private 并不表示此内存只能修改成员函数 - 意味着直接尝试访问此变量将导致编译错误。当你暴露对象的引用时，你已经有效地暴露了对象。

接收返回的私有变量参考？

不，这取决于你想要什么。如果 std :: vector< t> :: operator [] 如果不能返回非 - const 引用:)如果你想返回一个引用而不希望客户端能够修改它，只需使它成为一个 const 引用。 / p>

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.

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.