Re: What should function returning a reference return on failure?
"Alf P. Steinbach" <firstname.lastname@example.org> 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
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;
std::cout << j << std::endl;
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