Question

Is it possible to do something like this:

template <typename T, typename Comparator = std::less<typename T::Key>>
struct Container
{
    std::map<typename T::Key, T*, Comparator> m;
};

This is the problem part: typename Comparator = std::less<typename T::Key>

i.e. have a typedef within the implementation of the T be used for the specialization of the comparator. Is it always necessary that the thing that the comparator uses (T::Key) be specified as an explicit parameter to the template?

I realize there are other ways around this by changing the T class, (e.g. require an explicit T::compareKey method), but the question is more around whether template syntax can be used to achieve something like this.

Answer 1

I’m sure that’s valid, as long as all the necessary definitions are available when the template is instantiated, and it certainly works for me.

Are you including <map>? It’s possible that you might also need to include <functional> for std::less, although I’m sure that must be included by <map>.

Does your compiler support C++11, and have you enabled that support? Older compilers (or those configured for strict C++03 compliance) will get confused by your use of >> to close two template parameter lists; they will interpret it as the right-shift operator. In that case, you can fix the problem by adding a space between them.

If neither of these are the problem, please include a minimal, complete example that demonstrates the problem, the exact error message with an indication of which code lines it refers to, and let us know which compiler you’re using.