c++: How distant(in bytes) are class member in an array?

Question

I found no better way to formulate my question like this: Is the output below always true? is this portable?

struct Point
{
    int x;
    int y;
};

//...
std::vector<Point> points(3);
unsigned char* start = (unsigned char*)(&points[0]);
unsigned char* end = (unsigned char*)(&points[1]); 

std::cout << "is this the same ? " << std::distance(start,end) == sizeof(Point);

What if instead of a vector, points were defined as a raw array? Is the output still always true?

Point *points = new Point[3]; // instead of std::vector<Point> points(3);

Answer 1

For std::vector, [vector.overview]/1 (N3337) says:

The elements of a vector are stored contiguously, meaning that if v is a vector<T, Allocator> where T is some type other than bool, then it obeys the identity &v[n] == &v[0] + n for all 0 <= n < v.size().

So yes, your program's behaviour is portable and well-defined.

For arrays, [dcl.array]/1 states:

An object of array type contains a contiguously allocated non-empty set of N subobjects of type T.

This is not quite so explicit as the vector quote, but the co-usage of the word "contiguous" points to the fact that the std::vector storage identity also applies to arrays.

Answer 2

Yes, one of the requirement for the vector is that its elements must be store contiguously.

From n2798:

23.2.6 Class template vector [vector]

1 A vector is a sequence container that supports random access iterators. In addition, it supports (amortized) constant time insert and erase operations at the end; insert and erase in the middle take linear time. Storage management is handled automatically, though hints can be given to improve efficiency. The elements of a vector are stored contiguously, meaning that if v is a vector where T is some type other than bool, then it obeys the identity &v[n] == &v[0] + n for all 0 <= n < v.size().