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Rvalue References and Perfect Forwarding in C++0x

Wednesday, 03 December 2008

One of the new features in C++0x is the rvalue reference. Whereas the a "normal" lvalue reference is declared with a single ampersand &, an rvalue reference is declared with two ampersands: &&. The key difference is of course that an rvalue reference can bind to an rvalue, whereas a non-const lvalue reference cannot. This is primarily used to support move semantics for expensive-to-copy objects:

class X
{
    std::vector<double> data;
public:
    X():
        data(100000) // lots of data
    {}
    X(X const& other): // copy constructor
        data(other.data)   // duplicate all that data
    {}

    X(X&& other):  // move constructor
        data(std::move(other.data)) // move the data: no copies
    {}

    X& operator=(X const& other) // copy-assignment
    {
        data=other.data; // copy all the data
        return *this;
    }

    X& operator=(X && other) // move-assignment
    {
        data=std::move(other.data); // move the data: no copies
        return *this;
    }

};

X make_x(); // build an X with some data

int main()
{
    X x1;
    X x2(x1); // copy
    X x3(std::move(x1)); // move: x1 no longer has any data

    x1=make_x(); // return value is an rvalue, so move rather than copy
}

Though move semantics are powerful, rvalue references offer more than that.

Perfect Forwarding

When you combine rvalue references with function templates you get an interesting interaction: if the type of a function parameter is an rvalue reference to a template type parameter then the type parameter is deduce to be an lvalue reference if an lvalue is passed, and a plain type otherwise. This sounds complicated, so lets look at an example:

template<typename T>
void f(T&& t);

int main()
{
    X x;
    f(x);   // 1
    f(X()); // 2
}

The function template f meets our criterion above, so in the call f(x) at the line marked "1", the template parameter T is deduced to be X&, whereas in the line marked "2", the supplied parameter is an rvalue (because it's a temporary), so T is deduced to be X.

Why is this useful? Well, it means that a function template can pass its arguments through to another function whilst retaining the lvalue/rvalue nature of the function arguments by using std::forward. This is called "perfect forwarding", avoids excessive copying, and avoids the template author having to write multiple overloads for lvalue and rvalue references. Let's look at an example:

void g(X&& t); // A
void g(X& t);      // B

template<typename T>
void f(T&& t)
{
    g(std::forward<T>(t));
}

void h(X&& t)
{
    g(t);
}

int main()
{
    X x;
    f(x);   // 1
    f(X()); // 2
    h(x);
    h(X()); // 3
}

This time our function f forwards its argument to a function g which is overloaded for lvalue and rvalue references to an X object. g will therefore accept lvalues and rvalues alike, but overload resolution will bind to a different function in each case.

At line "1", we pass a named X object to f, so T is deduced to be an lvalue reference: X&, as we saw above. When T is an lvalue reference, std::forward<T> is a no-op: it just returns its argument. We therefore call the overload of g that takes an lvalue reference (line B).

At line "2", we pass a temporary to f, so T is just plain X. In this case, std::forward<T>(t) is equivalent to static_cast<T&&>(t): it ensures that the argument is forwarded as an rvalue reference. This means that the overload of g that takes an rvalue reference is selected (line A).

This is called perfect forwarding because the same overload of g is selected as if the same argument was supplied to g directly. It is essential for library features such as std::function and std::thread which pass arguments to another (user supplied) function.

Note that this is unique to template functions: we can't do this with a non-template function such as h, since we don't know whether the supplied argument is an lvalue or an rvalue. Within a function that takes its arguments as rvalue references, the named parameter is treated as an lvalue reference. Consequently the call to g(t) from h always calls the lvalue overload. If we changed the call to g(std::forward<X>(t)) then it would always call the rvalue-reference overload. The only way to do this with "normal" functions is to create two overloads: one for lvalues and one for rvalues.

Now imagine that we remove the overload of g for rvalue references (delete line A). Calling f with an rvalue (line 2) will now fail to compile because you can't call g with an rvalue. On the other hand, our call to h with an rvalue (line 3) will still compile however, since it always calls the lvalue-reference overload of g. This can lead to interesting problems if g stores the reference for later use.

Further Reading

For more information, I suggest reading the accepted rvalue reference paper and "A Brief Introduction to Rvalue References", as well as the current C++0x working draft.

Posted by Anthony Williams
[/ cplusplus /] permanent link
Tags: rvalue reference, cplusplus, C++0x, forwarding
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