我可以直接将地址分配给指针吗?如果是这样，该怎么做?

问题描述

int main()
{
  int a = 2; // address is 0x7ffeefbff58c
  int *b = &a;
  std::cout << "address of a: " << b << std::endl;
  return 0;
}

我在地址0x7ffeefbff58c中有一个int变量a，但是我可以直接为int * b分配0x7ffeefbff58c吗?

I have my int variable a at address 0x7ffeefbff58c, but can I directly assign int* b with 0x7ffeefbff58c?

我尝试了int * b = 0x7ffeefbff58c;但是有一个错误提示无法用'long'类型的右值初始化'int *'类型的变量"，所以我必须使用(& a)的地址来初始化指针吗?还是有其他方法可以做到?

I tried int * b = 0x7ffeefbff58c;But there is an error says "cannot initialize a variable of type 'int *' with an rvalue of type 'long'", so do I have to use the address of a (&a) to initialize the pointer? or there is other way to do it?

推荐答案

从技术上讲，是的.

带有重新解释的演员表.

With reinterpret cast.

但是要意识到，总体上绝对不能保证 a 位于地址0x7ffeefbff58c中.这样，将整数重新解释为指针就无能为力了.使用局部变量执行此操作将是毫无意义的.

But do realise that there is absolutely no guarantee in general that a would be in the address 0x7ffeefbff58c. As such, there isn't much that you can do with such integer reinterpreted as a pointer. Doing this with a local variable would be pointless.

将整数解释为指针很有用的情况是，某些嵌入式系统会保留一些恒定的内存地址以用于特殊目的.

A case where interpreting integer as a pointer is useful is some embedded systems that reserve some constant memory addresses for special purposes.

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