Question

我试图理解c ++临时对象生命周期扩展的语义。我尝试模拟简单的情况，感到有些惊讶。

下面我要提供代码。

#include <iostream>

struct C
{
    C(const int new_a) { a = new_a; };

    int a = 0;
};

C return_num()
{
    C num(20);

    std::cout << "From func(): num = " << num.a << ", by adress: " << &num.a << std::endl;

    return num;
}

void pass_num(const C& num)
{
    std::cout << "From func(): num = " << num.a << ", by adress: " << &num.a << std::endl;
}

int main()
{
    std::cout << "\nLifetime extention:" << std::endl;
    {
        const C& ext_num = return_num();

        std::cout << "From main(): num = " << ext_num.a << ", by adress: " << &ext_num.a << std::endl;
    }

    std::cout << "\nPassing by reference:" << std::endl;
    {
        C num(20);

        std::cout << "From main(): num = " << num.a << ", by adress: " << &num.a << std::endl;

        pass_num(num);
    }
}

主要问题是：return_num()从我的角度来看很奇怪，因为我希望在main中输出的变量地址与在return_num()内部。您能解释一下为什么不是吗？

例如，在pass_num()中的输出地址与我在main中获得的外部地址匹配。

以下是示例输出：

终身扩展：
从func（）：num = 20，按地址：0x7fff44fc8b4c
从main（）：num = 20，通过地址：0x7fff44fc8b70

通过引用传递：
从main（）：num = 20，按地址：0x7fff44fc8b6c
从func（）：num = 20，通过地址：0x7fff44fc8b6c

Answer 1

移动构造函数通常会“窃取”参数所拥有的资源（例如，指向动态分配的对象的指针，文件描述符，TCP套接字，I / O流，运行线程等），而不是复制它们并留下该参数处于某种有效但不确定的状态。

Please see Move Constructor

我在您的代码中更改了以下内容，希望它能按预期运行。我将int a更改为int* a

#include <iostream>

class C
{
   public:
   int *a;
   C( int new_a) 
   { 
      a = new int();
      *a = new_a;
   };
   C(const C& rhs) { std::cout << "Copy " << std::endl; this->a = rhs.a; }
   C(C&& rhs):a(std::move(rhs.a)) 
   {
      std::cout << "Move!!" <<"Address resource a " << &(*a) << ", Address of 
      resource rhs.a" << &(*rhs.a) << std::endl; rhs.a = nullptr;
      std::cout << "Value of a:: " << *a << std::endl;
   }  

  };

  C return_num()
  {
     C num(20);

     std::cout << "From return_num(): num = " << *num.a << ", Address of resource a : 
     "<< &(*num.a)<< std::endl;
    return (std::move(num));
  }

  void pass_num(const C& num)
  {
     std::cout << "From pass_num(): num = " << *num.a << ", by adress: " << &num.a << 
     std::endl;
  }

  int main()
  {
      std::cout << "\nLifetime extention:" << std::endl;
      {
         const C& ext_num = return_num();

         std::cout << "From main() 1 : num = " << *(ext_num.a) << ", by resource 
         adress: " << &(*ext_num.a) << std::endl;
      }

      std::cout << "\nPassing by reference:" << std::endl;
      {
         C num(20);

         std::cout << "From main() 2 : num = " << *num.a << ", by adress: " << &num.a 
         << std::endl;

         pass_num(num);
       }
       return 0;
    }

上面的代码产生以下输出：

Lifetime extention:
From return_num(): num = 20, Address of resource a : 0x7fffeca99280
Move!!Address resource a 0x7fffeca99280, Address of resource rhs.a0x7fffeca99280
Value of a:: 20
From main() 1 : num = 20, by resource adress: 0x7fffeca99280

Passing by reference:
From main() 2 : num = 20, by adress: 0x7ffff466f388
From pass_num(): num = 20, by adress: 0x7ffff466f388

希望对您有帮助！

Answer 2

想象一下这个功能：

int getNumber(){
    int num = 10;

    return num;
}

此函数不会将num作为对象返回，它会返回具有相同值的未命名副本（如果有，则为r值）。因此，它具有不同的地址。

return_num函数也会发生同样的事情。

Answer 3

我怀疑采用成员地址会抑制优化，因为编译器不知道如何处理所有可能的边缘情况。消除占用成员的地址即可优化工作。

.tolist()

终身扩展：
  从func（）：num = 20，按地址：0x7ffd61f48a50
  从main（）：num = 20，按地址：0x7ffd61f48a50

通过引用传递：
  从main（）：num = 20，按地址：0x7ffd61f48a90
  来自func（）：num = 20，按地址：0x7ffd61f48a90

c ++生命周期延长问题

3 个答案: