c++11 之emplace_back 与 push_back的区别

emplace_back函数原型：

template 
  
      void emplace_back (Args&&... args); 
  

#include 
  
    #include 
   
     #include 
    
      struct President { std::string name; std::string country; int year; President(std::string p_name, std::string p_country, int p_year) : name(std::move(p_name)), country(std::move(p_country)), year(p_year) { std::cout << "I am being constructed.\n"; } President(const President& other) : name(std::move(other.name)), country(std::move(other.country)), year(other.year) { std::cout << "I am being copy constructed.\n"; } President(President&& other) : name(std::move(other.name)), country(std::move(other.country)), year(other.year) { std::cout << "I am being moved.\n"; } President& operator=(const President& other); }; int main() { std::vector 
     
       elections; std::cout << "emplace_back:\n"; elections.emplace_back("Nelson Mandela", "South Africa", 1994); //没有类的创建 std::vector 
      
        reElections; std::cout << "\npush_back:\n"; reElections.push_back(President("Franklin Delano Roosevelt", "the USA", 1936)); std::cout << "\nContents:\n"; for (President const& president: elections) { std::cout << president.name << " was elected president of " << president.country << " in " << president.year << ".\n"; } for (President const& president: reElections) { std::cout << president.name << " was re-elected president of " << president.country << " in " << president.year << ".\n"; } } 
       
      
     
    
  

输出

emplace_back: I am being constructed. push_back: I am being constructed. I am being moved. Contents: Nelson Mandela was elected president of South Africa in 1994. 

网上有人说尽量使用emplace_back代替 push_back 有没有什么特例是不能替换的呢，搜了一下发现了一个例子：

emplace_back造成的引用失效 

勘误：window visual studio 2015 编译下面程序会出现 引用失效问题，而linux gcc 和qt 等编译环境中未出现下面问题。感谢大家的指正。

#include 
  
    #include 
   
     #include 
    
      using namespace std; int main() { vector 
     
       ivec; ivec.emplace_back(1); ivec.emplace_back(ivec.back()); for (auto it = ivec.begin(); it != ivec.end(); ++it) cout << *it << " "; return 0; } //输出： 1 - 
      
     
    
  

尝试1：不直接给emplace_back传递ivec.back()：

#include 
  
    #include 
   
     #include 
    
      using namespace std; int main() { vector 
     
       ivec; ivec.emplace_back(1); auto &it = ivec.back(); ivec.emplace_back(it); for (auto it = ivec.begin(); it != ivec.end(); ++it) cout << *it << " "; return 0; } 输出： 1 - 
      
     
    
  

尝试2：不给emplace_back传递引用：

#include 
  
    #include 
   
     #include 
    
      using namespace std; int main() { vector 
     
       ivec; ivec.emplace_back(1); auto it = ivec.back(); ivec.emplace_back(it); for (auto it = ivec.begin(); it != ivec.end(); ++it) cout << *it << " "; return 0; } 输出： 1 1 
      
     
    
  

我们如愿以偿，这时候应该可以得到结论了，ivec.back()返回的是引用，但是这个引用失效了，所以才会输出不正确；我们之前也提到过，重新分配内存会造成迭代器的失效，这里是造成了引用的失效。

#include 
  
    #include 
   
     #include 
    
      using namespace std; int main() { vector 
     
       ivec; ivec.reserve(4); ivec.emplace_back(1); ivec.emplace_back(ivec.back()); for (auto it = ivec.begin(); it != ivec.end(); ++it) cout << *it << " "; return 0; } 输出： 1 1 
      
     
    
  

参考链接：

https://blog.csdn.net/windpenguin/article/details/ 

https://blog.csdn.net/xiaolewennofollow/article/details/ 

https://blog.csdn.net/wangshubo1989/article/details/