Are there any alternatives to making const version of class? - c++

Question

In C++ I'm often facing a situation when I need to prepare const and non-const version of class in analogy to const_iterator and iterator from standard library.
class const_MyClass
{
public:
const_MyClass(const int * arr):
m_arr(arr)
{
}
int method() const; //does something with m_arr without modifying it
private:
const int * m_arr;
}
class MyClass
{
public:
MyClass(int * arr):
m_arr(arr)
{
}
int method() const; //does something with m_arr without modifying it
void modify(int i); //modify m_arr
private:
int * m_arr;
}
The problem with this is that I need to repeat whole code of const_MyClass in MyClass and distribute any changes in API to both classes. Thus sometimes I inherit const_MyClass and do some const_casts, which also isn't perfect and pretty solution. Still when I want to pass const_MyClass instance by reference it looks moronic:
void func(const const_MyClass & param)
Instance param is marked with two "consts", and it has only const methods...
This is where const constructors would be handy, but are there any existing alternatives?
Some use examples to explain problem better:
//ok to modify data
void f(int * data)
{
MyClass my(data);
my.modify();
...
}
//cant modify data, cant use MyClass
void fc(const int * data)
{
const_MyClass my(data);
int i = my.method();
...
}

Answer 1

You can make a template class to act as a base, like this:
template<typename T>
class basic_MyClass
{
public:
basic_MyClass(T * arr) :m_arr(arr) {}
int method() const; //does something with m_arr without modifying it
private:
T * m_arr;
};
Then, for your const version, since it doesn't add anything, you can just use a typedef:
typedef basic_MyClass<const int> const_MyClass;
For your non-const version, you can inherit:
class MyClass : public basic_MyClass<int>
{
public:
using basic_MyClass::basic_MyClass; // inherit all the constructors
void modify(int i); //modify m_arr
};

Answer 2

Have you considered simply tracking two pointers and raising exceptions from the mutable operations when no mutable value is available? Maybe an example will help describe what I am thinking of.
class MyClass
{
public:
MyClass(int *mutable_data):
m_mutable_view(mutable_data), m_readonly_view(mutable_data)
{
}
MyClass(const int *immutable_data):
m_mutable_view(NULL), m_readonly_view(immutable_data)
{
}
int retrieve_value(int index) {
return m_readonly_view[index];
}
void set_value(int index, int value) {
require_mutable();
m_mutable_view[index] = value;
}
protected:
void require_mutable() {
throw std::runtime_error("immutable view not available");
}
private:
const int *m_readonly_view;
int *m_mutable_view;
};
The idea is pretty simple here - use a sentinel value to indicate whether modifications are possible or not instead of depending on the type system to do that for you. Personally, I would think about doing the inheritance based approach that #BenjaminLindley suggested but I wanted to present a slightly different solution that might not have occurred to you.

Answer 3

After talk with Neil Kirk I realized what I was doing wrong. I started by separating data from logic as he suggested.
This attempt resulted in two classes MyClassPtr and const_MyClassPtr. They only provide functions for data access (like iterators) and may look like that:
class const_MyClassPtr
{
public:
const_MyClassPtr(const int * arr);
int operator [](int i) const;
const int * ptr() const;
private:
const int * m_arr;
}
class MyClassPtr
{
public:
MyClassPtr(int * arr);
int operator [](int i) const;
int & operator [](int i);
const int * ptr() const;
int * ptr();
//promotion to const pointer
const_MyClassPtr () const {return const_MyClassPtr(m_arr);}
private:
int * m_arr;
}
Now it is clear that objects of these classes should be treated like pointers, so when I use them as function parameters I pass them by value!
void func(const_MyClassPtr param) //instead of void func(const const_MyClass & param)
To provide methods I have created MyClassOp class template and used static polymorphism.
template <class DERIVED>
class MyClassOp
{
public:
const DERIVED & derived() const {return static_cast<const DERIVED &>(*this)}
DERIVED & derived() {return static_cast<DERIVED &>(*this)}
int method() const; //operates on derived() const
void modify(int i); //operates on derived()
}
MyClassOp is a collection of methods. It does not have state. In general it is a trait. To make these methods accessible I overloaded -> and * operators
class const_MyClassPtr : private MyClassOp<const_MyClassPtr>
{
public:
const MyClassOp<MyClassPtr> * operator ->() const {return this;}
const MyClassOp<MyClassPtr> & operator *() const {return *this;}
...
}
class MyClassPtr : private MyClassOp<MyClassPtr>
{
public:
MyClassOp<MyClassPtr> * operator ->() {return this;}
MyClassOp<MyClassPtr> & operator *() {return *this;}
...
}
This works O.K., but is a bit cumbersome. If I have for example equality operator I need to write something like *myptr1 == myptr2 to compare values kept by two MyClassPtr objects (it's easy to make a mistake and compare myptr1 == myptr2 or expect that something like *myptr1 == *myptr2 could work). Also when I have allocating type:
class MyClass : public MyClassOp<MyClass>
{
MyClass(int x, int y, int z);
...
int m_arr[3];
}
I would want to be able to use temporaries as function arguments.
void f(const_MyClassPtr my);
//use temporary when calling f()
f(MyClass(1, 2, 3));
I can do this by providing conversion operators or conversion constructors (that convert MyClass to const_MyClassPtr). But then const_MyClassPtr behaves more like reference than pointer. If iterators are generalization of pointers then why one could not imitate reference? Therefore I divided MyClassOp into two parts (const and non const) and replaced -> and * operators implemented by const_MyClassPtr and MyClassPtr with public inheritance and changed their names to ressemble reference. I ended up with following structures.
MyClassOp : public const_MyClassOp
const_MyClassRef : public const_MyClassOp<const_MyClassRef>
MyClassRef : public MyClassOp<MyClassRef>
MyClass : public MyClassOp<MyClass>
However const_MyClassRef and MyClassRef are not perfect generalization of reference as it impossible to imitate some of C++ reference properties, so Ref suffix is there to denote reference-like structure.

Answer 4

Maybe you can find some hints in effective c++ item 4 "Avoid duplication in const and non-const Member function"
I may summarize like following ( it makes you avoid code duplication even if using somewhat ugly cast ):
struct my_class
{
my_class(int x):_x(x){};
const int& method(void) const;
int& method(void);
int _x;
};
const int& my_class::method(void) const //func for const instance
{
return _x;
}
int& my_class::method(void) //func for normal instance
{
return const_cast<int& >(static_cast<const my_class& >(*this).method()) ;
}
int main()
{
my_class a(1);
const my_class b(2);
a.method() = 5;
cout << a.method() << endl;
//b.method() = 4; //b is const, wont compile
cout << b.method() << endl;
return 0;
}