Forcing class instances to be const - c++

Question

Is there any way to force to only allow const instances of class to be instantiated, and have non-const instances be detected as an error by the compiler?

Answer 1

is there any generic way to take an existing class, and "constify" it by removing all non-const functionality?
One possible workaround is to create a wrapper class that holds an instance of the class and only gives access to a const reference to it.
template<class T>
class Immutable {
public:
template<typename... Args>
Immutable(Args&&... args) : instance(forward<Args>(args)...) {
}
operator const T&() {
return instance;
}
const T& get() const {
return instance;
}
private:
Immutable& operator=(const Immutable& other) = delete;
T instance;
};
Suppose you have a mutable class Class:
class Class {
public:
Class() : m_value(0) {
}
Class(const Class& other) : m_value(other.m_value) {
}
Class(int value) : m_value(value) {
}
Class(int x, int y) : m_value(x + y) {
}
void change(int value) {
m_value = value;
}
int value() const {
return m_value;
}
private:
int m_value;
};
Here is how Immutable<Class> can be used:
void functionTakingConstReference(const Class& x) {
}
void functionTakingNonConstReference(Class& x) {
}
void functionTakingImmutableClass(Immutable<Class>& x) {
}
void functionTakingValue(Class x) {
}
int main(int argc, char *argv[])
{
// Any constructor of Class can also be used with Immutable<Class>.
Immutable<Class> a;
Immutable<Class> b(1);
Immutable<Class> c(2, 3);
Immutable<Class> d(c);
// Compiles and works as expected.
functionTakingConstReference(a);
functionTakingImmutableClass(a);
functionTakingValue(a);
cout << a.get().value() << endl;
// Doesn't compile because operator= is deleted.
// b = a;
// Doesn't compile because "change" is a non-const method.
// a.get().change(4);
// Doesn't compile because the function takes a non-const reference to Class as an argument.
// functionTakingNonConstReference(a);
return 0;
}

Answer 2

Is there any way to force to only allow const instances of class to be instantiated, and have non-const instances be detected as an error by the compiler?
No.
But you can declare all members as const. Then both const and non-const instances would behave largely in the same way and it shouldn't matter whether the instances are const.

Answer 3

I think you are looking for an immutable class. An easy way to get immutability is to declare all member variables as const. This way you ensure that the state of your objects will not change after construction.
This guarantee is independent of whether your object is const and even if you have non-const member functions for that class.
For example:
class Foo
{
public:
Foo(int id, double value) : m_id(id), m_value(value) { }
int Id() const { return m_id; }
double Value() const { return m_value; }
private:
const int m_id;
const double m_value;
};
Another way you could generate an immutable object of any type is through a template class. Something like this:
class Bar
{
public:
Bar(int id, double value) : m_id(id), m_value(value) { }
int Id() const { return m_id; }
double Value() const { return m_value; }
void SetId(int id) { m_id = id; }
private:
int m_id;
double m_value;
};
template<typename T>
class MyImmutable
{
public:
const T m_obj;
MyImmutable(const T& obj) : m_obj(obj)
{ }
};
int main()
{
cout << "Hello World!" << endl;
Foo a(1,2.0);
Bar x(2,3.0);
MyImmutable<Bar> y(x);
cout << "a.id = " << a.Id() << endl;
cout << "a.value = " << a.Value() << endl;
cout << "y.id = " << y.m_obj.Id() << endl;
cout << "y.value = " << y.m_obj.Value() << endl;
y.m_obj.SetId(2); // calling non-const member throws an error.
return 0;
}