shallow or deep copy or the array

Question

I am trying to solve a problem. I have a class with an int array prix. If I copy the object Test with the copy constructor. Will it make a deep or a shallow copy of the int array prix?

I cannot use any stl containers ((

Test
{
    Test(const char *id), id(id);
    {
        prix=new int[1];
    }
    Test(const Test & rhs): 
        id(rhs.id),
        prix(rhs.prix)
        {}
    //...
    const char *id;
    int * prix;
};

edit: I was wrong then, it is just a pointer. How can I copy the array which is pointed?

Answer 1

If the allocated array always has size equal to 1 then the constructor will look as

Test(const Test & rhs) : id( rhs.id )
{
    prix = new int[1];
    *prix = *rhs.prix;
}

Or if the compiler supports initializer lists then the constructor can be written as

Test(const Test & rhs) : id( rhs.id )
{
    prix = new int[1] { *rhs.prix };
}

Otherwise the class has to have an additional data member that will contain the number of elements in the array. Let assume that size_t size is such a data member. Then the constructor will look as

#include <algorithm>

//... 

Test(const Test & rhs) : id( rhs.id ), size( rhs.size )
{
    prix = new int[size];
    std::copy( rhs.prix, rhs.prix + rhs.size, prix );
}

You could write for example as

Test(const Test & rhs) : id( rhs.id ), size( rhs.size ), prix( new int[size] )
{
    std::copy( rhs.prix, rhs.prix + rhs.size, prix );
}

but in this case data member size has to be defined before data member prix in the class definition.