30

In c/c++ (I am assuming they are the same in this regard), if I have the following:

struct S {
  T a;
  .
  .
  .
} s;

Is the following guaranteed to be true?

(void*)&s == (void*)&s.a;

Or in other words, is there any kind of guarantee that there will be no padding before the first member?

Cole Tobin
  • 9,206
  • 15
  • 49
  • 74
Baruch
  • 20,590
  • 28
  • 126
  • 201
  • 1
    They aren't the same in this regard – Mooing Duck Jul 24 '13 at 21:34
  • 3
    @MooingDuck: Well, it depends on what the `...` is. For the same struct definition, C++ will treat it the same way as C (C structures always meet the C++ *standard-layout* requirement). – Ben Voigt Jul 24 '13 at 21:51

2 Answers2

43

In C, yes, they're the same address. Simple, and straightforward.

In C++, no, they're not the same address. Base classes can (and I would suspect, do) come before all members, and virtual member functions usually add hidden data to the struct somewhere. Even more confusing, a C++ compiler may also rearrange members at will, unless the class is a standard layout type (though I don't know that any compiler does so)

Finally, if the C++ struct is composed of standard layout types, contains no base classes nor virtual functions and all members have the same visibility, and possibly other limitations I forgot, then it falls back on the C rules, and requires the first member to be at the same address as the object itself.

§ 9.2/7

A standard-layout class is a class that:
— has no non-static data members of type non-standard-layout class (or array of such types) or reference,
— has no virtual functions (10.3) and no virtual base classes (10.1),
— has the same access control (Clause 11) for all non-static data members,
— has no non-standard-layout base classes,
— either has no non-static data members in the most derived class and at most one base class with non-static data members, or has no base classes with non-static data members, and
— has no base classes of the same type as the first non-static data member.

§ 9.2/20

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 ]

Mooing Duck
  • 64,318
  • 19
  • 100
  • 158
  • "standard-layout class" is also known as POD ("plain old data type"), and you can verify it using std::is_pod. – marcinj Jul 24 '13 at 21:48
  • 10
    @marcin_j: No. POD requires both *standard-layout* and *trivial construction/copy/destruction*. A class can have non-defaulted constructors and destructors and other special members and still be *standard-layout*. – Ben Voigt Jul 24 '13 at 21:48
  • thanks for clarification, I see that it is possible to use std::is_standard_layout to verify if such comparison is correct. – marcinj Jul 24 '13 at 21:58
  • "Base classes can (and I would suspect, do) come before all members" -- It's undefined where bases are in layout; some compilers (none of the "mainstream" ones) do put them at the end of the class, iirc. – Sean Middleditch Jul 24 '13 at 23:21
  • @SeanMiddleditch: I was aware it was IB, but I have heard that most compilers put _nonvirtual_ base classes before members, and I've heard mixed things about the placement of virtual base classes. – Mooing Duck Jul 25 '13 at 05:01
  • @SeanMiddleditch Of course it is, that's why it is a *"can"* and not a *"do"*. – Christian Rau Jul 25 '13 at 07:29
16

Yes, it is.

It is guaranteed there is no padding before the first struct member in C and in C++ (if it is a POD).

C quote:

(C11, 6.7.2.1p15) "There may be unnamed padding within a structure object, but not at its beginning."

C++ quote:

(C++11, 9.2p20) "There might therefore be unnamed padding within a standard-layout struct object, but not at its beginning, as necessary to achieve appropriate alignment"

ouah
  • 142,963
  • 15
  • 272
  • 331