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Is it allowed to use a lambda function as the default value of a template argument? The code fragment in question would be:

template <typename K, typename E, typename C = [](const K& l, const K& r) { return (l < r); }> class FooBar { typedef C compare_fn; };

If it's not, why so, and what could be the most passable alternative, aside from pointers to functions? TIA.

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1 Answer 1

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No. A lambda is an object, not a type.

One way to fix this is to use a named class, as in

template<typename K, typename E, typename C = std::less<K>> class FooBar { typedef C compare_fn; };
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