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

Question

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 pract

Answer 1

This code:

 int& methodTwo() { return x; }

Means that the function returns a reference to an integer. Just like when passing a value by reference to a function, if the return value of methodTwo gets changed, so does the value that methodTwo returned. In this case, class field x.

In the code you have written, this means that you are letting the private variable x escape its scope (a class field) and be passed around in the outside world. This certainly is a bad practice (because x can be changed in ways that may break class foo, but it is certainly allowable.

Remember public/private/protected are compile-time only. Once your application gets compiled, private fields sit next to public fields and there is no protection against modification. The same is true for managed languages like C# and Java.

You should generally avoid returning references because it makes it crazy-hard to understand when constructors/destructors get called. However, returning a reference can be faster. If your method returned a struct type that was HUGE, returning a const reference to that same struct type should only take four-to-eight-bytes (a pointer to that object). However, there are better ways to optimize for this sort of thing.