Question

If I have a code like the following:

try {
  doSomething();
} catch (...) {
  noteError();
}

void noteError() {
  try {
    throw;
  } catch (std::exception &err) {
    std::cerr << "Note known error here: " << err.what();
  } catch (...) {
    std::cerr << "Note unknown error here.";
  }
  throw;
}

Will the original exceptions get thrown from both places inside the lower frame of noteError()?

Answer 1

The wording in the standard (§15.1/2) is (emphasis mine):

When an exception is thrown, control is transferred to the nearest handler with a matching type (15.3); “nearest” means the handler for which the compound-statement, ctor-initializer, or function-body following the try keyword was most recently entered by the thread of control and not yet exited.

When has a try block “exited”? According to the grammar (§15/1), try blocks end with a sequence of handlers, so the block ends when the last handler ends. In other words:

try // <- start of try block
{
}
catch (whatever) // <- first handler
{
}
// ... more handlers
catch (whatever_again) // <- last handler
{
} // <- end of try block

So yes, your code is fine. When re-thrown, the nearest try block has a matching handler (namely catch (...)), so that handler is entered.