emplace_back 一对对象时如何避免临时变量创建和死亡？

Question

我有一个

vector

容器，类型名称是

std::pair

，但是我发现添加元素时有很多构造/解构操作。

这是一个演示代码：

#include <iostream>
#include <vector>

using namespace std;

struct A { 
  explicit A(int a) : a_(a) {
    printf("%d created at %p\n", a_, (void *)this);
  }
  A(const A &other) : a_(other.a_) { printf("%d copied\n", a_); }
  A(A &&other) : a_(other.a_) {
    printf("%d moved into %p\n", a_, (void *)this);
  }
  A &operator=(const A &other) {
    printf("%d copy assigned\n", a_);
    a_ = other.a_;
    return *this;
  }
  A &operator=(A &&other) {
    printf("%d move assigned to %p\n", a_, (void *)this);
    a_ = other.a_;
    return *this;
  }
  ~A() { printf("%d %p dead\n", a_, (void *)this); }
  int a_; 
  void show() const { printf("%d\n", a_); }
};

int main() {
  std::vector<std::pair<A, A>> v;
  v.emplace_back(std::make_pair(A(1), A(2)));
  while(1);
}

g++ a.cpp -O2
./a.out

2 created at 0x7fffb413ca84
1 created at 0x7fffb413ca80
1 moved into 0x7fffb413ca88
2 moved into 0x7fffb413ca8c
1 moved into 0x1cf72c0
2 moved into 0x1cf72c4
2 0x7fffb413ca8c dead
1 0x7fffb413ca88 dead
1 0x7fffb413ca80 dead
2 0x7fffb413ca84 dead

如输出所示：

创建了四个临时变量，如何减少它？

Answer 1

为什么不直接写

v.emplace_back(1, 2);

而不是你目前拥有的：

v.emplace_back(std::make_pair(A(1), A(2)));

？

运行这段代码，我只得到：

1 created at 0x586bf7ac12b0
2 created at 0x586bf7ac12b4