Is not catching an exception undefined behavior?

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Consider the following code:

#include <iostream>

class Widget {
public:
    ~Widget() {
        std::cout << "Destructor Called!";
    }
};

void doStuff() {
    Widget w;
    throw 1;
}

int main() {
    doStuff();
}

Because the exception is not caught in main, the compiler can arrange for the program to call terminate immediately after the throw with ~Widget not called.

Since the standard does not specify whether or not the destructors are called, is this undefined behavior?

Note: The example is from this CppCon 2015 talk by Fedor Pikus.

Edit: This question asks about the compiler's freedom for destructor elision, however it is in a special, different situation. The answer to that question does not apply to this question.

undefined-behavior language-lawyer destructor exception c++
1 Answers

Is not catching an exception undefined behavior?

No, it is well-defined that this will result in a call to std::terminate (with no UB), albeit whether the stack is unwound before the call is implementation-defined, as per [except.terminate]:

/1 In some situations exception handling is abandoned for less subtle error handling techniques.

  • [...]
  • when the exception handling mechanism cannot find a handler for a thrown exception ([except.handle]), or

/2 In such cases, the function std​::​terminate is called. In the situation where no matching handler is found, it is implementation-defined whether or not the stack is unwound before std​::​terminate is called. [...] An implementation is not permitted to finish stack unwinding prematurely based on a determination that the unwind process will eventually cause a call to the function std​::​terminate

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