对复制构造函数和析构函数的无关调用

Question

[的后续行动]

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
    }
}

产出如下：

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

为了清楚地了解发生了什么，我建议在输出中包含this指针，以确定哪个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

因此，流可以解释为：

1. 临时A (…)
2. 临时A (…)cf)被复制到向量(…)中。60.临时A(
3. ) (…)是destroyed.
4. Another临时A (…)对created.
5. The向量进行扩展，并对原A (…)进行扩展。在该向量中复制到新位置(…)70)
6. The其他临时A (…)将ce复制到向量(…)中。
7. 所有不必要的A (…)拷贝60，…现在destroyed.
8. The向量被破坏了，所以A的(…)70，…71)内部也被摧毁。--

如果你这样做了，第5步就会消失。

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

产出将成为：

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