Re: What should function returning a reference return on failure?
Jim Langston wrote:
Dang, there's one problem with the try...catch.
try
{
CMap& ThisMap = FindMap( MapNumber );
}
catch ( int )
{
LogError("Could not find map");
}
That ThisMap is only going to exist during the lifetime of the try block.
And I can't create it outside the block because it's a reference and has to
be initialized.
Don't try to use exceptions as return codes, See 17.12 and 17.13 from
the FAQ for some suggestions about the proper ways to use exceptions.
Now this means I'll have to put whole blocks of code inside the try block,
but I don't want to catch errors in a block for all the code, and a lot of
the code should execute anyway even if they can't find the map.
You won't be catching errors for all the code. You will only catch
exceptions for which you have a corresponding catch
block. Consider the following:
// Always inherit from std::exception to make your life easier.
#include <stdexcept>
class KeyNotFound : public std::exception
{
// Various members that might make diagnosing the error easier.
};
try
{
CMap& ThisMap = FindMap( MapNumber );
ThisMap.DoSomething();
}
catch (KeyNotFound &e)
{
// Handle this error.
}
Here, if FindMap throws a KeyNotFound exception, it will be caught and
handled appropriately. If any other exception is thrown, from
"ThisMap.DoSomething();" for example, then it is not caught (not by this
code, anyhow).
--
Alan Johnson