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vector<string> buf_string;
buf_string.reserve(256);
vector<int> buf_mat_prices;
buf_mat_prices.reserve(1000);

During loops I fill the vectors and then I empty them with commands
like buf_string.clear();
buf_mat_prices.clear();

Does this mean that the memory allocation returns to default or is my
original reserve still in place?

pkirk25 wrote in message
< .com>...
> vector<string> buf_string;
> buf_string.reserve(256);
> vector<int> buf_mat_prices;
> buf_mat_prices.reserve(1000);
>
>During loops I fill the vectors and then I empty them with commands
>like buf_string.clear();
>buf_mat_prices.clear();
>
>Does this mean that the memory allocation returns to default or is my
>original reserve still in place?


// Prove it for yourself: // using std::cout;
cout << buf_string.size() << std::endl; // S/B '0'
cout << buf_string.capacity() << std::endl; // clear() doesn't resize

cout << buf_mat_prices.size() << std::endl; // S/B '0'
cout << buf_mat_prices.capacity() << std::endl; // clear() doesn't
resize

You can re-size the vector by using (surprise!):

buf_string.resize(25);
cout << buf_string.size() << std::endl;
cout << buf_string.capacity() << std::endl;

That help?
--
Bob R
POVrookie

Very neat - many thanks! I didn't see the capacity member

pkirk25 wrote:
> vector<int> buf_mat_prices;
> buf_mat_prices.reserve(1000);
> buf_mat_prices.clear();
>
> Does this mean that the memory allocation returns to default or is my
> original reserve still in place?

No, you still have the reservation. To "unreserve" you need to do a
swap-with-new:

vector<int>( ).swap(buf_mat_prices).

The vector<int>( ) is a temporary that gets the 1000 reservation, and
then dies.
buf_mat_prices gets the capacity of the freshly constructed temporary.

HTH,
Michiel Salters