Question

我编写了以下代码来调查一个内部捕获的地址 lambda表达式

 # 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);
  }

当此代码运行时，＆＃34;典型输出＆＃34;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

在那个输出中我很容易理解：

1）main

中地址的相对值

2）通过引用d

捕获的地址的标识

3）按值i和a

捕获的不同地址

4）本地参数y

的完全不同的地址

5）捕获的i和a

的地址的相对值

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

这是什么意思？ lambda范围内的i和d（以及a）的地址是否重叠...？

提前感谢您的回答。

Answer 1

因为 0x30 - 0x28 = 0x8 。

这是十六进制而不是十进制，所以数字序列应为

0x28 0x29 0x2a 0x2b 0x2c 0x2d 0x2e 0x2f 0x30 ...

如果是十进制数，那就是

28 29 30 ...

但这不是。

Answer 2

但对于我来说，很难理解捕获的i的地址相对于捕获的d的相对值：它们似乎是，如果我没有错，除了每个之外只有两个字节其他

你错了，十六进制中的28和30是相隔8个字节 - 当你用十进制减去你借十（十进制的基数），十六进制你借16（十六进制的基数）所以16 - 8 == 8

哪些地址在C ++ lambda表达式中有捕获？

2 个答案: