c++ - 对复制构造函数和析构函数的无关调用

Question

[跟进 this question ]

class A
{
    public:
         A()          {cout<<"A Construction"     <<endl;}
         A(A const& a){cout<<"A Copy Construction"<<endl;}
        ~A()          {cout<<"A Destruction"      <<endl;}
};

int main() {
    {
        vector<A> t;
        t.push_back(A());
        t.push_back(A());   // once more
    }
}

输出是:

A Construction        // 1
A Copy Construction   // 1
A Destruction         // 1
A Construction        // 2
A Copy Construction   // 2
A Copy Construction   // WHY THIS?
A Destruction         // 2
A Destruction         // deleting element from t
A Destruction         // deleting element from t
A Destruction         // WHY THIS?

Answer 1

为了清楚地看到发生了什么，我建议在输出中包含 this 指针以识别哪个 A 正在调用该方法。

     A()          {cout<<"A (" << this << ") Construction"     <<endl;}
     A(A const& a){cout<<"A (" << &a << "->" << this << ") Copy Construction"<<endl;}
    ~A()          {cout<<"A (" << this << ") Destruction"      <<endl;}

我得到的输出是

A (0xbffff8cf) Construction
A (0xbffff8cf->0x100160) Copy Construction
A (0xbffff8cf) Destruction
A (0xbffff8ce) Construction
A (0x100160->0x100170) Copy Construction
A (0xbffff8ce->0x100171) Copy Construction
A (0x100160) Destruction
A (0xbffff8ce) Destruction
A (0x100170) Destruction
A (0x100171) Destruction

所以流程可以解释为:

1. 创建了临时 A (...cf)。
2. 临时 A (…cf) 被复制到 vector (…60) 中。
3. 临时A(…cf)被销毁。
4. 创建了另一个临时 A (...ce)。
5. vector被展开，那个vector中的旧A(…60)被复制到新的地方(…70)
6. 将另一个临时 A (…ce) 复制到 vector (…71) 中。
7. A (…60, …ce) 的所有不必要的拷贝现在都被销毁了。
8. vector 被破坏，所以里面的A(…70, …71)也被破坏。

如果你这样做，第 5 步将消失

    vector<A> t;
    t.reserve(2); // <-- reserve space for 2 items.
    t.push_back(A());
    t.push_back(A());

输出将变为:

A (0xbffff8cf) Construction
A (0xbffff8cf->0x100160) Copy Construction
A (0xbffff8cf) Destruction
A (0xbffff8ce) Construction
A (0xbffff8ce->0x100161) Copy Construction
A (0xbffff8ce) Destruction
A (0x100160) Destruction
A (0x100161) Destruction