是unique_ptr< Derived>到unique_ptr< Base>自动上投? [英] Is unique_ptr<Derived> to unique_ptr<Base> up-casting automatic?

问题描述

我知道派生类unique_ptr可能发生在多态类型需要基类unique_ptr的地方.例如，从函数返回时

I know it is possible that a derived class unique_ptr can take place where base class unique_ptr is required for polymorphic types. For example, while returning from function

unique_ptr<Base> someFunction()
{
     return make_unique<Derived>(new Derived());
}

或传递给函数作为参数.

or passing to function as argument.

// Function taking unique pointer
void someOtherFunction(unique_ptr<Base>&& ptr)
// Code calling this function
someOtherFunction(std::move(ptrToDerived));

我的问题是:这种转换总是自动的吗?还是我们需要使用dynamic_cast明确执行它?

My question is: Is this upcasting always automatic? Or do we need to explicitly perform it using dynamic_cast?

推荐答案

(草稿)标准说:

// 20.8.1.2.1, constructors
...
template <class U, class E>
  unique_ptr(unique_ptr<U, E>&& u) noexcept;
template <class U>
  unique_ptr(auto_ptr<U>&& u) noexcept;

这些是 any unique_ptr中的构造函数.该标准通过如下子句进一步限制了它们的使用:

Those are constructors from any unique_ptr. The standard further restricts their usage by clauses like this:

24 备注:该构造函数不得参与重载解析，除非U*可隐式转换为T*并且D与default_delete<T>

24 Remarks: This constructor shall not participate in overload resolution unless U* is implicitly convertible to T* and D is the same type as default_delete<T>

此注释的作用是，unique_ptr<T>可以从unique_ptr<U>构造，而U*可以转换为T*(并且满足所有删除器要求).特别是当T是U的明确公共基类时.

The effect of this remark is that unique_ptr<T> is constructible from unique_ptr<U> precisely U* is convertible to T* (and all deleter requirements are met). In particular, when T is an unambiguous public base class of U.

由于构造函数不是explicit，因此它用作从unique_ptr<U>到unique_ptr<T>的隐式转换器.

Since the constructor is not explicit, it serves as an implicit converter from unique_ptr<U> to unique_ptr<T>.

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