Why can't you use offsetof on non-POD strucutures in C++?

Question

I was researching how to get the memory offset of a member to a class in C++ and came across this on wikipedia:

In C++ code, you can not use offsetof to access members of structures or classes that are not Plain Old Data Structures.

I tried it out and it seems to work fine.

class Foo
{
private:
    int z;
    int func() {cout << "this is just filler" << endl; return 0;}

public: 
    int x;
    int y;
    Foo* f;

    bool returnTrue() { return false; }
};

int main()
{
    cout << offsetof(Foo, x)  << " " << offsetof(Foo, y) << " " << offsetof(Foo, f);
    return 0;
}

I got a few warnings, but it compiled and when run it gave reasonable output:

Laptop:test alex$ ./test
4 8 12

I think I'm either misunderstanding what a POD data structure is or I'm missing some other piece of the puzzle. I don't see what the problem is.

Answer 1

You're still a bit POD. Try adding constructors and public methods that may be over-ridden.

Your object will need exta (hidden) data to manage virtual dispatch, hence it's bigger than it appears to be and hence offsetof has trouble,

Answer 2

For the definition of POD data structure,here you go with the explanation [ already posted in another post in Stack Overflow ]

http://stackoverflow.com/questions/146452/what-are-pod-types-in-c/146589

Now, coming to your code, it is working fine as expected. This is because, you are trying to find the offsetof(), for the public members of your class, which is valid.

Please let me know, the correct question, if my viewpoint above, doesnot clarify your doubt.

Answer 3

If you add, for instance, a virtual empty destructor:

virtual ~Foo() {}

Your class will become "polymorphic", i.e. it will have a hidden member field which is a pointer to a "vtable" that contains pointers to virtual functions.

Due to the hidden member field, the size of an object, and offset of members, will not be trivial. Thus, you should get trouble using offsetof.

Answer 4

In C++ you can get the relative offset like this:

class A {
public:
  int i;
};

class B : public A {
public:
  int i;
};

void test()
{
  printf("%p, %p\n", &A::i, &B::i); // edit: changed %x to %p
}

Answer 5

I bet you compile this with VC++. Now try it with g++, and see how it works...

Long story short, it's undefined, but some compilers may allow it. Others do not. In any case, it's non-portable.

Answer 6

Short answer: offsetof is a feature that is only in the C++ standard for legacy C compatibility. Therefore it is basically restricted to the stuff than can be done in C. C++ supports only what it must for C compatibility.

As offsetof is basically a hack (implemented as macro) that relies on the simple memory-model supporting C, it would take a lot of freedom away from C++ compiler implementors how to organize class instance layout.

The effect is that offsetof will often work (depending on source code and compiler used) in C++ even where not backed by the standard - except where it doesn't. So you should be very careful with offsetof usage in C++, especially since I do not know a single compiler that will generate a warning for non-POD use...

Edit: As you asked for example, the following might clarify the problem:

#include <iostream>
using namespace std;

struct A { int a; };
struct B : public virtual A   { int b; };
struct C : public virtual A   { int c; };
struct D : public B, public C { int d; };

#define offset_d(i,f)    (long(&(i)->f) - long(i))
#define offset_s(t,f)    offset_d((t*)1000, f)

#define dyn(inst,field) {\
    cout << "Dynamic offset of " #field " in " #inst ": "; \
    cout << offset_d(&i##inst, field) << endl; }

#define stat(type,field) {\
    cout << "Static offset of " #field " in " #type ": "; \
    cout.flush(); \
    cout << offset_s(type, field) << endl; }

int main() {
    A iA; B iB; C iC; D iD;
    dyn(A, a); dyn(B, a); dyn(C, a); dyn(D, a);
    stat(A, a); stat(B, a); stat(C, a); stat(D, a);
    return 0;
}

This will crash when trying to locate the field a inside type B statically, while it works when an instance is available. This is because of the virtual inheritance, where the location of the base class is stored into a lookup table.

While this is a contrived example, an implementation could use a lookup table also to find the public, protected and private sections of a class instance. Or make the lookup completely dynamic (use a hash table for fields), etc.

The standard just leaves all possibilities open by restricting offsetof to POD (IOW: no way to use a hash table for POD structs... :)

Just another note: I had to reimplement offsetof (here: offset_s) for this example as GCC actually errors out when I call offsetof for a field of a virtual base class.

Answer 7

This may answer your question.

Answer 8

I think your class fits the c++0x definition of a POD. g++ has implemented some of c++0x in their latest releases. I think that VS2008 also has some c++0x bits in it.

From wikipedia's c++0x article

C++0x will relax several rules with regard to the POD definition.

A class/struct is considered a POD if it is trivial, standard-layout, and if all of its non-static members are PODs.

A trivial class or struct is defined as one that:

1. Has a trivial default constructor. This may use the default constructor syntax (SomeConstructor() = default;).
2. Has a trivial copy constructor, which may use the default syntax.
3. Has a trivial copy assignment operator, which may use the default syntax.
4. Has a trivial destructor, which must not be virtual.

A standard-layout class or struct is defined as one that:

1. Has only non-static data members that are of standard-layout type
2. Has the same access control (public, private, protected) for all non-static members
3. Has no virtual functions
4. Has no virtual base classes
5. Has only base classes that are of standard-layout type
6. Has no base classes of the same type as the first defined non-static member
7. Either has no base classes with non-static members, or has no non-static data members in the most derived class and at most one base class with non-static members. In essence, there may be only one class in this class's hierarchy that has non-static members.

Answer 9

In general, when you ask "why is something undefined", the answer is "because the standard says so". Usually, the rational is along one or more reasons like:

• it is difficult to detect statically in which case you are.

• corned cases are difficult to define and nobody took the pain of defining special cases;

• its use is mostly covered by other features;

• existing practices at the time of standardization varied and breaking existing implementation and programs depending on them was deemed more harmful that standardization,

Back to offsetof, the second reason is probably a dominant one. If you look at C++0X, where the standard was previously using POD, it is now using "standard layout", "layout compatible", "POD" allowing more refined cases. And offsetof now needs "standard layout" classes, which are the cases where the committee didn't want to force a layout.

You have also to consider the common use of offsetof(), which is to get the value of a field when you have a void* pointer to the object. Multiple inheritance -- virtual or not -- is problematic for that use.

Answer 10

Bluehorn's answer is correct, but for me it doesn't explain the reason for the problem in simplest terms. The way I understand it is as follows:

If NonPOD is a non-POD class, then when you do:

NonPOD np;
np.field;

the compiler does not necessarily access the field by adding some offset to the base pointer and dereferencing. For a POD class, the C++ Standard constrains it to do that(or something equivalent), but for a non-POD class it does not. The compiler might instead read a pointer out of the object, add an offset to that value to give the storage location of the field, and then dereference. This is a common mechanism with virtual inheritance if the field is a member of a virtual base of NonPOD. But it is not restricted to that case. The compiler can do pretty much anything it likes. It could call a hidden compiler-generated virtual method if it wants.

In the complex cases, it is obviously not possible to represent the location of the field as an integer offset. So offsetof is not valid on non-POD classes.

In cases where your compiler just so happens to store the object in a simple way (such as single inheritance, and normally even non-virtual multiple inheritance, and normally fields defined right in the class that you're referencing the object by as opposed to in some base class), then it will just so happen to work. There are probably cases which just so happen to work on every single compiler there is. This doesn't make it valid.

Appendix: how does virtual inheritance work?

With simple inheritance, if B is derived from A, the usual implementation is that a pointer to B is just a pointer to A, with B's additional data stuck on the end:

A* ---> field of A  <--- B*
        field of A
        field of B

With simple multiple inheritance, you generally assume that B's base classes (call 'em A1 and A2) are arranged in some order peculiar to B. But the same trick with the pointers can't work:

A1* ---> field of A1
         field of A1
A2* ---> field of A2
         field of A2

A1 and A2 "know" nothing about the fact that they're both base classes of B. So if you cast a B* to A1*, it has to point to the fields of A1, and if you cast it to A2* it has to point to the fields of A2. The pointer conversion operator applies an offset. So you might end up with this:

A1* ---> field of A1 <---- B*
         field of A1
A2* ---> field of A2
         field of A2
         field of B
         field of B

Then casting a B* to A1* doesn't change the pointer value, but casting it to A2* adds sizeof(A1) bytes. This is the "other" reason why, in the absence of a virtual destructor, deleting B through a pointer to A2 goes wrong. It doesn't just fail to call the destructor of B and A1, it doesn't even free the right address.

Anyway, B "knows" where all its base classes are, there always stored at the same offsets. So in this arrangement offsetof would still work. The standard doesn't require implementations to do multiple inheritance this way, but they often do (or something like it). So offsetof might work in this case on your implementation, but it is not guaranteed to.

Now, what about virtual inheritance? Suppose B1 and B2 both have A as a virtual base. This makes them single-inheritance classes, so you might think that the first trick will work again:

A* ---> field of A   <--- B1* A* ---> field of A   <--- B2* 
        field of A                    field of A
        field of B1                   field of B2

But hang on. What happens when C derives (non-virtually, for simplicity) from both B1 and B2? C must only contain 1 copy of the fields of A. Those fields can't immediately precede the fields of B1, and also immediately precede the fields of B2. We're in trouble.

So what implementations might do instead is:

// an instance of B1 looks like this, and B2 similar
A* --->  field of A
         field of A
B1* ---> pointer to A
         field of B1

The "pointer" to A could be a pointer or an offset, it doesn't really matter. In an instance of B1, created as a B1, it points to (char*)this - sizeof(A), and the same in an instance of B2. But if we create a C, it will look like this:

A* --->  field of A
         field of A
B1* ---> pointer to A    // points to (char*)(this) - sizeof(A) as before
         field of B1
B2* ---> pointer to A    // points to (char*)(this) - sizeof(A) - sizeof(B1)
         field of B2
C* ----> pointer to A    // points to (char*)(this) - sizeof(A) - sizeof(B1) - sizeof(B2)
         field of C
         field of C

So to access a field of A using a pointer or reference to B2 requires more than just applying an offset. We must read the "pointer to A" field of B2, follow it, and only then apply an offset, because depending what class B2 is a base of, that pointer will have different values. There is no such thing as offsetof(B2,field of A): there can't be. offsetof will never work with virtual inheritance, on any implementation.