When extending a padded struct, why can’t extra fields be placed in the tail padding?

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Short answer (for the C++ part of the question): The Itanium ABI for C++ prohibits, for historical reasons, using the tail padding of a base subobject of POD type. Note that C++11 does not have such a prohibition. The relevant rule 3.9/2 that allows trivially-copyable types to be copied via their underlying representation explicitly excludes base subobjects.

Long answer: I will try and treat C++11 and C at once.

  1. The layout of S1 must include padding, since S1::a must be aligned for int, and an array S1[N] consists of contiguously allocated objects of type S1, each of whose a member must be so aligned.
  2. In C++, objects of a trivially-copyable type T that are not base subobjects can be treated as arrays of sizeof(T) bytes (i.e. you can cast an object pointer to an unsigned char * and treat the result as a pointer to the first element of a unsigned char[sizeof(T)], and the value of this array determines the object). Since all objects in C are of this kind, this explains S2 for C and C++.
  3. The interesting cases remaining for C++ are:
    1. base subobjects, which are not subject to the above rule (cf. C++11 3.9/2), and
    2. any object that is not of trivially-copyable type.

For 3.1, there are indeed common, popular “base layout optimizations” in which compilers “compress” the data members of a class into the base subobjects. This is most striking when the base class is empty (∞% size reduction!), but applies more generally. However, the Itanium ABI for C++ which I linked above and which many compilers implement forbids such tail padding compression when the respective base type is POD (and POD means trivially-copyable and standard-layout).

For 3.2 the same part of the Itanium ABI applies, though I don’t currently believe that the C++11 standard actually mandates that arbitrary, non-trivially-copyable member objects must have the same size as a complete object of the same type.

Previous answer kept for reference.

I believe this is because S1 is standard-layout, and so for some reason the S1-subobject of S3 remains untouched. I’m not sure if that’s mandated by the standard.

However, if we turn S1 into non-standard layout, we observe a layout optimization:

struct EB { };

struct S1 : EB {   // not standard-layout
    EB eb;
    int a;
    char b;
};

struct S3 : S1 {
    char c;
};

Now sizeof(S1) == sizeof(S3) == 12 on my platform. Live demo.

And here is a simpler example:

struct S1 {
private:
    int a;
public:
    char b;
};

struct S3 : S1 {
    char c;
};

The mixed access makes S1 non-standard-layout. (Now sizeof(S1) == sizeof(S3) == 8.)

Update: The defining factor seems to be triviality as well as standard-layoutness, i.e. the class must be POD. The following non-POD standard-layout class is base-layout optimizable:

struct S1 {
    ~S1(){}
    int a;
    char b;
};

struct S3 : S1 {
    char c;
};

Again sizeof(S1) == sizeof(S3) == 8. Demo

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