关于“循环引用”，我使用了weak_ptr，但仍然发生内存泄漏

Question

我读到了：

How to avoid memory leak with shared_ptr?

我知道我需要使用 weak_ptr 来避免 循环引用。

所以我创建了一个小程序来播放循环引用。

以下对象（spyder）将被调用

class spyder {
 public:
  spyder(std::string _name): 
    m_name(_name), finger(nullptr)
  {  }
  inline const std::string ask_name() const{
    return m_name;
  }
  std::shared_ptr<spyder> finger;
 private:
  std::string m_name;
};

我在主代码中使用 shared_ptr 和 weak_ptr 调用 spyder：

int main(){
  auto sA = std::make_shared<spyder>("A");
  auto sB = std::make_shared<spyder>("B");
  std::weak_ptr<spyder> wp_sA(sA);
  std::weak_ptr<spyder> wp_sB(sB);  
  sA->finger = wp_sB.lock();
  sB->finger = wp_sA.lock();
}

以上代码发生内存泄漏（使用valgrind检查）。

==20753== Memcheck, a memory error detector
==20753== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==20753== Using Valgrind-3.14.0 and LibVEX; rerun with -h for copyright info
==20753== Command: ./t
==20753== 
==20753== 
==20753== HEAP SUMMARY:
==20753==     in use at exit: 128 bytes in 2 blocks
==20753==   total heap usage: 3 allocs, 1 frees, 72,832 bytes allocated
==20753== 
==20753== LEAK SUMMARY:
==20753==    definitely lost: 64 bytes in 1 blocks
==20753==    indirectly lost: 64 bytes in 1 blocks
==20753==      possibly lost: 0 bytes in 0 blocks
==20753==    still reachable: 0 bytes in 0 blocks
==20753==         suppressed: 0 bytes in 0 blocks
==20753== Rerun with --leak-check=full to see details of leaked memory
==20753== 
==20753== For counts of detected and suppressed errors, rerun with: -v
==20753== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

但是在我将上面的代码修改为：

int main(){
  spyder sA("A"), sB("B");
  std::weak_ptr<spyder> wp_sA( std::make_shared<spyder>("A") ) ;
  std::weak_ptr<spyder> wp_sB( std::make_shared<spyder>("B") );
  sA.finger = wp_sB.lock();
  sB.finger = wp_sA.lock();
}

内存泄漏还没有发生，

==20695== Memcheck, a memory error detector
==20695== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==20695== Using Valgrind-3.14.0 and LibVEX; rerun with -h for copyright info
==20695== Command: ./t
==20695== 
==20695== 
==20695== HEAP SUMMARY:
==20695==     in use at exit: 0 bytes in 0 blocks
==20695==   total heap usage: 3 allocs, 3 frees, 72,832 bytes allocated
==20695== 
==20695== All heap blocks were freed -- no leaks are possible
==20695== 
==20695== For counts of detected and suppressed errors, rerun with: -v
==20695== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

我对此感到困惑。

Answer 1

你根本没有解决循环依赖。

在您的第一个版本中，

sA 由shared_ptr 管理，并将shared_ptr 存储到sB。 sB 又由shared_ptr 管理，并将shared_ptr 存储到sA。这意味着它们的引用计数永远不会为 0。

相反，finger 应该是 std::weak_ptr 类型，并且您应该在需要使用它之前只使用 lock()。

class spyder {
 public:
  spyder(std::string _name): 
    m_name(_name), finger()
  {  }
  inline const std::string ask_name() const{
    return m_name;
  }
  std::weak_ptr<spyder> finger;
 private:
  std::string m_name;
};

int main(){
  auto sA = std::make_shared<spyder>("A");
  auto sB = std::make_shared<spyder>("B");
  sA->finger = sB;
  sB->finger = sA;
}

---

你的第二个版本

从一个临时的shared_ptr 构造weak_ptr，这意味着在构造之后你的weak_ptr 是expired()。在这个版本中sA.finger 和sB.finger 都存储nullptr（但没有内存泄漏）。

Answer 2

您的代码泄漏，因为它仍然具有循环引用。 sA -> sA::finger -> sB -> sB::finger -> sA std::weak_ptr 的要点是存储对指针的引用。为避免内存泄漏，您的类源应如下所示：

class spyder {
 public:
  spyder(std::string _name): 
    m_name(_name), finger(nullptr)
  {  }
  inline const std::string ask_name() const{
    return m_name;
  }
  std::weak_ptr<spyder> finger;
 private:
  std::string m_name;
};