It is possible to deduce arity of a non-generic lambda by accessing its operator().
template <typename F>
struct fInfo : fInfo<decltype(&F::operator())> { };
template <typename F, typename Ret, typename... Args>
struct fInfo<Ret(F::*)(Args...)const> { static const int arity = sizeof...(Args); };
This is nice and dandy for something like [](int x){ return x; } as the operator() is not templated.
However, generic lambdas do template the operator() and it is only possible to access a concrete instantiation of the template - which is slightly problematic because I can't manually provide template arguments for the operator() as I don't know what its arity is.
So, of course, something like
auto lambda = [](auto x){ return x; };
auto arity = fInfo<decltype(lambda)>::arity;
doesn't work.
I don't know what to cast to nor do I know what template arguments to provide (or how many) (operator()<??>).
Any ideas how to do this?
It's impossible, as the function call operator can be a variadic template. It's been impossible to do this forever for function objects in general, and special-casing lambdas because they happened to not be equally powerful was always going to be a bad idea. Now it's just time for that bad idea to come home to roost.