Flaming Dangerzone

Some pitfalls with forwarding constructors

Let’s say we’re implementing a class template similar to boost::optional. Unlike the one in boost, which was implemented with C++03 features, we’re adding a constructor that forwards arguments to the constructor of the value type.

A naïve initial definition of that class template could look like this:

template <typename T>
class optional {
    // forwarding constructor
    template <typename... U>
    optional(U&&... u);

    // copy constructor
    optional(optional<T> const& that);

    // move constructor
    optional(optional<T>&& that);

    // rest of implementation omitted

The forwarding constructor has two issues: it allows implicit conversions from any other type, and it is a copy constructor. Let’s go over each one in turn.

Implicit conversions

An implicit constructor with a variadic argument pack can be called with any number of arguments. This means it can work as a default constructor if called with zero arguments, or as a conversion constructor if called with one argument. Because it is not explicit, this means it can be called to perform implicit conversions. And it can accept any argument type! This isn’t a very good idea. It should be marked explicit to avoid accidental conversions.

    template <typename... U>
    explicit optional(U&&...);

Copy construction

How can that constructor work as a copy constructor if the class already has one? Consider the following example:

optional<int> o1;
optional<int> o2 { o1 };

Both the optional<T> const& constructor and the forwarding constructor are candidates in this case. The forwarding constructor is a candidate because the U parameter pack is deduced as a pack with optional<T>& as its single element. And here is the problem: optional<T>& is a better match because it doesn’t require a conversion. Instead of making a copy, that code tries to construct an int from an optional<int>.

How can this be solved?


We want to prevent the forwarding constructor from being selected by the overload resolution process when we want the “normal” copy constructor instead. We can achieve this by causing a substitution failure in that situation. All we need is to specify a trait that detects that situation.

We want the copy constructor to be picked any time the type involved is optional<T>, regardless of references or const-qualification. In other words, any time the bare type involved is optional<T>.

template <typename T, typename U>
struct is_related : std::is_same<Bare<T>, Bare<U>> {};

(I’m not very fond of the current name I use for this trait, but I haven’t seen a better one yet.)

Now we can use DisableIf to cause the substitution failure:

    // forwarding constructor
    template <typename... U,
              DisableIf<is_related<U, optional<T>> // oops...

But there’s a problem. U is a parameter pack. This raises issues: we only want this check to be true if the pack has size one. If it has size zero substitution will fail because is_related expects two arguments; this is undesirable. A solution would be to have is_related take one type parameter and a type parameter pack, along with a specialization for when two parameters are given.

template <typename T, typename... U>
struct is_related : std::false_type {};

template <typename T, typename U>
struct is_related<T, U> : std::is_same<Bare<T>, Bare<U>> {};

And now the forwarding constructor can be written as follows:

    // forwarding constructor
    template <typename... U,
              DisableIf<is_related<optional<T>, U...>>...>
    optional(U&&... u);