Why can I expose private members when I return a reference from a public member function?

c++ function return-type
Mahesh picture Mahesh · Jan 16, 2011 · Viewed 14.8k times · Source

In the code snippet, I am able to access the private member variable outside the class scope. Though this should never be done, why is it allowed in this case? Is it a bad practice to receive a returned private variable by reference ? 

#include <iostream>
#include <cstdlib>

class foo
{
    int x;
    public:
        foo(int a):x(a){}
        int methodOne() { return x; }
        int& methodTwo() { return x; }
};

int main()
{
    foo obj(10);
    int& x = obj.methodTwo();
    x = 20;              // With this statement, modifying the state of obj::x

    std::cout << obj.methodOne();
    getchar();
    return 0;
}

And regarding this method, what does the return type convey ? And also when should I have return type of this kind ?

int& methodTwo() { return x; }

PS: I am sorry if the subject line is vague. Can someone change it to the content relevant here. Thanks.

Answer

Billy ONeal picture Billy ONeal · Jan 16, 2011

private does not mean "this memory may only be modified by member functions" -- it means "direct attempts to access this variable will result in a compile error". When you expose a reference to the object, you have effectively exposed the object.

Is it a bad practice to receive a returned private variable by reference ?

No, it depends on what you want. Things like std::vector<t>::operator[] would be quite difficult to implement if they couldn't return a non-const reference :) If you want to return a reference and don't want clients to be able to modify it, simply make it a const reference.

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