AMDG On 06/23/2012 02:25 PM, Robert Ramey wrote:
Steven Watanabe wrote:
So - boost::throw_except is changed and.... what? I already included all the useful information I can add. How can boost::throw_exception add to this on it's own? It can't. Boost.Exception allows /callers/ of your library to add their own information to the exception if they have more context than the Serialization library itself does.
agreed - but one doesn't need boost.exception to do that.
It isn't strictly necessary, but it makes things a lot easier.
try { throw enable_current_exception(x); } catch(...) { exception_ptr e = current_exception(); } // move e to another thread rethrow_exception(e);
OK - get it - but I don't see that it adds any value compared to
try { throw x; } catch(...){ exception_ptr e = std::current_exception(); std::rethrow_exception(e); // n3242 18.8} }
It doesn't. For compilers that support std::current_exception, boost::enable_current_exception can be a no-op.
The code using Boost.Exception would look like: typedef boost::error_info< struct tag_archive_filename, std::string> archive_filename; ...
try { ifstream is(filename); binary_archive ia(is); my_data d; ia >> d; } catch (boost::exception& e) { e << archive_filename(filename); throw; }
lol - now you've thrown away all the information in archive_exception!
Nope. throw without an argument rethrows the original exception object.
Use this instead.
try { ifstream is(filename); binary_archive ia(is); my_data d; ia >> d; } catch (boost::archive::exception& ae) { // assuming you love boost exception boost::exception e(ae); ae << archive_filename(filename); throw(ae); } catch(std::exception & se){ // special sauce for standard library exceptions throw ... }
Nothing in this requires any special help from boost::throw_exception. It's all in he user's program.
To make this work you have to enumerate all the possible exception types that you might have to deal with. You may not even know all the possible exception types. Also, knowing that you have something that derives from std::exception isn't enough. To preserve all information, you have to know the full dynamic type of the exception.
If you really like boost::exception your code when you eventually can eventually rethrow
catch (std::exception e){ BOOST_EXCEPTION_THROW_EXCEPTION(e) }
This of course loses the original type of the exception.
I don't think that's true. It's not clear to me from the documentation. I get this idea from looking at the code. Of course if it were true it would be huge blunder in the design of the library.
If you throw something with a static type of std::exception, it's simply not possible to know what its dynamic type is. A throw with no argument can preserve the original exception, but there's no way to use "throw;" to inject Boost.Exception.
How is this different from using std::exception ?
std::exception doesn't allow any addition annotation. Using std::exception, there's simply no way to add any additional information without either a) Knowing the type of the original exception, or b) Discarding all the type information of the original exception.
A user can create an object at the catch site to add any information he wants. He can do this using boost exception if he wants or any other system he prefers. This doesn't require any thing special from the throw site. That is, there is no reason not to use anything but simple throw which of course is the original definition of boost::throw_exception.
You're ignoring my point. Would you mind explaining how to avoid both (a) and (b)? Or pick one and explain why you think it isn't a problem? As I see it, (a) is extremely verbose if there are multiple possible exception types, and (b) can obviously lose important information. In Christ, Steven Watanabe