Recursive references

Recursive data structures are data types that refers to themselves. Chances are you’re already familiar with linked lists:

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struct ListNode {
    int data;
    ListNode* next;
};

A ListNode owns some data, and contains a (non-owning, nullable) reference to the next node.

Why does this work? Because C/C++ allows to define a reference to an incomplete type. If we leave out the pointer part of next, the compiler should complain about something like this:

error: field ‘next’ has incomplete type ‘ListNode’
note: definition of ‘struct ListNode’ is not complete until the closing brace

…, which is self-explanatory: During the definition of a struct, the struct itself is seen as an incomplete type. An incomplete type has unknown size, so the compiler couldn’t figure out how much storage it’ll use.

However, references/pointers to incomplete types are allowed here, because they have fixed size, in terms of inplementations.

Ownership

For simple one-way linked lists, it is possible to replace the raw pointer with std::unique_ptr<ListNode>, so the first node automatically owns the entire list by recursion.

Recursive variant

std::variant is often employed to represent data with runtime-variable type. A recursive variant is a variant type that is able to store itself, which is useful for hierarchical data structures. For example, a property list structure could represent its node using using Node = std::variant<t1, t2, t3, ... , Self>, where Self is some type that contains another (child) Node. An instance of Node is a leaf node if it’s not of type Self.

We cannot simply write using Node = std::variant<t1, ..., Node>, since recursive using is simply not allowed. Some sort of indirection struct is required here. We could forward declare a struct that contains Node, and use it with the variant. What if:

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using Node = std::variant<t1, t2, t3, ... , struct NodeHolder>;
struct NodeHolder {
    Node node;
}; 

Again, this won’t work either. To determine sizeof(Node) would be an infinition recursion.

Simply wrap the forward declared struct inside a smart pointer:

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using Node = std::variant<t1, ..., std::unique_ptr<struct NodeHolder>>;
struct NodeHolder {
    Node node;
};

This way, the parent node has ownership over the child node (if it has one). The additional price we pay is dynamic memory allocation.

Let’s test this with some examples:

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using VPtr = std::unique_ptr<struct S>;
using V = std::variant<int, float, VPtr>;

struct S {
    V v;
};

struct Visitor {
    using std::cout;
    void operator()(const int& i) const {
        cout << "int: " << i << '\n';
    }

    void operator()(const float& f) const {
        cout << "float: " << f << '\n';
    }

    void operator()(const VPtr& p) const {
        cout << "VPtr: ";
        std::visit(*this, p->v); // recurse visitation
    }
};

int main()
{
    V v1{1};
    V v2{std::unique_ptr<S>{new S{std::move(v1)}}}; // v2 now owns v1
    V v3{std::unique_ptr<S>{new S{std::move(v2)}}}; // v3 now owns v2
    // make_unique? See: https://wg21.link/p0960

    std::visit(Visitor{}, v1);
    std::visit(Visitor{}, v3);
}

This should print:

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int: 1
VPtr: VPtr: int: 1

Inside main(), v1 is constructed, then moved into v2 as a child. v2 is again transferred to v3. Since v1 contains a plain int, the first move is equivalent to a copy. Semantically, v3 owns all the data, and is automatically released when dropped. The Visitor prints data contained in V, and if it has a child node, visit that child node recursively.

V isn’t particularly useful, it only serves as a demostration. In practice, we’re more likely to place the variant in some container, e.g.:

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using Map = std::map<std::string, std::variant<t1, t2, ..., std::unique_ptr<struct MapHolder>>>;

struct MapHolder {
    Map map;
};

Map is capable of containing instances of itself!