Can a template function deduce a lambda's arguments?

Question

A templated function that takes a function pointer can deduce the arguments of that function pointer like this:

template<class... Args>
void func(void (*ptr)(Args&& ...)) {
    //Do something useful knowing the Args...
}

Can you do the same with a lambda as the argument; without resorting to std::function or writing a metaprogramming traits class like a function_traits? i.e. deduce the arguments just using a function.

Answer 1

No, that's not possible. Template argument deduction can only deduce types, constants, and templates that are "compositionally" part of the argument types, for example, deducing void and Args... from void(*)(Args&&...) as in your example, or deducing T and N from T(&)[N]. It cannot deduce anything that doesn't appear in the type.

For a non-polymorphic lambda type T, the type of &T::operator() contains deducible information about the lambda's argument types. But T itself does not.