Impossible to use virtual keyword on destructor

user_10185183 0
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Is there any valid and usable case, that will force you not to use the virtual keyword before the destructor.

class Base {
public:
  virtual ~Base() { ... } // `virtual` causes error (not compile time or syntax) or wrong behaviour
  // could contain other fields

};

// some example
c++
2 Answers

It's possible that making the destructor virtual could convert your class from a non-polymorphic type to a polymorphic type.

Note that a polymorphic type is never trivially copyable. So, for example, you could break any use of std::memcpy by the introduction of a virtual destructor.

In some situations - particularly when interoperating with C - that can wreak havoc, as you can no longer assume that the address of an instance of your class is the same as the address of the first member.

Reference: https://en.cppreference.com/w/cpp/types/is_trivially_copyable

user_23808300

Is there any valid and usable case, that will force you not to use the virtual keyword before the destructor.

Yes. If you ever use std::memcpy or std::memcmp with instances of the class or its members, or rely on a pointer to/from an instance being convertible to the first member of the class,or examine common initial sequence of an inactive union member of the class type.

In general: If you rely on the class being a standard-layout type or trivially copyable, then the destructor (as well as all other member functions) must be non-virtual. Most cases of erroneously assuming standard-layoutness or triviality have undefined behaviour. Those properties can be enforced with type traits and static asserts, so that you get a nice compilation error instead.

user_20793030
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