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?


1/17/2015 10:43:01 AM

Accepted Answer

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

To make it clear you can rewrite

int a = (*vecPtr)[i];


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.

10/18/2011 1:27:59 PM

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:


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)[].

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