Are std::vector elements guaranteed to be contiguous?

This was missed from C++98 standard proper but later added as part of a TR. The forthcoming C++0x standard will of course contain this as a requirement.

From n2798 (draft of C++0x):

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().

As other answers have pointed out, the contents of a vector is guaranteed to be continuous (excepting bool's weirdness).

The comment that I wanted to add, is that if you do an insertion or a deletion on the vector, which could cause the vector to reallocate it's memory, then you will cause all of your saved pointers and iterators to be invalidated.

The standard does in fact guarantee that a vector is continuous in memory and that &a[0] can be passed to a C function that expects an array.

The exception to this rule is vector<bool> which only uses one bit per bool thus although it does have continuous memory it can't be used as a bool* (this is widely considered to be a false optimization and a mistake).

BTW, why don't you use iterators? That's what they're for.

As other's have already said, vector internally uses a contiguous array of objects. Pointers into that array should be treated as invalid whenever any non-const member function is called IIRC.

However, there is an exception!!

vector<bool> has a specialised implementation designed to save space, so that each bool only uses one bit. The underlying array is not a contiguous array of bool and array arithmetic on vector<bool> doesn't work like vector<T> would.

(I suppose it's also possible that this may be true of any specialisation of vector, since we can always implement a new one. However, std::vector<bool> is the only, err, standard specialisation upon which simple pointer arithmetic won't work.)