从getter方法返回const和非const值

我在此处问了一个问题，未能获得令人满意的答复，我仍在思考这个话题，因此开始这个新问题，希望我可以更好地描述我的想知道.

I asked one question here which did not get desirable response, I am still pondering on the topic, so start this new question, hopefully I can describe my wondering better.

我有这样的代码:

struct A
{
  P* get_p() const;
};

void dump_p(const A *a)
{
  const P* p = a->get_p();
  ...... dump out p's content for debugging purpose ......
}

void re-format(A *a)
{
   P* p = a->get_p();
   ...... update p's content ......
}

您可以看到dump_p()实际上是一个只读函数，但是get_p()返回一个自由指针，该指针允许调用者自由更新数据.这让我很na，我认为解决方案编译器可以帮助执行语义检查.

You can see dump_p() is really a read-only function, but get_p() returns a free pointer which allows caller to update the data freely. This is nagging to me, I am thinking a solution compiler can help enforcing the semantic check.

这就是我的想法:

struct A
{
  P* get_p() const;
  const P* get_rdonly_p() const;
};

void dump_p(const A *a)
{
  const P* p = a->get_rdonly_p();
}

void re-format(A *a)
{
   P* p = a->get_p();
}

添加此get_rdonly_p()是因为c ++不允许仅使用具有不同返回类型的重载方法.

Add this get_rdonly_p() because c++ does not allow overloading method with different return type only.

我疯了吗?或您有什么解决方案?

Am I going nuts? Or what is your solution?

[更新]感谢Adrian，这可能是解决方案.但是我仍然会碰墙-您提到的非const版本get_p()不能100％满足我，该函数本身不会更改任何struct A，但是我没有将其声明为const，这是唠叨.

[UPDATE] Thanks Adrian, that might be the solution. But I am still banging my head against wall - the non-const version get_p() you mentioned does not 100% satisfies me, the function itself does not change any of the struct A, but I do not declare it as const, this is nagging.

让我坚持检查您的答案，很可能是这样，也许c ++在语言级别上缺少一些构造.

Let me hold checking yours as the answer, very likely it is, maybe c++ lacks some construct at language level.