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copied!<p>I don't think you can write the actual type in a fully type-safe way in C++, because all of the types used when instantiating a template must be complete when you instantiate the template, and what you basically need to do is something like:</p> <pre><code>typedef boost::variant<Value, std::map<Key, ExtendedValue> > ExtendedValue; typedef std::map<Key, ExtendedValue> MyMap; </code></pre> <p>This causes a recursion in the template dependencies: in order to instantiate <code>std::map</code>, <code>ExtendedValue</code> must be complete, but in order to instantiate <code>ExtendedValue</code>, <code>std::map</code> must be complete.</p> <p>You should be able to create a less type-safe version using: </p> <pre><code>typedef std::map<Key, boost::any> MyMap; </code></pre> <p>As long as all you put into the map have type <code>Value</code> or <code>MyMap</code>, this should work. (I've done something similar in Java, where the mapped type was <code>Object</code>.)</p> <p>Alternatively, you could map to a boost::variant which contained a pointer to the map, rather than the map itself. Even this is difficult to name unless you you wrap the map in a class, however. You can easily forward declare a class, and use pointers to it, but you can't declare an instantiation of a template until the types used in the template are at least known. So you'ld have to write something like:</p> <pre><code>class MyMap; typedef boost::variant<Value, MyMap*> ExtendedValue; class MyMap : private std::map<Key, ExtendedValue> { public: using ...; }; </code></pre> <p>Since you have to use pointers, this might be the way to go anyway; you could then wrap the member functions you need to ensure correct member management as well. (In which case, a function like <code>operator[]</code> would probably have to return a proxy, so that you could intercept writes, and do the allocation.)</p>

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