How to deal with initialization of non-const member in const object?

Question

Let's say you have a class

class C 
{
  int * i;

  public:

     C(int * v):i(v) {};

     void method() const;  //this method does not change i
     void method();       //this method changes i
}

Now you may want to define const instance of this class

const int * k = whatever;
const C c1(k); //this will fail

but this will fail because of non-const int C's constructor C(int * v)

so you define a const int constructor

C(const int * v):i(v) {}; //this will fail also

But this will fail also since C's member "int * i" is non-const.

What to do in such cases? Use mutable? Casting? Prepare const version of class?

edit: After discussion with Pavel (below) I investigated this problem a bit. To me what C++ does is not correct. Pointer target should be a strict type, that means that you could not for example do the following:

int i;
const int * ptr;
ptr = & i;

In this case language grammar treats const as a promise not to change pointer's target. In addition int * const ptr is a promise not to change pointer value itself. Thus you have two places where const can be applied. Then you may want your class to model a pointer (why not). And here things are falling into pieces. C++ grammar provides const methods which are able to promise not to change field's values itself but there is no grammar to point out that your method will not change targets of your in-class pointers.

A workaround is to define two classes const_C and C for example. It isn't a royal road however. With templates, their partial specializations it's hard not to stuck into a mess. Also all possible arguments variations like const const_C & arg, const C & arg, const_C & arg, C & arg don't look pretty. I really don't know what to do. Use separate classes or const_casts, each way seems to be wrong.

In both cases should I mark methods which don't modify pointer's target as const? Or just follow traditional path that const method doesn't change object's state itself (const method don't care about pointer target). Then in my case all methods would be const, because class is modelling a pointer thus pointer itself is T * const. But clearly some of them modify pointer's target and others do not.

Answer 1

Your example doesn't fail, k is passed by value. The member i is 'implicitly constant' as direct members of C can't be changed when the instance is constant.
Constness says that you can't change members after initialization, but initializing them with values in the initialization list is of course allowed - how else would you give them a value?

What doesn't work is invoking the constructor without making it public though ;)

update addressing updated question:

Yes, C++ forces you into some verboseness sometimes, but const correctness is a common standard behaviour that you can't just redefine without breaking expectations. Pavels answer already explains one common idiom, which is used in proven libraries like the STL, for working around this situation.

Sometimes you have to just accept that languages have limitations and still deal with the expectations of the users of the interface, even if that means applying an apparently sub-optimal solution.

Answer 2

Your question does not make sense. Where did you get all these "this will fail" predictions? None of them are even remotely true.

Firstly, it is completely irrelevant whether the constructor's parameter is declared const or not. When you are passing by value (as in your case) you can pass a const object as an argument in any case, regardless of whether the parameter is declared as const or not.

Secondly, from the constructor's point of view, the object is NOT constant. Regardless of what kind of object you are constructing (constant or not), from within the constructor the object is never constant. So there's no need for mutable or anything.

Why don't you just try compiling your code (to see that nothing will fail), instead of making strange ungrounded predictions that something "will fail"?

Answer 3

Hi,

I didnt understand your solution. I am very much new to stackoverflow. I dont know where to post the replies. so i am posting it in the Answers section. sorry for that.

Although the constructor is made public, VS2008 still gives compiler error for the above program. Can you explain me in detail, if you dont mind?

Answer 4

Sounds like you want an object that can wrap either int* (and then behave as non-const), or int const* (and then behave as const). You can't really do it properly with a single class.

In fact, the very notion that const applied to your class should change its semantics like that is wrong - if your class models a pointer or an iterator (if it wraps a pointer, it's likely to be the case), then const applied to it should only mean that it cannot be changed itself, and should not imply anything regarding the value pointed to. You should consider following what STL does for its containers - it's precisely why it has distinct iterator and const_iterator classes, with both being distinct, but the former being implicitly convertible to the latter. As well, in STL, const iterator isn't the same as const_iterator! So just do the same.

[EDIT] Here's a tricky way to maximally reuse code between C and const_C while ensuring const-correctness throughout, and not delving into U.B. (with const_cast):

template<class T, bool IsConst>
struct pointer_to_maybe_const;

template<class T>
struct pointer_to_maybe_const<T, true> { typedef const T* type; };

template<class T>
struct pointer_to_maybe_const<T, false> { typedef T* type; };

template<bool IsConst>
struct C_fields {
   typename pointer_to_maybe_const<int, IsConst>::type i;
   // repeat for all fields
};


template<class Derived>
class const_C_base {
public:
    int method() const { // non-mutating method example
        return *self().i;
    }
private:
    const Derived& self() const { return *static_cast<const Derived*>(this); }
};

template<class Derived>
class C_base : public const_C_base<Derived> {
public:
    int method() { // mutating method example
        return ++*self().i;
    }
private:
    Derived& self() { return *static_cast<Derived*>(this); }
};


class const_C : public const_C_base<const_C>, private C_fields<true> {
    friend class const_C_base<const_C>;
};

class C : public C_base<C>, private C_fields<false> {
    friend class C_base<C>;
};

If you actually have few fields, it may be easier to duplicate them in both classes rather than going for a struct. If there are many, but they are all of the same type, then it is simpler to pass that type as a type parameter directly, and not bother with const wrapper template.

Answer 5

A const int* is not the same as a int* const. When your class is const, you have the latter (constant pointer to mutable integer). What you're passing is the former (mutable pointer to constant integer). The two are not interchangeable, for obvious reasons.

Answer 6

When you instantiate

const C c1(...)

Because c1 is const, its member i turns in to:

int* const i;

As someone else mentioned, this is called implicit const.

Now, later in your example, you attempt to pass a const int*. So your constructor is basically doing this:

const int* whatever = ...;
int* const i = whatever; // error

The reason you get an error is because you can't cast const to non-const. The 'whatever' pointer is not allowed to change the thing it points to (the int part is const). The 'i' pointer is allowed to change what it points to, but cannot itself be changed (the pointer part is const).

You also mention wanting your class to model a pointer. The STL does this with iterators. The model some implementations use is having a class called 'const_iterator' which hides the real pointer and only supplies const methods to access the pointed-to data. Then there's also an 'iterator' class which inherits from 'const_iterator', adding non-const overloads. This works nicely - it's a custom class which allows the same constness as pointers, where the types mirror pointers like so:

• iterator -> T*
• const iterator -> T* const
• const_iterator -> const T*
• const const_iterator -> const T* const

Hopefully that makes sense :)

Answer 7

OK here's what I have done so far. To allow inheritance after const version of class without const_casts or additional space overhead I created an union which basically looks like ths:

template <typename T>
union MutatedPtr
{
protected:
 const T * const_ptr;
 T * ptr;

public:
 /**
  * Conversion constructor.
  * @param ptr pointer.
  */
 MutatedPtr(const T * ptr): const_ptr(ptr) {};

 /**
  * Conversion to T *.
  */
 operator T *() {return ptr;}

 /**
  * Conversion to const T *.
  */
 operator const T *() const {return const_ptr;}
};

When MutatedPtr field is declared, it ends up so that in const methods const_ptr is returned, while non-const ones get plain ptr. It delegates method's const-ness to pointer target which makes sense in my case.

Any comments?

BTW you can of course do similar thing with non-pointer types or even methods, so it looks that introducing mutable keyword wasn't necessary(?)