Dynamically allocate a vector using new keyword-CodePudding

I was wondering if it is possible to dynamically allocate a vector using new keyword, similar to an array.

what I mean is this:

vector<int> *vptr = new vector<int>;

I could not find proper reference about this problem over the internet. I would like to know what the below statement means. Both are valid syntax. Also how to dereference the pointer here.

vector<int> *vptr = new vector<int>[10];
vector<int> *vptr1 = new vector<int>(2,5);

CodePudding user response：

First things first, creating std::vector dynamically(like you do in the example) should not be your first choice.

I would like to know what the below statement means

Let's look at them one by one.

Case 1

Here we consider the statement:

vector<int> *vptr = new vector<int>[10];

This creates 10 empty std::vector<int>s and initialize the pointer vptr with the pointer to the very first of those dynamically created vectors. You can compare this with int *ptr = new int[10];.

Case 2

Here we consider the statment:

vector<int> *vptr1 = new vector<int>(2,5);

This creates a single std::vector<int> of size 2 with each of its elements initialized to 5 and then the pointer vptr is initialized with the pointer to that dynamically created vector.

CodePudding user response：

#include <iostream>
#include <vector>


using namespace std;

int main() {
    // v is pointer toward a vector<int>
    vector<int> *v = new vector<int>();
    // So we should use '->' to dereference
    // and use the method push_back
    v->push_back(1);
    v->push_back(2);
    v->push_back(3);
    for (auto i = 0u; i < v->size(); i  ) {
        // we are using at method to access 
        // elements as it is more readable 
        // than (*v)[i]
        cout << v->at(i) << endl;
    }
    // v1 is an array of vector<int>
    vector<int> *v1 = new vector<int>[10];
    for (int i = 0; i < 10; i  ) {
        // we access to the ith vector with 
        // the [] operator same way we deal 
        // with other arrays. v[i] is of type 
        // vector so we use directly the push_back method
        v1[i].push_back(i);
    }
    for (int i = 0; i < 10; i  ) {
        cout << v1[i][0] << endl;
    }
    delete [] v1;
    
    // same as v
    vector<int> *v2 = new vector<int>(2, 3);
    for (auto i = 0u; i < v2->size(); i  ) {
        cout << v2->at(i) << endl;
    }

    return 0;
}