Re: What should function returning a reference return on failure?

"Jim Langston" <>
Mon, 8 May 2006 08:43:37 -0700
"Alf P. Steinbach" <> wrote in message

* Jim Langston:

I'm sure this has been asked a few times, but I'm still not sure.

If the function returns a reference, it's guaranteeing that if it returns,
the result is a valid reference.

You have the choice of using (1) a reference to some special object
denoting "no object" or "failure", or (2) throwing an exception.

(2) is most clean, most reliable wrt. client code checking, and may help
avoid constructing a large dummy object.

Dang, there's one problem with the try...catch.

   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.

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.

That is, this code won't compile:

    int i = 1;
        int& j = i;
    catch (...)

    std::cout << j << std::endl;

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