我有这样的代码：

#include <iostream>
#include <exception>
#include <stdexcept>
#include <string>
#include <vector>
#include <utility>
     
struct Foo {
    Foo(std::string t):text(t){}
    //destructor deleted
        
    std::string text;
};

int main()
{
    std::vector<Foo> v;
        
    v.push_back(Foo("foo 1"));
    v.push_back(Foo("foo 2"));
    v.push_back(Foo("foo 3"));
      
    for(auto& foo : v){
        std::cout<<foo.text<<"\n";
    }
      
    Foo fooBack = std::move(v.back());
      
    std::cout<<fooBack.text<<"\n";
      
    for(auto& foo : v){
        std::cout<<foo.text<<"\n";
    }
}

在这种情况下，std::move()不会窃取返回变量。有没有办法避免过度复制？
这就是结果：

foo 1
foo 2
foo 3
foo 3
foo 1
foo 2
foo 3

基于sugestions的新代码（不知道是否继续在同一个线程上问不知道这是否是我所期望的，我做这一切的原因是为了减少代码）：

#include <iostream>
#include <exception>
#include <stdexcept>

#include<string>
#include<vector>
#include<utility>
 
struct Foo {
        Foo(std::string t):text(t){
            std::cout<<text<<" construct foo called \n";
        }
        Foo(const Foo& f){
            text = f.text;
            std::cout<<text<<"copy construct foo called \n";
        }
        Foo(Foo&& f){
            text = std::move(f.text);
            std::cout<<text<<"move construct foo called \n";
        
        }
        /*
    ~Foo() {
    }
    */
    
    std::string text;
};
int main()
{
    
    std::vector<Foo> v;
    
  v.emplace_back("foo 1");
  v.emplace_back("foo 2");
  v.emplace_back("foo 3");
  
  for(auto&& foo : v){
    std::cout<<foo.text<<"\n";
  }
  
  Foo fooBack = std::move(v.back());
  v.pop_back();
  
  std::cout<<fooBack.text<<"\n";
  
  for(auto&& foo : v){
    std::cout<<foo.text<<"\n";
  }
  
}

新的结果来看看它的行动

foo 1 construct foo called 
foo 2 construct foo called 
foo 1copy construct foo called 
foo 3 construct foo called 
foo 1copy construct foo called 
foo 2copy construct foo called 
foo 1
foo 2
foo 3
foo 3move construct foo called 
foo 3
foo 1
foo 2