Is there a std::function type or similar for lambda with auto parameter?

Question

When I assign a lambda to an explicitly typed variable (for example when it is recursive, to capture the function in itself), I use std::function.

Consider this silly "bit counting" function as an example:

std::function<int(int)> f;
f = [&f](int x){ return x ? f(x/2)+1 : 0; };

What about the case when we use an auto parameter to generalize x, as introduced in C++14 generic lambda?

std::function<int(???)> f;
f = [&f](auto x){ return x ? f(x/2)+1 : 0; };

Obviously, I can't place auto in the function type parameters.

Is there a possibility to define a functor class generically enough to cover the exact case above, but still using lambda for the function definition?

(Don't over-generalize this, only accept a single auto parameter and hard-code the return value.) The use case would be for the scenario like above: capturing the function in itself by reference for recursive calls.

Answer 1

You can create a lambda that calls itself by passing it to itself as a parameter:

auto f = [](auto self, auto x) -> int {
    return x ? self(self, x / 2) + 1 : 0;
};

std::cout << f(f, 10);

You can then capture that lambda in another lambda, so you don't have to worry about passing it to itself:

auto f2 = [&f](auto x) {
    return f(f, x);
};

std::cout << f2(10);