What throw statement throws if i do not tell what type of object is to be thrown in c++?

Question

The following code terminates abnormally as no object is explicitly thrown. What is thrown by throw statement in the following code?

int main()
{
  try{
  cout<<"try";
  throw ;
}
catch(...){
 cout<<"catch";
 }
return 0;
}

Answer 1

throw without an argument should only be used inside a catch statement, to rethrow the caught exception object. You code tries to use it outside the catch statement - instead you should pick a type to throw, if in doubt it's not unreasonable to start with std::runtime_error. For more options, see here. You can also throw your own types, but it's usually a good idea to derive them from one of the Standard-library provided types so client code has a better chance at specifying appropriate handling for all logically similar errors, rather than having to catch and handle them separately and being constantly updated for each new possible error.

FWIW, the Standard says in 15.1/9:

If no exception is presently being handled, executing a throw-expression with no operand calls std::terminate().

So very explicitly, the answer to "What is thrown..." is that no throwing is done, and std::terminate is called instead.

Answer 2

So the question is: "What happens when I throw outside a catch block?" The answer to this can be found in its documentation:

Rethrows the currently handled exception. Abandons the execution of the current catch block and passes control to the next matching exception handler (but not to another catch clause after the same try block: its compound-statement is considered to have been 'exited'), reusing the existing exception object: no new objects are made. This form is only allowed when an exception is presently being handled (it calls std::terminate if used otherwise). The catch clause associated with a function-try-block must exit via rethrowing if used on a constructor.

Emphasize mine.