C++对象池,它提供项目作为智能指针,在删除时返回到池

Question

我正在玩c ++ - 想法,并且对这个问题有点困惑.

我想要一个LIFO管理资源池的类.当请求资源(通过acquire())时,它返回对象作为unique_ptr删除后的对象,导致资源返回到池中.

单元测试将是:

// Create the pool, that holds (for simplicity, int objects)
SharedPool<int> pool;
TS_ASSERT(pool.empty());

// Add an object to the pool, which is now, no longer empty
pool.add(std::unique_ptr<int>(new int(42)));
TS_ASSERT(!pool.empty());

// Pop this object within its own scope, causing the pool to be empty
{
  auto v = pool.acquire();
  TS_ASSERT_EQUALS(*v, 42);
  TS_ASSERT(pool.empty());
}

// Object should now have returned to the pool
TS_ASSERT(!pool.empty())

---

基本实现,除了重要的最终测试外,将通过测试:

template <class T>
class SharedPool
{
 public:
  SharedPool(){}
  virtual ~SharedPool(){}

  void add(std::unique_ptr<T> t) {
    pool_.push(std::move(t));
  }

  std::unique_ptr<T> acquire() {
    assert(!pool_.empty());
    std::unique_ptr<T> tmp(std::move(pool_.top()));
    pool_.pop();
    return std::move(tmp);
  }

  bool empty() const {
    return pool_.empty();
  }

 private:
  std::stack<std::unique_ptr<T> > pool_;
};

---

问题:如何进行以便acquire()返回一个unique_ptr类型,使得删除器知道this并调用this->add(...)将资源返回池中.

Answer 1

天真的实施

该实现使用unique_ptr自定义删除器将对象返回到池中.这两个acquire和release的O(1).此外,unique_ptr自定义删除器可以隐式转换为shared_ptr.

template <class T>
class SharedPool
{
 public:
  using ptr_type = std::unique_ptr<T, std::function<void(T*)> >;

  SharedPool() {}
  virtual ~SharedPool(){}

  void add(std::unique_ptr<T> t) {
    pool_.push(std::move(t));
  }

  ptr_type acquire() {
    assert(!pool_.empty());
    ptr_type tmp(pool_.top().release(),
                 [this](T* ptr) {
                   this->add(std::unique_ptr<T>(ptr));
                 });
    pool_.pop();
    return std::move(tmp);
  }

  bool empty() const {
    return pool_.empty();
  }

  size_t size() const {
    return pool_.size();
  }

 private:
  std::stack<std::unique_ptr<T> > pool_;
};

用法示例:

SharedPool<int> pool;
pool.add(std::unique_ptr<int>(new int(42)));
pool.add(std::unique_ptr<int>(new int(84)));
pool.add(std::unique_ptr<int>(new int(1024)));
pool.add(std::unique_ptr<int>(new int(1337)));

// Three ways to express the unique_ptr object
auto v1 = pool.acquire();
SharedPool<int>::ptr_type v2 = pool.acquire();    
std::unique_ptr<int, std::function<void(int*)> > v3 = pool.acquire();

// Implicitly converted shared_ptr with correct deleter
std::shared_ptr<int> v4 = pool.acquire();

// Note that adding an acquired object is (correctly) disallowed:
// pool.add(v1);  // compiler error

您可能已经遇到了此实现的严重问题.以下用法并非不可想象:

  std::unique_ptr< SharedPool<Widget> > pool( new SharedPool<Widget> );
  pool->add(std::unique_ptr<Widget>(new Widget(42)));
  pool->add(std::unique_ptr<Widget>(new Widget(84)));

  // [Widget,42] acquired(), and released from pool
  auto v1 = pool->acquire();

  // [Widget,84] is destroyed properly, together with pool
  pool.reset(nullptr);

  // [Widget,42] is not destroyed, pool no longer exists.
  v1.reset(nullptr);
  // Memory leak

我们需要一种方法来保存删除器所需的活动信息以进行区分

1. 我应该将对象归还池吗？
2. 我应该删除实际对象吗？

这样做的一种方式(通过TC所建议的),在具有每个删除器保持weak_ptr到shared_ptr构件SharedPool.这使得删除器知道池是否已被销毁.

正确实施:

template <class T>
class SharedPool
{
 private:
  struct External_Deleter {
    explicit External_Deleter(std::weak_ptr<SharedPool<T>* > pool)
        : pool_(pool) {}

    void operator()(T* ptr) {
      if (auto pool_ptr = pool_.lock()) {
        try {
          (*pool_ptr.get())->add(std::unique_ptr<T>{ptr});
          return;
        } catch(...) {}
      }
      std::default_delete<T>{}(ptr);
    }
   private:
    std::weak_ptr<SharedPool<T>* > pool_;
  };

 public:
  using ptr_type = std::unique_ptr<T, External_Deleter >;

  SharedPool() : this_ptr_(new SharedPool<T>*(this)) {}
  virtual ~SharedPool(){}

  void add(std::unique_ptr<T> t) {
    pool_.push(std::move(t));
  }

  ptr_type acquire() {
    assert(!pool_.empty());
    ptr_type tmp(pool_.top().release(),
                 External_Deleter{std::weak_ptr<SharedPool<T>*>{this_ptr_}});
    pool_.pop();
    return std::move(tmp);
  }

  bool empty() const {
    return pool_.empty();
  }

  size_t size() const {
    return pool_.size();
  }

 private:
  std::shared_ptr<SharedPool<T>* > this_ptr_;
  std::stack<std::unique_ptr<T> > pool_;
};

Answer 2

这是一个自定义删除程序,用于检查池是否仍处于活动状态.

template<typename T>
class return_to_pool
{
  std::weak_ptr<SharedPool<T>> pool

public:
  return_to_pool(const shared_ptr<SharedPool<T>>& sp) : pool(sp) { }

  void operator()(T* p) const
  {
    if (auto sp = pool.lock())
    {
      try {
        sp->add(std::unique_ptr<T>(p));
        return;
      } catch (const std::bad_alloc&) {
      }
    }
    std::default_delete<T>{}(p);
  }
};

template <class T>
class SharedPool : std::enable_shared_from_this<SharedPool<T>>
{
public:
  using ptr_type = std::unique_ptr<T, return_to_pool<T>>;
  ...
  ptr_type acquire()
  {
    if (pool_.empty())
      throw std::logic_error("pool closed");
    ptr_type tmp{pool_.top().release(), this->shared_from_this()};
    pool_.pop();
    return tmp;
  }
  ...
};

// SharedPool must be owned by a shared_ptr for enable_shared_from_this to work
auto pool = std::make_shared<SharedPool<int>>();

Answer 3

虽然这个问题很老并且已经得到了回答，但我对@swalog 提出的解决方案有一个小小的评论。

由于双重删除，Deleter 函子可能会导致内存损坏：

void operator()(T* ptr) {
  if (auto pool_ptr = pool_.lock()) {
    try {
      (*pool_ptr.get())->add(std::unique_ptr<T>{ptr});
      return;
    } catch(...) {}
  }
  std::default_delete<T>{}(ptr);
}

unique_ptr当异常被捕获时，这里创建的将被销毁。因此，

std::default_delete<T>{}(ptr);

将导致双重删除。

它可以通过更改从 T* 创建 unique_ptr 的位置来修复：

void operator()(T* ptr) {
  std::unique_ptr<T> uptr(ptr);
  if (auto pool_ptr = pool_.lock()) {
    try {
      (*pool_ptr.get())->add(std::move(uptr));
      return;
    } catch(...) {}
  }
}