Must elements of a std::vector be contiguous?

[MacApp]

I read in
http://www.open-std.org/jtc1/sc22/wg...efects.html#69
they proposed:

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()
 

My question is: does the above implies the:

(Type *) &v<char>[n]==(Type *)(&v<char>[0] + n*sizeof (Type));
 

?
where v is a std::vector of char, and Type is in general a simple structured type.
Thanks and sorry for my bad english

[Banfa]

No, in fact it implies precisely not that as long as sizeof(Type) != 1.

for

vector<char> v;

as you have already stated

&v[n] == &v[0] + n

for for sizeof(Type) != 1 then n != n * sizeof(Type)

so

&v[n] != &v[0] + n*sizeof (Type);

However elements of a vector must be contiguous, that is what the standard says and that is true.

[MacApp]

Thanks for the fast reply, and you are right... i made a mistake in my question sorry.

I meant:

&v[n] == &v[0] + n;
 

implies:

(Type *) &v<char>[n*sizeof (Type)]==(Type *)(&v<char>[0] + n*sizeof (Type));
 

?

where, in pratic v<char> is used as a C buffer of chars containing infact an array of Type.

[Banfa]

Well in that case you are correct. However I wouldn't necessarily call that good programing practice particularly in a C++ program. On the whole if the object has a type it should be held as that type.