Question

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.

Answer 1

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;
};