C++生命周期延长的问题

Question

我试图理解 C++ 临时对象生命周期延长的语义。我试图模拟简单的情况，但有点惊讶。

下面我提供我的代码。

#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);
    }
}

这里是主要问题：从我的角度来看，return_num() 工作起来很奇怪，因为我预计我试图在 main 中输出的变量地址将是与 return_num() 中的内部相同。你能解释一下为什么不是吗？

例如 pass_num() 中的输出地址匹配我在 main 中得到的外部地址。

这是示例输出：

寿命延长：
从 func(): num = 20, by address: 0x7fff44fc8b4c
来自 main(): num = 20, by address: 0x7fff44fc8b70

通过引用：
从 main(): num = 20, by address: 0x7fff44fc8b6c
从 func(): num = 20, by address: 0x7fff44fc8b6c

Answer 1

想象一下这个函数：

int getNumber(){
    int num = 10;

    return num;
}

此函数不 return num 作为对象，它 return 是它的 no-named 副本（r-value，如果你愿意的话）相同的值。因此，它有一个不同的地址。

您的 return_num 函数也会发生同样的事情。

Answer 2

我怀疑获取成员的地址会抑制优化，因为编译器不知道如何处理所有可能的边缘情况。消除取会员地址使优化工作。

#include <iostream>

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

    int a = 0;
    struct C* t = this;
};

C return_num()
{
    C num(20);

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

    return num;
}

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

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

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

        pass_num(num);
    }
}

Lifetime extention:
From func(): num = 20, by adress: 0x7ffd61f48a50
From main(): num = 20, by adress: 0x7ffd61f48a50

Passing by reference:
From main(): num = 20, by adress: 0x7ffd61f48a90
From func(): num = 20, by adress: 0x7ffd61f48a90

Answer 3

移动构造函数通常 "steal" 参数持有的资源（例如指向 dynamically-allocated 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，并将论点保留在一些有效但不确定的状态。

Please see Move Constructor

我更改了您代码中的以下内容，希望它能按预期工作。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;
    }

以上代码产生以下输出：

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

希望对您有所帮助！

C++生命周期延长的问题

Problems with c++ lifetime extension

c++

object-lifetime