前向声明类的 unique_ptr 的 C++11 容器

Question

g++ -std=c++11 不会编译包含容器的类，该容器包含指向前向声明类的唯一指针。问题:

• 为什么？
• 是否有合理的解决方法？

代码示例:

#include <vector>
#include <memory>

// variant 1 (with full class definition): compiles
class Bar { [..] }; 
using BarPtr = std::unique_ptr<Bar>;

// variant 2 (with shared instead of unique pointers): compiles
using BarPtr = std::shared_ptr<class Bar>;

// variant 0 (which is what we want): compilation fails below
using BarPtr = std::unique_ptr<class Bar>;

// end of variants

class Foo {
    std::vector<BarPtr> vec;
 public:
    Foo() {} // compilation of variant 0 fails here:
             // In instantiation of ‘void std::default_delete<Bar>::operator()(Bar*) const
             // ... 
             // invalid application of ‘sizeof’ to incomplete type ‘Bar’
};

我看过How to forward declare a class to be used in a standard container of unique_ptr和 Is std::unique_ptr<T> required to know the full definition of T? , 但没有找到令人信服的答案来回答我的上述问题。

Answer 1

您需要将 Foo 的那些部分移动到实现文件中，这需要 Bar 的完整定义(请参阅 Howard Hinnant 的表格:https://stackoverflow.com/a/6089065/2173029)。按照这个指南，编译:

#include <vector>
#include <memory>

class Bar;
using BarPtr = std::unique_ptr<Bar>;

class Foo {
    std::vector<BarPtr> vec;
 public:
    Foo(); // Needs Bar in implementation
    ~Foo();// Needs Bar in implementation
};