Is the practice of returning a C++ reference variable evil?

Solution 1:

In general, returning a reference is perfectly normal and happens all the time.

If you mean:

int& getInt() {
    int i;
    return i;  // DON'T DO THIS.
}

That is all sorts of evil. The stack-allocated i will go away and you are referring to nothing. This is also evil:

int& getInt() {
    int* i = new int;
    return *i;  // DON'T DO THIS.
}

Because now the client has to eventually do the strange:

int& myInt = getInt(); // note the &, we cannot lose this reference!
delete &myInt;         // must delete...totally weird and  evil

int oops = getInt(); 
delete &oops; // undefined behavior, we're wrongly deleting a copy, not the original

Note that rvalue references are still just references, so all the evil applications remain the same.

If you want to allocate something that lives beyond the scope of the function, use a smart pointer (or in general, a container):

std::unique_ptr<int> getInt() {
    return std::make_unique<int>(0);
}

And now the client stores a smart pointer:

std::unique_ptr<int> x = getInt();

References are also okay for accessing things where you know the lifetime is being kept open on a higher-level, e.g.:

struct immutableint {
    immutableint(int i) : i_(i) {}

    const int& get() const { return i_; }
private:
    int i_;
};

Here we know it's okay to return a reference to i_ because whatever is calling us manages the lifetime of the class instance, so i_ will live at least that long.

And of course, there's nothing wrong with just:

int getInt() {
   return 0;
}

If the lifetime should be left up to the caller, and you're just computing the value.

Summary: it's okay to return a reference if the lifetime of the object won't end after the call.

Solution 2:

No. No, no, a thousand times no.

What is evil is making a reference to a dynamically allocated object and losing the original pointer. When you new an object you assume an obligation to have a guaranteed delete.

But have a look at, eg, operator<<: that must return a reference, or

cout << "foo" << "bar" << "bletch" << endl ;

won't work.