behaviour and resource leaking.
struct A {
virtual ~A() {}
};
struct B : A {
std::string us_constitution;
};
B* pb = new B();
A* pa = pb;
delete pa; // without the virtual d'tor in the base class, 'B::us_constitution' would never be freed.
来源:https://stackoverflow.com/questions/3336499/virtual-desctructor-on-pure-abstract-base-class