为什么 std::function 的初始化程序必须是可复制构造的?
根据http://en.cppreference.com/w/cpp/utility/functional/function/function，初始化器的类型，即形式(5)中的F，应满足CopyConstructible的要求.我不太明白这个.为什么 F 不能只是 MoveConstructible?
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?
推荐答案
std::function 在内部使用类型擦除，因此 F 必须是可复制构造的，即使您使用的特定 std::function 对象从未被复制.
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.
类型擦除工作原理的简化:
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()) {}
//...
};
您必须能够生成 Model，这会强制 F 为 CopyConstructible.
You have to be able to generate Model<F>::clone(), which forces F to be CopyConstructible.
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