c++ - 为什么 push_back 和 emplace_back 组合在一起时会有不同的行为

Question

在下面的代码中:

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. 当只使用push_back时，结果是预期的(移动构造函数，构造函数和析构函数被调用。)

结果:

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

1. 当只使用emplace_back时，结果是预期的(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和emplace_back结合使用时，为什么emplace_back调用了移动构造函数，而单独使用时却没有调用它) ?

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

因为当您同时使用 push_back 和 emplace_back 时，您将向 v1 添加两个元素。在第二次重新分配发生时，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。

您可以提前使用reserve来防止重新分配。

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