Dereference vector pointer to access element

If i have in C++ a pointer to a vector:

vector<int>* vecPtr;

And i'd like to access an element of the vector, then i can do this by dereferncing the vector:

int a = (*vecPtr)[i];

but will this dereferencing actually create a copy of my vector on the stack? let's say the vector stores 10000 ints, will by dereferencing the vecPtr 10000 ints be copied?

Thanks!

10000 ints will not be copied. Dereferencing is very cheap.

To make it clear you can rewrite

int a = (*vecPtr)[i];

as

vector<int>& vecRef = *vecPtr; // vector is not copied here
int a = vecRef[i];

In addition, if you are afraid that the whole data stored in vector will be located on the stack and you use vector<int>* instead of vector<int> to avoid this: this is not the case. Actually only a fixed amount of memory is used on the stack (about 16-20 bytes depending on the implementation), independently of the number of elements stored in the vector. The vector itself allocates memory and stores elements on the heap.

No, nothing will be copied; dereferencing just tells C++ that you want to invoke operator[] on the vector, not on your pointer, vecPtr. If you didn't dereference, C++ would try to look for an operator[] defined on the std::vector<int>* type.

This can get really confusing, since operator[] is defined for all pointer types, but it amounts to offsetting the pointer as though it pointed to an array of vector<int>. If you'd really only allocated a single vector there, then for any index other than 0, the expression evaluates to a reference to garbage, so you'll get either a segfault or something you did not expect.

In general, accessing vectors through a pointer is a pain, and the (*vecPtr)[index] syntax is awkward (but better than vecPtr->operator[](index)). Instead, you can use:

vecPtr->at(index)

This actually checks ranges, unlike operator[], so if you don't want to pay the price for checking if index is in bounds, you're stuck with (*vecPtr)[].