How should I define a std::function variable with default arguments?

Question

To set a std::function variable to a lambda function with default argument I can use auto as in:

auto foo = [](int x = 10){cout << x <<          


        

Answer 1

In C++20 you will be able to do this:

struct foo {
    decltype([](int a = 10){std::cout << a << '\n';}) my_lambda{};
};

int main() {
    foo f;
    f.my_lambda();
    f.my_lambda(5);
}

Live on Godbolt

It does look a bit strange, but it works just fine.

What makes this possible is the ability to use lambdas in unevaluated contexts and default construct stateless lambdas.