为什么push_back和emplace_back结合在一起时会有不同的行为

Question

在以下代码中：

struct aa
{
   int a;
   aa( int x):a(x){ cout<<"const is called"<<endl;}
   ~aa(){cout <<"dest is called"<<endl;}
   aa( aa&& obj)
   {
       a = obj.a;
       cout<<"move const is called"<<endl;
   }
};
int main ()
{
  vector<aa> v1;
  v1.push_back(aa(9)); 
  v1.emplace_back(9);
}

1. 仅使用push_back时，可以预期结果（调用Move构造函数，构造函数和析构函数。）

结果：

const is called
move const is called
dest is called
dest is called

1. 仅使用emplace_back时，可以预期结果（不调用Move构造函数，只调用构造函数和析构函数。）

结果：

const is called
dest is called

1. 但是当我将push_back与emplace_back结合使用时，为什么要为emplace_back调用move构造函数，而在单独使用时却没有对其调用）？

结果：

const is called
move const is called
dest is called
const is called
move const is called
dest is called
dest is called
dest is called

Answer 1

因为同时使用push_back和emplace_back，所以要向v1添加两个元素。在第二次重新分配时，将创建新的内部缓冲区，并将现有元素复制/移动到新的内部缓冲区。

您可以预先使用reserve来防止重新分配。

vector<aa> v1;
v1.reserve(2);
v1.push_back(aa(9)); 
v1.emplace_back(9);