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

One way you could solve this would be to wrap your std::function in a functor object which implements the default arguments for you:

struct MyFunc
{
    void operator()(int x = 10) { f(x); }
    std::function f;
};

struct Bar
{
    MyFunc foo = {[](int x){std::cout << x << "\n";}};
};

int main() {
    Bar bar;
    bar.foo();
}