Best way to return early from a function returning a reference

Question

Let us say we have a function of the form:

const SomeObject& SomeScope::ReturnOurObject()
{
    if( ! SomeCondition )
    {
        // return early
        return  ;
    }

    return ourObject;
}

Clearly the code above has an issue, if the condition fails then we have a problem as to how to return from this function. The crux of my question is what is the best way to handle such a situation?

Answer 1

This isn't a syntactic issue, but a design issue. You have to specify what ReturnOurObject() is supposed to return when SomeCondition is true. That depends mainly on what the function is going to be used for. And that you haven't told us.

Depending on the design issues, I see a few possible syntactic ways out of this:

• return a reference to some other object; you would have to have some ersatz object somewhere
• have a special "no-object-to-return" object somewhere that you return a reference to; clients can check for this; if they don't check they get reasonable default behavior
• return a pointer, not a reference; clients would have to always check the return value of the function
• throw an exception; if SomeCondition is something exceptional which clients can not deal with that would be appropriate
• assert; if SomeCondition should always hold, it should be asserted