std vector C++ — deep or shallow copy

Question

I wonder whether copying a vector I am copying the vector with its values (whereas this is not working with array, and deep copy need a loop or memcpy).

Could you hi

Answer 1

Vector will resize to have enough space for the objects. It will then iterate through the objects and call the default copy operator for every object.

In this way, the copy of the vector is 'deep'. The copy of each object in the vector is whatever is defined for the default copy operator.

In examples... this is BAD code:

#include <iostream>
#include <vector>

using namespace std;

class my_array{
public:
    int *array;
    int size;
    my_array(int size, int init_val):size(size){
        array = new int[size];
        for(int i=0; i<size; ++i)
            array[i]=init_val;
    }
    ~my_array(){
        cout<<"Destructed "<<array[0]<<endl;
        if(array != NULL)
            delete []array;
        array = NULL;
        size = 0;
    }

};

void add_to(vector<my_array> &container){
    container.push_back(my_array(4,1));
}

int main(){

    vector<my_array> c;
    {
        my_array a(5,0);
        c.push_back(a);
    }
    add_to(c);
    //At this point the destructor of c[0] and c[1] has been called.
    //However vector still holds their 'remains'
    cout<<c[0].size<<endl; //should be fine, as it copies over with the = operator
    cout<<c[0].array[0]<<endl;//undefined behavior, the pointer will get copied, but the data is not valid
    return 0;
}

This is BETTER code:

#include <iostream>
#include <vector>

using namespace std;

class my_array{
public:
    int *array;
    int size;
    my_array(int size, int init_val):size(size){
        cout<<"contsructed "<<init_val<<endl;
        array = new int[size];
        for(int i=0; i<size; ++i)
            array[i]=init_val;
    }
    my_array(const my_array &to_copy){
        cout<<"deep copied "<<to_copy.array[0]<<endl;
        array = new int[to_copy.size];
        size = to_copy.size;
        for(int i=0; i<to_copy.size; i++)
            array[i]=to_copy.array[i];
    }

    ~my_array(){
        cout<<"Destructed "<<array[0]<<endl;
        if(array != NULL)
            delete []array;
        array = NULL;
        size = 0;
    }

};

void add_to(vector<my_array> &container){
    container.push_back(my_array(4,1));
}

int main(){

    vector<my_array> c;
    {
        my_array a(5,0);
        c.push_back(a);
    }
    add_to(c);
    //At this point the destructor of c[0] and c[1] has been called.
    //However vector holds a deep copy'
    cout<<c[0].size<<endl; //This is FINE
    cout<<c[0].array[0]<<endl;//This is FINE
    return 0;
}

Answer 2

You are making a deep copy any time you copy a vector. But if your vector is a vector of pointers you are getting the copy of pointers, not the values are pointed to

For example:

std::vector<Foo> f;
std::vector<Foo> cp = f; //deep copy. All Foo copied

std::vector<Foo*> f;
std::vector<Foo*> cp = f; //deep copy (of pointers), or shallow copy (of objects).
//All pointers to Foo are copied, but not Foo themselves