Lambda expression as a class attribute?

Question

Is it possible to use a lambda expression as a class attribute? I'm working on a little game in C++ where all bots have the same update routine, but everyone should have his own optional extra update routine.

I thought something like that

class Bot
{
private:
    Lambdatype lambda;

public:
    Bot(Lambda l) {lambda = l;} 
    update() { dosomething(); lambda(); }
};

Answer 1

You can use std::function, for example assume it's void function and gets two int:

class Bot
{
private:
    using Lambda = std::function<void(int, int) >;
    Lambda lambda;

public:

    Bot(const Lambda &l) : lambda(l)
    {
    }

    void update()
    {
        //dosomething...;
        lambda(1, 2);
    }
};

int main()
{
   Bot bot ([](int x, int y){ cout << x+y << endl; });

   bot.update();
}

More generic:

template <typename L>
class Bot
{
private:
    L lambda;

public:

    Bot(const L &l) : lambda(l)
    {
    }

    void update()
    {
        //dosomething...;
        lambda(1, 2);
    }
};

int main()
{
   Bot<std::function<void(int,int)>> bot (
                         [](int x, int y){ cout << x+y << endl; }
   );
   bot.update();
}

Template based:

template <typename L>
struct Bot
{
private:
   L lambda;

public:
   Bot(const L &l) : lambda{l}  {}
   void update() {  lambda(1,2); }
};

int main()
{
    auto l = [](int x, int y){ std::cout << x + y << std::endl; };

    Bot<decltype(l)> bot(l);

    bot.update();
}

Answer 2

Addional you may use make_bot

template< class L >
struct bot
{
   bot( L l ) : lambda{l}  {}
   void update() {  lambda(1,2); }
private:
   L lambda;
};

template< class L > bot<L>  make_bot(L l ) { return {l}; }

int main()
{
   auto my_bot = make_bot( [](int x, int y){ std::cout << x + y << std::endl;} ) ;

   my_bot.update();
}