Is vector in c++ a pointer?

Question

So lets say we have a vector v and I pass this to a function f

void f(vector<int> &v) {
    cout<<&v;
}

int main() {
    vector<int> v;
    v.push_back(10);  
    v.push_back(20);
    v.push_back(30);

    cout<<&v[0]<<"\n"<<&v<<"\n";
    f(v);
}

For this I get the outputs as:

0x55c757b81300
0x7ffded6b8c70
0x7ffded6b8c70

We can see that &v and &v[0] have different address. So is va pointer that points to the start of the vector? If so, why can't I access *v ? Is there a mistake I'm making?

Thanks

Answer 1

Is vector in c++ a pointer?

No. std::vector is not a pointer. It is a class.

So is va pointer that points to the start of the vector?

No, but v (an instance of the vector class) does happen to contain a pointer to the start of the array.

why can't I access *v ?

Because there is no overload for the unary operator* for the class std::vector .

Answer 2

The std::vector structure may contain additional data, such as the length. The contiguous elements are stored in a separate location and that location changes as the array is resized. &v should remain the same so long as v isn't moved, but &v[0] can be invalidated for a number of reasons.

Imagine, in very rough terms, it's:

struct vector {
  size_t size;
  T* elements;
};

The operator[] implementation takes over when you employ &v[0] .

It's not a pointer, and it behaves very differently from new[] .

Answer 3

std::vector is a sequence container that encapsulates dynamic size arrays. So definately, it is not a pointer.

As @tadman said, v (an instance of the vector class) does happen to contain a pointer to the start of the array.

You cannot access *v because there is no overload for the unary operator* for the class std::vector .

Answer 4

No the vector in c++ is not a pointer. The vector is a class in c++