Below is my code. What i want to do is to assign a value to variable data. I'm not allowed to write any method in the public area of XYZ class. But a pointer to an object of XYZ class is allowed but in the private area of XYZ class only but public portion of ABC class is allowed. I can write the methods into this.
Any idea to do this...???
RegardsCode:class ABC { private: class XYZ { private: int data; }; public: ABC(); };
Lucky
Why do you want to do this? If this is for an interview question, I suggest that you move on, because a company that expects its developers to know how to cheat the access control mechanism intended to help developers is probably not worth working for.Originally Posted by luckyali444
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thanks for your reply Laserlight, i am not going to do any kind of job. I am a student and one of my friend asked me that if it is possible to assign any value to data but without making any methods in Public part of XYZ class.
No, it is not possible. However, it is possible to make it appear as if you have assigned a value to that private member variable, though what you really would be doing is just writing to the memory that is used by the member variable, by making assumptions about the member layout of the object. This should not be done if the object is not of a POD type (a class with private members is not a POD type).
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I think perhaps the trick is that friends can help you do this.
Well at least it doesn't violate the rules laid out so far.
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Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
Well, to exploit the letter of the rule, we can define a non-virtual public member function in XYZ, since that would not be a method, in C++ terminology.
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Actually, in C++11 XYZ would be POD. Data member don't have to be public, they just have to have the same access level.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
Here is a *dangerous* experiment I once tried:
(Feel free to delete this post if considered harmful)
Anyway.. here it is..
Code:#include<iostream> class foo1 { int x; public: foo1(int x_){x=x_;}; }; class foo2 { public: int x; }; int main() { foo1* f1 = new foo1(5); foo2* f2; f2= (foo2*)f1; //Shooting downwards, hoping not to hurt the legs ! std::cout<<f2->x; return 0; }
That leaks memory, and you don't need to do that.
This should suffice:
Of course, don't use this in production code. This is for experimental purposes only!Code:foo1 f1(5); foo2* f2 = reinterpret_cast<foo2*>(&f1); //Shooting downwards, hoping not to hurt the legs !
Unless you're aiming for this.Of course, don't use this in production code. This is for experimental purposes only!
Anyway, my point&& question was whether the integer is guaranteed to remain in the same position for the two classes .
If they are compiled with the same compiler and the same padding settings, then it is likely. However, it is still undefined. Any attempt to circumvent access control will result in undefined behavior AFAIK.
The standard says "A pointer to a standard-layout struct object, suitably converted using a reinterpret_cast, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa. [ Note: There might therefore be unnamed padding within a standard-layout struct object, but not at its beginning,
as necessary to achieve appropriate alignment. —end note ]"
So reinterpret_cast<int*>(&xyz) == &xyz.data.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
