Under what circumstances are C++ destructors not going to be called?

WilliamKF picture WilliamKF · Jul 5, 2010 · Viewed 15.5k times · Source

I know that my destructors are called on normal unwind of stack and when exceptions are thrown, but not when exit() is called.

Are there any other cases where my destructors are not going to get called? What about signals such as SIGINT or SIGSEGV? I presume that for SIGSEGV, they are not called, but for SIGNINT they are, how do I know which signals will unwind the stack?

Are there any other circumstances where they will not be called?

Answer

stinky472 picture stinky472 · Jul 5, 2010

Are there any other circumstances where they[destructors] will not be called?

  1. Long jumps: these interfere with the natural stack unwinding process and often lead to undefined behavior in C++.
  2. Premature exits (you already pointed these out, though it's worth noting that throwing while already stack unwinding as a result of an exception being thrown leads to undefined behavior and this is why we should never throw out of dtors)
  3. Throwing from a constructor does not invoke the dtor for a class. This is why, if you allocate multiple memory blocks managed by several different pointers (and not smart pointers) in a ctor, you need to use function-level try blocks or avoid using the initializer list and have a try/catch block in the ctor body (or better yet, just use a smart pointer like scoped_ptr since any member successfully initialized so far in an initializer list will be destroyed even though the class dtor will not be called).
  4. As pointed out, failing to make a dtor virtual when a class is deleted through a base pointer could fail to invoke the subclass dtors (undefined behavior).
  5. Failing to call matching operator delete/delete[] for an operator new/new[] call (undefined behavior - may fail to invoke dtor).
  6. Failing to manually invoke the dtor when using placement new with a custom memory allocator in the deallocate section.
  7. Using functions like memcpy which only copies one memory block to another without invoking copy ctors. mem* functions are deadly in C++ as they bulldoze over the private data of a class, overwrite vtables, etc. The result is typically undefined behavior.
  8. Instantiation of some of smart pointers (auto_ptr) on an incomplete type, see this discussion