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Why is C++11's POD "standard layout" definition the way it is?

Asked 2011-08-23T12:16:54.870
56

I'm looking into the new, relaxed POD definition in C++11 (section 9.7)

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.

I've highlighted the bits that surprised me.

What would go wrong if we tolerated data members with varying access controls?

What would go wrong if the first data member was also a base class? i.e.

struct Foo {};
struct Good : Foo {int x; Foo y;};
struct Bad  : Foo {Foo y; int x;};

I admit it's a weird construction, but why should Bad be prohibited but not Good?

Finally, what would go wrong if more than one constituent class had data members?

c++ c++11 standard-layout
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1 Answer

9

What would go wrong if we tolerated data members with varying access controls?

The current language says that the compiler cannot reorder members under the same access control. Like:

struct x
{
public:
    int x;
    int y;
private:
    int z;
};

Here x must be allocated before y, but there is no restriction on z relative to x and y.

struct y
{
public:
    int x;
public:
    int y;
};

The new wording says that y is still a POD despite the two publics. This is actually a relaxation of the rules.

answered 2011-08-23T12:50:54.227

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