地址确实在C lambda表达式中具有捕获

我编写了以下代码，以调查在捕获中的地址lambda表达

 # include <iostream>
 # include <functional>
 using fu = std::function<void(int)>;
 void f(fu l, int x)
  { l(x); }
 int main()
  {
   double d{1.17};
   int i = 12;
   char a = 'a';
   // write addresses on stdout:
   std::cout << "address of d in main is " << &d << std::endl;
   std::cout << "address of i in main is " << &i << std::endl;
   std::cout << "address of a in main is 0x" 
             << std::hex << reinterpret_cast<long>(&a) << std::dec 
             << std::endl;
   // now let us introduce a lambda expr as follows
   auto l = [&, i, a](int y)
    {
     std::cout<<"captured d by reference " << d
              <<" at address " << &d << std::endl;
     std::cout<<"captured i by value " << i
              <<" at address " << &i << std::endl;
     std::cout<<"captured a by value " << a
              <<" at address 0x" << std::hex
              << reinterpret_cast<long>(&a) << std::dec
              << std::endl;
     std::cout<<"got parameter " << y
              <<" at address " << &y << std::endl;
    };
  // now send the lambda to f which will execute it
  f(l, i);
  }

运行此代码时，"典型输出"看起来像：

   address of d in main is 0x7ffd96bc4628
   address of i in main is 0x7ffd96bc4624
   address of a in main is 0x7ffd96bc4623
   captured d by reference 1.17 at address 0x7ffd96bc4628
   captured i by value 12 at address 0x7ffd96bc4630
   captured a by value a at address 0x7ffd96bc4634
   got parameter 12 at address 0x7ffd96bc4564

在该输出中，我可以轻松理解：

1）main中的地址的相对值

2）通过参考d捕获的地址的身份

3）值i和a

4）本地参数y

5）捕获的i和a

，但对我来说，很难理解地址的相对价值相对于捕获的d捕获的i：它们似乎是如果我没有错，只有两个字节彼此相距。

这是什么意思？lambda范围内的i和d（以及a）的地址重叠...？

预先感谢您的任何答案。