c++ - 为什么类中的静态数据成员在将其发送到函数时更新不正确？

Question

执行后 Goomba::liveGoombas 等于某个负值。我调试了它但不明白为什么它比构造函数启动更多次析构函数。为什么这里不能正常工作？

// Here is a simple Goomba class. It just keeps track of how many Goombas are alive.

class Goomba
{
public:
  static int liveGoombas;

  Goomba() { liveGoombas++; }
  ~Goomba() { liveGoombas--; }
};

int Goomba::liveGoombas = 0;

// And a Goomba legion class. Please don't change this class.
class GoombaLegion
{
public:
  void add(Goomba goomba)
  {
    goombas.push_back(goomba); //it seems that something wrong in this function
  }

private:
  std::vector<Goomba> goombas;
};

void goombas()
{
  {
    GoombaLegion legion;
  }

  // The legion went out of scope and was destroyed. But how many Goombas are alive?
  std::cout << "There are " << Goomba::liveGoombas << " live goombas" << std::endl;
}



int main()
{
  goombas();

}

Answer 1

如果您不指定自己的实现，编译器将创建其他构造函数。这些是复制构造函数，对于 C++11 和更新版本，它们是移动构造函数。

当您看到负计数时，可以通过使用编译器生成的构造函数之一来解释。因此，实例数量增加了，但 liveGoombas 没有增加。

要获得准确的计数，您应该将 Goomba 更改为如下所示。

如果您使用启用了移动语义的编译器(C++0x/C++11 和更新版本):

class Goomba
{
public:
  static int liveGoombas;

  Goomba() { liveGoombas++; }
  Goomba(const Goomba&) { liveGoombas++; }
  Goomba(Goomba&&) { liveGoombas++; }

  // Need to explicitly state we want default or it will be deleted
  // due to the above move constructor having been defined.
  Goomba & operator = (const Goomba&) = default;

  // Not really essential but including for completeness 
  Goomba & operator = (const Goomba&&) = default;

  ~Goomba() { liveGoombas--; }
};

否则:

class Goomba
{
public:
  static int liveGoombas;

  Goomba() { liveGoombas++; }
  Goomba(const Goomba&) { liveGoombas++; }

  ~Goomba() { liveGoombas--; }
};

默认赋值运算符很好，因为它们不会创建新实例，而只会修改现有实例。因此，他们不应更改实例计数。