Question

I would like to be able to convert between std::vector and its underlying C array int* without explicitly copying the data.

Does std::vector provide access to the underlying C array? I am looking for something like this

vector<int> v (4,100)
int* pv = v.c_array();

EDIT:

Also, is it possible to do the converse, i.e. how would I initialize an std::vector from a C array without copying?

int pv[4] = { 4, 4, 4, 4};
vector<int> v (pv);

Answer 1

You can get a pointer to the first element as follows:

int* pv = &v[0];

This pointer is only valid as long as the vector is not reallocated. Reallocation happens automatically if you insert more elements than will fit in the vector s remaining capacity (that is, if v.size() + NumberOfNewElements > v.capacity(). You can use v.reserve(NewCapacity) to ensure the vector has a capacity of at least NewCapacity.

Also remember that when the vector gets destroyed, the underlying array gets deleted as well.

Answer 2

In c++11, you can use vector::data() to get C array pointer.

Answer 3

int* pv = &v[0]

Note that this is only the case for std::vector<>, you can not do the same with other standard containers.

Scott Meyers covers this topic extensively in his books.

Answer 4

If you have very controlled conditions, you can just do:

std::vector<int> v(4,100);
int* pv = &v[0];

Be warned that this will only work as long as the vector doesn t have to grow, and the vector will still manage the lifetime of the underlying array (that is to say, don t delete pv). This is not an uncommon thing to do when calling underlying C APIs, but it s usually done with an unnamed temporary rather than by creating an explicit int* variable.

Answer 5

One way of protecting yourself against size changes is to reserve the maximal space (or larger) that you will need:

std::vector<int> v(4,100); //Maybe need 
v.reserve(40);             //reallocate to block out space for 40 elements

This will ensure that push_backs won t cause reallocation of the existing data.

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