Question

I’ve got a C++ data-structure that is a required “scratchpad” for other computations. It’s not long-lived, and it’s not frequently used so not performance critical. However, it includes a random number generator amongst other updatable tracking fields, and while the actual value of the generator isn’t important, it is important that the value is updated rather than copied and reused. This means that in general, objects of this class are passed by reference.

If an instance is only needed once, the most natural approach is to construct them whereever needed (perhaps using a factory method or a constructor), and then passing the scratchpad to the consuming method. Consumers’ method signatures use pass by reference since they don’t know this is the only use, but factory methods and constructors return by value – and you can’t pass unnamed temporaries by reference.

Is there a way to avoid clogging the code with nasty temporary variables? I’d like to avoid things like the following:

scratchpad_t<typeX<typeY,potentially::messy>, typename T> useless_temp = factory(rng_parm);
xyz.initialize_computation(useless_temp);

I could make the scratchpad intrinsically mutable and just label all parameters const &, but that doesn’t strike me as best-practice since it’s misleading, and I can’t do this for classes I don’t fully control. Passing by rvalue reference would require adding overloads to all consumers of scratchpad, which kind of defeats the purpose – having clear and concise code.

Given the fact that performance is not critical (but code size and readability are), what’s the best-practice approach to passing in such a scratchpad? Using C++0x features is OK if required but preferably C++03-only features should suffice.

Edit: To be clear, using a temporary is doable, it’s just unfortunate clutter in code I’d like to avoid. If you never give the temporary a name, it’s clearly only used once, and the fewer lines of code to read, the better. Also, in constructors’ initializers, it’s impossible to declare temporaries.

Answer 1

While it is not okay to pass rvalues to functions accepting non-const references, it is okay to call member functions on rvalues, but the member function does not know how it was called. If you return a reference to the current object, you can convert rvalues to lvalues:

class scratchpad_t
{
    // ...

public:

    scratchpad_t& self()
    {
        return *this;
    }
};

void foo(scratchpad_t& r)
{
}

int main()
{
    foo(scratchpad_t().self());
}

Note how the call to self() yields an lvalue expression even though scratchpad_t is an rvalue.

Please correct me if I’m wrong, but Rvalue reference parameters don’t accept lvalue references so using them would require adding overloads to all consumers of scratchpad, which is also unfortunate.

Well, you could use templates…

template <typename Scratch> void foo(Scratch&& scratchpad)
{
    // ...
}

If you call foo with an rvalue parameter, Scratch will be deduced to scratchpad_t, and thus Scratch&& will be scratchpad_t&&.

And if you call foo with an lvalue parameter, Scratch will be deduced to scratchpad_t&, and because of reference collapsing rules, Scratch&& will also be scratchpad_t&.

Note that the formal parameter scratchpad is a name and thus an lvalue, no matter if its type is an lvalue reference or an rvalue reference. If you want to pass scratchpad on to other functions, you don’t need the template trick for those functions anymore, just use an lvalue reference parameter.

By the way, you do realize that the temporary scratchpad involved in xyz.initialize_computation(scratchpad_t(1, 2, 3)); will be destroyed as soon as initialize_computation is done, right? Storing the reference inside the xyz object for later user would be an extremely bad idea.

self() doesn’t need to be a member method, it can be a templated function

Yes, that is also possible, although I would rename it to make the intention clearer:

template <typename T>
T& as_lvalue(T&& x)
{
    return x;
}