Preserve access privileges when using '->' operator

user_11228029 0
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I have two classes,

template<class Type>
class SafePtr {
public:
    SafePtr() {}
    ~SafePtr() {}

    void Lock(Type* data, void* key)
    {
        if (!pKey)
        {
            pKey = key;
            pData = data;
        }
    }

    Type* Unlock(void* key) const
    {
        if (key == pKey)
            return pData;
    }

    Type* operator->() 
    {
        return pData;
    }

private:
    Type* pData = nullptr;
    void* pKey = nullptr;
};

template<class Type>
class SafePtrArray {
public:
    SafePtrArray() {}
    ~SafePtrArray() {}

    template<class... Args>
    SafePtr<Type> CreatePtr(Args&&... args)
    {
        Type* data = new Type(args...);
        ptrs.insert(ptrs.end(), data);

        SafePtr<Type> ptr;
        ptr.Lock(data, this);
        return ptr;
    }

    Type* UnlockPtr(const SafePtr<int>& ptr)
    {
        return ptr.Unlock(this);
    }

    void Destroy(const SafePtr<int>& ptr)
    {
        Type* pointer = ptr.Unlock(this);

        for (auto itr = ptrs.begin(); itr != ptrs.end(); itr++)
        {
            if ((*itr) == pointer)
            {
                delete pointer;
                ptrs.erase(itr);
            }
        }
    }

private:
    std::vector<Type*> ptrs;
};

The goal is to protect a pointer so that the user can access its members but not get to manipulate its actual pointer (mainly delete it prematurely). And also I need to store all the pointers in an array so that when the parent object destroys, I can automatically destroy all the allocated pointers

For this I use two classes, SafePtr and SafePtrArray. SafePtrArray creates and stores the pointers and wraps them in the SafePtr and returns it to the user. SafePtr is just a wrapper and should not let the user get access to the underlying pointer but will allow them to access its members.

It works fine at first but soon I found this error,

int main()
{
    SafePtrArray<int> ptr;
    auto pInt = ptr.CreatePtr();
    int* i = pInt.operator->();     // Users can get access to the underlying pointer using this.

    ptr.Destroy(pInt);
}

Is there a way to prevent users from getting access to the underlying type and prevent them from manipulating the pointer while having the privilege to access its members?

c++
2 Answers

I still think you try to solve a problem that has more to do with possible flaws in the design of the API/of the code, the documentation, or with the lack of C++ knowledge of the one using it, with a "solution" that has more cons than pros.

If a C++ programmer does not know what ownership is or does not respect it and blindly deletes objects or frees the memory of pointers, then there will be much bigger concerns. You likely will move the problem just to a different part of the code.

Having that said, the closest you can do to not expose the pointer right now is something like this:

(The code is just a proof of concept, so things like call might need to be improved)

#include <iostream>
#include <string>

struct Test {
  void foo(int x, int y, std::string str) {
    std::cout << x << " " << y << " " << str << std::endl;
  }

  double test = 0.5;
};

template <typename T>
struct Ptr {

  template <auto M, typename... Args>
  auto call(Args... args) {
    return (obj.*M)(std::forward<Args>(args)...);
  }

  template <auto M>
  auto get() { 
    return (obj.*M);
  }

protected:
  T obj;
};

int main() {
  Ptr<Test> p;

  p.call<&Test::foo>(1, 2, "hello");
  std::cout << p.get<&Test::test>() << std::endl;

  return 0;
}

But I still don't think that this is a good approach.

And a user can still mess around with the code and do something bad like:

int main() {
  Ptr<Test> p;
  
  delete &p;

  return 0;
}

Or this, which for sure is undefined behavior, but that does not really matter as deleting a not owned object will also result in undefined behavior at some point:

template<typename T>
struct Ptr {
protected:
    T *obj;
}

template<typename T>
struct Ptr2 {
public:
    T *obj;
};

int main()
{
    Ptr<Test> p;
    Ptr2<Test> *p2 = reinterpret_cast<Ptr2<Test>*>(&p);
    
    std::cout << p2->obj << std::endl;
}

So there is no protection again such things.

Besides the shown code, there is a proposal for reflection that is feature complete now, which would allow getting information about the members of a type, but this was not added to c++20, and one for metaclasses which is also not in the standard yet.

With these two proposals, you might be able to implement something better useable. But my concerns about the benefits of this remain.

user_19604550

Is there a way to prevent users from getting access to the underlying type and prevent them from manipulating the pointer while having the privilege to access its members?

Under certain conditions, no, this is not possible. If the underlying Type is a standard layout class then providing access to the first non-static non-bitfield data member breaks your goal. (Caveat: providing access to just the value of the member is a different story.) The address of that member can be converted to a pointer to the underlying object via reinterpret_cast, which allows calling delete on that pointer. (Well, "allows" in the sense that the call is syntactically valid. Not much else matters for "allows" since we are headed into undefined behavior anyway.)

For classes that are not standard layout, there are probably compiler-specific (non-portable) methods to achieve the same effect (converting the address of a data member into a pointer to the underlying object). There is no reason for a compiler to try actively to thwart such things.

If a programmer is determined to invoke undefined behavior, there is little you can do to stop it.

user_98373010
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