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According to http://en.cppreference.com/w/cpp/utility/functional/function/function, the type of the initializer, i.e., F in form (5), should meet the requirements of CopyConstructible. I don't quite get this. Why is it not OK for F to be just MoveConstructible?
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It lets you store function pointers, lambdas, or classes with operator() . It will do conversion of compatible types (so std::function<double(double)> will take int(int) callable things) but that is secondary to its primary purpose.
Instances of std::function can store, copy, and invoke any CopyConstructible Callable target -- functions (via pointers thereto), lambda expressions, bind expressions, or other function objects, as well as pointers to member functions and pointers to data members.
std::function is a type erasure object. That means it erases the details of how some operations happen, and provides a uniform run time interface to them. For std::function , the primary1 operations are copy/move, destruction, and 'invocation' with operator() -- the 'function like call operator'.
std::function uses type erasure internally, so F has to be CopyConstructible even if the particular std::function object you are using is never copied.
A simplification on how type erasure works:
class Function
{
struct Concept {
virtual ~Concept() = default;
virtual Concept* clone() const = 0;
//...
}
template<typename F>
struct Model final : Concept {
explicit Model(F f) : data(std::move(f)) {}
Model* clone() const override { return new Model(*this); }
//...
F data;
};
std::unique_ptr<Concept> object;
public:
template<typename F>
explicit Function(F f) : object(new Model<F>(std::move(f))) {}
Function(Function const& that) : object(that.object->clone()) {}
//...
};
You have to be able to generate Model<F>::clone(), which forces F to be CopyConstructible.
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