Class variables: public access read-only, but private access read/write

Question

Whoopee, not working on that socket library for the moment. I'm trying to educate myself a little more in C++.

With classes, is there a way to make a variable read-only to the public, but read+write when accessed privately? e.g. something like this:

class myClass {
    private:
    int x; // this could be any type, hypothetically

    public:
    void f() {
        x = 10; // this is OK
    }
}

int main() {
    myClass temp;

    // I want this, but with private: it's not allowed
    cout << temp.x << endl;


    // this is what I want:

    // this to be allowed
    temp.f(); // this sets x...

    // this to be allowed
    int myint = temp.x;

    // this NOT to be allowed
    temp.x = myint;
}

My question, condensed, is how to allow full access to x from within f() but read-only access from anywhere else, i.e. int newint = temp.x; allowed, but temp.x = 5; not allowed? like a const variable, but writable from f()...

EDIT: I forgot to mention that I plan to be returning a large vector instance, using a getX() function would only make a copy of that and it isn't really optimal. I could return a pointer to it, but that's bad practice iirc.

P.S.: Where would I post if I just want to basically show my knowledge of pointers and ask if it's complete or not? Thanks!

Answer 1

Of course you can:

class MyClass
{
    int x_;

public:
    int x() const { return x_; }
};

If you don't want to make a copy (for integers, there is no overhead), do the following:

class MyClass
{
    std::vector<double> v_;

public:
    decltype(v)& v() const { return v_; }
};

or with C++98:

class MyClass
{
    std::vector<double> v_;

public:
    const std::vector<double>& v() const { return v_; }
};

This does not make any copy. It returns a reference to const.

Answer 2

While I think a getter function that returns const T& is the better solution, you can have almost precisely the syntax you asked for:

class myClass {
    private:
    int x_; // Note: different name than public, read-only interface

    public:
    void f() {
        x_ = 10; // Note use of private var
    }
    const int& x;
    myClass() : x_(42), x(x_) {} // must have constructor to initialize reference
};

int main() {
    myClass temp;

    // temp.x is const, so ...
    cout << temp.x << endl; // works
    // temp.x = 57;  // fails

}

EDIT: With a proxy class, you can get precisely the syntax you asked for:

class myClass {
public:

    template <class T>
    class proxy {
        friend class myClass;
    private:
        T data;
        T operator=(const T& arg) { data = arg; return data; }
    public:
        operator const T&() const { return data; }
    };

    proxy<int> x;
    // proxy<std::vector<double> > y;


    public:
    void f() {
        x = 10; // Note use of private var
    }
};

temp.x appears to be a read-write int in the class, but a read-only int in main.

Answer 3

A simple solution, like Rob's, but without constructor:

class myClass {
private:
    int m_x = 10; // Note: name modified from read-only reference in public interface
public:
    const int& x = m_x;
};

int main() {
    myClass temp;

    cout << temp.x << endl; //works.
    //temp.x = 57;  //fails.
}

It is more like get method, but shorter.

Constant pointer is simple, and should work at all types you can make pointer to.

Answer 4

This may do what you want.

If you want a readonly variable but don't want the client to have to change the way they access it, try this templated class:

template<typename MemberOfWhichClass, typename primative>                                       
class ReadOnly {
    friend MemberOfWhichClass;
public:
    inline operator primative() const                 { return x; }

    template<typename number> inline bool   operator==(const number& y) const { return x == y; } 
    template<typename number> inline number operator+ (const number& y) const { return x + y; } 
    template<typename number> inline number operator- (const number& y) const { return x - y; } 
    template<typename number> inline number operator* (const number& y) const { return x * y; }  
    template<typename number> inline number operator/ (const number& y) const { return x / y; } 
    template<typename number> inline number operator<<(const number& y) const { return x <<y; }
    template<typename number> inline number operator>>(const number& y) const { return x >> y; }
    template<typename number> inline number operator^ (const number& y) const { return x ^ y; }
    template<typename number> inline number operator| (const number& y) const { return x | y; }
    template<typename number> inline number operator& (const number& y) const { return x & y; }
    template<typename number> inline number operator&&(const number& y) const { return x &&y; }
    template<typename number> inline number operator||(const number& y) const { return x ||y; }
    template<typename number> inline number operator~() const                 { return ~x; }

protected:
    template<typename number> inline number operator= (const number& y) { return x = y; }       
    template<typename number> inline number operator+=(const number& y) { return x += y; }      
    template<typename number> inline number operator-=(const number& y) { return x -= y; }      
    template<typename number> inline number operator*=(const number& y) { return x *= y; }      
    template<typename number> inline number operator/=(const number& y) { return x /= y; }      
    template<typename number> inline number operator&=(const number& y) { return x &= y; }
    template<typename number> inline number operator|=(const number& y) { return x |= y; }
    primative x;                                                                                
};      

Example Use:

class Foo {
public:
    ReadOnly<Foo, int> x;
};

Now you can access Foo.x, but you can't change Foo.x! Remember you'll need to add bitwise and unary operators as well! This is just an example to get you started

Answer 5

There is a way to do it with a member variable, but it is probably not the advisable way of doing it.

Have a private member that is writable, and a const reference public member variable that aliases a member of its own class.

class Foo
{
  private:
      Bar private_bar;

  public:
      const Bar& readonly_bar; // must appear after private_bar
                              // in the class definition

  Foo() :
       readonly_bar( private_bar )
  {
  }
};

That will give you what you want.

void Foo::someNonConstmethod()
{
    private_bar.modifyTo( value );
}

void freeMethod()
{
    readonly_bar.getSomeAttribute();
}

What you can do, and what you should do are different matters. I'm not sure the method I just outlined is popular and would pass many code reviews. It also unnecessarily increases sizeof(Foo) (albeit by a small amount) whereas a simple accessor "getter" would not, and can be inlined, so it won't generate more code either.

Answer 6

You may want to mimic C# properties for access (depending what you're going for, intended environment, etc.).

class Foo
{
  private:
    int bar;

  public:
    __declspec( property( get = Getter ) ) int Bar;

    void Getter() const
    {
      return bar;
    }
}

Answer 7

You would have to leave it private and then make a function to access the value;

private:

    int x;

public:

    int X()
    {
        return x;
    }

Answer 8

As mentioned in other answers, you can create read only functionality for a class member by making it private and defining a getter function but no setter. But that's a lot of work to do for every class member.

You can also use macros to generate getter functions automatically:

#define get_trick(...) get_
#define readonly(type, name) \
private: type name; \
public: type get_trick()name() {\
    return name;\
}

Then you can make the class this way:

class myClass {
    readonly(int, x)
}

which expands to

class myClass {
    private: int x;
    public: int get_x() {
        return x;
    }
}

Answer 9

Write a public getter function.

int getX(){ return x; }

Answer 10

You need to make the member private and provide a public getter method.

Answer 11

The only way I know of granting read-only access to private data members in a c++ class is to have a public function. In your case, it will like:

int getx() const { return x; }

or

int x() const { return x; }.

By making a data member private you are by default making it invisible (a.k.a no access) to the scope outside of the class. In essence, the members of the class have read/write access to the private data member (assuming you are not specifying it to be const). friends of the class get access to the private data members.

Refer here and/or any good C++ book on access specifiers.

Answer 12

I had a similiar problem. Here is my solution:

enum access_type{readonly, read_and_write};

template<access_type access>
class software_parameter;

template<>
class software_parameter<read_and_write>{
protected:
    static unsigned int test_data;
};

template<>
class software_parameter<readonly> : public software_parameter<read_and_write>{
public:
    static const unsigned int & test_data;
};

class read_and_write_access_manager : public software_parameter<read_and_write>{
    friend class example_class_with_write_permission;
};

class example_class_with_write_permission{
public:
    void set_value(unsigned int value);
};

And the .cpp file:

unsigned int software_parameter<read_and_write>::test_data=1;
const unsigned int & software_parameter<readonly>::test_data=software_parameter<read_and_write>::test_data;

void example_class_with_write_permission::set_value(unsigned int value){software_parameter<read_and_write>::test_data=value;};

The idea is, that everyone can read the software parameter, but only the friends of the class read_and_write_access_manager are allowed to change software parameter.

Answer 13

but temp.x = 5; not allowed?

This is any how not allowed in the snippet posted because it is anyhow declared as private and can be accessed in the class scope only.

Here are asking for accessing

cout << temp.x << endl;

but here not for-

int myint = temp.x;

This sounds very contradictory.