使用unique_ptr转发声明？ [英] Forward declaration with unique_ptr?

问题描述

我发现使用类的正向声明与 std :: unique_ptr 结合是有用的，如下面的代码。它编译和工作与GCC，但整个事情看起来很奇怪，我不知道这是否是标准的行为（即标准要求）？因为当我声明 unique_ptr 时，B不是完整的类型。

I have found it useful to use forward declaration of classes in combination with std::unique_ptr as in the code below. It compiles and works with GCC, but the whole thing seem kind of strange, and I wonder if this is standard behaviour (i.e. required by the standard)? Since B isn't a complete type when I declare the unique_ptr.

#include <memory>

class B;

class A {
    std::unique_ptr<B> myptr;
    // B::~B() can't be seen from here
public:
    ~A();
};

A.cpp

A.cpp

#include "B.hpp"
//B.hpp has to be included, otherwise it doesn't work.

A::~A() = default; // without this line, it won't compile 
// however, any destructor definiton will do.

我怀疑这与析构函数有关（因此需要调用<$

I suspect this has to do with the destructor (and therefore the need to call the destructor of unique_ptr<B>) is defined in a specific compilation unit (A.cpp).

推荐答案

这是明确合法的。规则是，在标准库中用于实例化
模板的类型必须完成，除非指定
，否则。在 unique_ptr 的情况下，§20.7.1/ 5说[...]
模板参数T的unique_ptr可能是一个不完整的类型。 ;

It's explicitly legal. The rule is that the types used to instantiate a template in the standard library must be complete, unless otherwise specified. In the case of unique_ptr, §20.7.1/5 says “[...] The template parameter T of unique_ptr may be an incomplete type.”

指针上有一些操作需要一个完整的
类型;特别是当对象实际上将被破坏时（在
处，最小值为缺省的删除者）。在你的例子中，例如，如果
A ::〜A（）是内联的，这可能会导致问题。 （注意，如果
不自己声明析构函数，它将是inline的。其中
部分地违反了使用 std :: unique_ptr 。）

There are certain operations on the pointer which require a complete type; in particular, when the object will actually be destructed (at least with the default deleter). In your example, for example, if A::~A() were inline, this might cause problems. (Note that if you don't declare the destructor yourself, it will be inline. Which partially defeats the purpose of using std::unique_ptr.)

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