Constructor as a function try block - Exception aborts program

There's a relevant gotw

http://gotw.ca/gotw/066.htm

Basically even if you don't throw in your catch block, the exception will automatically be rethrown

If the handler body contained the statement "throw;" then the catch block would obviously rethrow whatever exception A::A() or B::B() had emitted. What's less obvious, but clearly stated in the standard, is that if the catch block does not throw (either rethrow the original exception, or throw something new), and control reaches the end of the catch block of a constructor or destructor, then the original exception is automatically rethrown.

This is normal behaviour according to the cppreference.com documentation for function-try blocks: a so-called function-try-block on a constructor or destructor must throw from its catch-clause or else there is an implicit rethrow after the catch-clause.

This makes perfect sense: the object A has not been properly constructed and hence is not in a state fit for use: it must throw an exception. You have to ensure whether the construction succeeded at the place where the object is constructed, i.e. in the case of your example in main().

Exception cannot be caught in constructor function-try-block.

n3376 15.2/15

The currently handled exception is rethrown if control reaches the end of a handler of the function-try-block of a constructor or destructor.

You should catch it in object creation place.