In my opinion, the following code (from some C++ question) should lead to UB, but the it seems it is not. Here is the code:
#include <iostream>
using namespace std;
class some{ public: ~some() { cout<<"some's destructor"<<endl; } };
int main() { some s; s.~some(); }
and the answer is:
some's destructor
some's destructor
I learned form c++ faq lite that we should not explicitly call destructor. I think after the explicitly call to the destructor, the object s should be deleted. The program automatically calls the destructor again when it’s finished, it should be UB. However, I tried it on g++, and get the same result as the above answer.
Is it because the class is too simple (no new/delete involved)? Or it’s not UB at all in this case?
The behavior is undefined because the destructor is invoked twice for the same object:
Invoking the destructor on an object whose lifetime has ended results in undefined behavior per C++03 §12.4/6:
An object’s lifetime ends when its destructor is called per §3.8/1:
Note that this means if your class has a trivial destructor, the behavior is well-defined because the lifetime of an object of such a type does not end until its storage is released, which for automatic variables does not happen until the end of the function. Of course, I don’t know why you would explicitly invoke the destructor if it is trivial.
What is a trivial destructor? §12.4/3 says:
As others have mentioned, one possible result of undefined behavior is your program appearing to continue running correctly; another possible result is your program crashing. Anything can happen and there are no guarantees whatsoever.