How should I define a std::function variable with default arguments?
The signature in std::function
is based on how you plan to call it and not on how you construct/assign it. Since you want to call it two different ways, you'll need to store to different std::function
objects, as in:
struct Call
{
template<typename F>
explicit Call(F f) : zero_(f), one_(std::move(f)) {}
void operator()() { zero_(); }
void operator()(int i) { one_(i); }
std::function<void()> zero_;
std::function<void(int)> one_;
};
Alternatively, you can do the type erasure yourself (what std::function
does behind the scenes) to only store the lambda once, as in:
class TECall
{
struct Concept
{
Concept() = default;
Concept(Concept const&) = default;
virtual ~Concept() = default;
virtual Concept* clone() const = 0;
virtual void operator()() = 0;
virtual void operator()(int) = 0;
};
template<typename T>
struct Model final : Concept
{
explicit Model(T t) : data(std::move(t)) {}
Model* clone() const override { return new Model(*this); }
void operator()() override { data(); }
void operator()(int i) override { data(i); }
T data;
};
std::unique_ptr<Concept> object;
public:
template<typename F>
TECall(F f) : object(new Model<F>(std::move(f))) {}
TECall(TECall const& that) : object(that.object ? that.object->clone() : nullptr) {}
TECall(TECall&& that) = default;
TECall& operator=(TECall that) { object = std::move(that.object); return *this; }
void operator()() { (*object)(); }
void operator()(int i) { (*object)(i); }
};
Don't know if that will help you, but you can store a lambda in a templated struct.
template <typename F>
struct Bar {
F foo;
Bar (F fun): foo (std::move (fun)) {}
};
auto f = [](int x = 10) {cout << x << endl;};
Bar<decltype (f)> bar (f);
bar.foo();
auto makeFun = [](){return [](int x = 10) {cout << x << endl;};};
Bar<decltype (makeFun())> bar2 (makeFun());
bar2.foo();