如何找到变量分配的内存？

发布于 2024-12-17 11:24:58 字数 264 浏览 0 评论 0原文

class app {
        public int x = 3; 

        static void Main(string[] args)
        {
        }
}

是否可以获取x变量分配的内存地址？

该示例可以采用 C、C++、C# 或 D 语言。我希望它是清楚的提前致谢

原文

class app {
        public int x = 3; 

        static void Main(string[] args)
        {
        }
}

it's possible get the memory address allocated by x variable?

the example can be in C, C++, C# or D.
I hope it is clear
Thanks in advance

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够运 2024-12-24 11:24:58

在大多数类 C 语言中，与号 (&) 是“取址”运算符：

int x;
printf("Address of x is %p\n", &x);

& 的返回值实际上是指向其操作数的指针。

The ampersand (&) is the "address-of" operator in most C-like languages:

int x;
printf("Address of x is %p\n", &x);

The return value of & is effectively a pointer to its operand.

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病女 2024-12-24 11:24:58

在 C 和 C++ 中，这是相当简单的。我将给出 C++ 中的示例：

struct App
{
   int x;
   App() : x(3) { }
};

int main()
{
  App a;
  int * p = &a.x; // address goes here
}

当然不存在“变量 App::x”之类的东西，因为 App 只是类型< /em>。这种类型的每个实例（例如示例中的a）都带有自己的一组成员变量，并且很容易获得指向该成员变量的指针。（C 中的普通数据结构也是如此。）

_{请注意，C++ 还有另一个相关功能：成员指针。这使我们能够形成不透明值 int App::*pm = &App::x ，它本身不指向任何内容，而仅携带有关偏移量的信息类内的 App::x ，如果你愿意的话。该动物可以与实例一起使用来获取实际值，例如a.*pm.}

In C and in C++ this is fairly straight-forward. I'll give the example in C++:

struct App
{
   int x;
   App() : x(3) { }
};

int main()
{
  App a;
  int * p = &a.x; // address goes here
}

There is of course no such thing as "the variable App::x", since App is only the type. Each instance of this type, such as a in the example, carries its own set of member variables, and a pointer to the member variable is readily obtained. (The same is true for plain data structs in C.)

_{Note that C++ has another, related feature: Member pointers. This allows us to form the opaque value int App::*pm = &App::x which by itself doesn't point to anything, but only carries information about the offset of App::x inside the class, if you will. This animal can be used together with an instance to obtain the actual value, e.g. a.*pm.}

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