C/C++ Struct memory layout equivalency

Question

Consider the following C struct and C++ struct declarations:

extern "C" { // if this matters
typedef struct Rect1 {
  int x, y;
  int w, h;
} Rect1;
}

struct Vector {
  int x;
  int y;
}

struct Rect2 {
  Vector pos;
  Vector size;
}

• Are the memory layouts of Rect1 and Rect2 objects always identical?

• Specifically, can I safely reinterpret_cast from Rect2* to Rect1* and assume that all four int values in the Rect2 object are matched one on one to the four ints in Rect1?

• Does it make a difference if I change Rect2 to a non-POD type, e.g. by adding a constructor?

Answer 1

• I would think so, but I also think there could (legally) be padding between Rect2::pos and Rect2::size. So to make sure, I would add compiler-specific attributes to "pack" the fields, thereby guaranteeing all the ints are adjacent and compact. This is less about C vs. C++ and more about the fact that you are likely using two "different" compilers when compiling in the two languages, even if those compilers come from a single vendor.
• Using reinterpret_cast to convert a pointer to one type to a pointer to another, you are likely to violate "strict aliasing" rules. Assuming you do dereference the pointer afterward, which you would in this case.
• Adding a constructor will not change the layout (though it will make the class non-POD), but adding access specifiers like private between the two fields may change the layout (in practice, not only in theory).