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Is this possible:

class base;
class derived; //ublic base

vector <base*> bList;
vector<derived*> dList;

//add some derived class pointer entries to dList;

bList = dList;
//OR bList = (vector <base*>) dList;

//use the entries as base class pointers


- CD

"CD" <> wrote in message
news: om...
> Is this possible:
>
> class base;
> class derived; //ublic base
>
> vector <base*> bList;
> vector<derived*> dList;
>
> //add some derived class pointer entries to dList;
>
> bList = dList;
No: assignment will only work if both vectors have the same type.
> //OR bList = (vector <base*>) dList;
No: something similar *might* work in practice, but would be UB
(undefined behavior).

The modified code below will work, however:

#include <vector>
using namespace std;

class Base {};
class Derived : public Base {};

void f( vector<Derived*> const d, vector<Base*>& b )
{
b.assign( d.begin(), d.end() ); // ok, copies & converts the pointers
// b now has a (converted) copy of the contents of b
}


hth,
Ivan
--
http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form
Brainbench MVP for C++ <> http://www.brainbench.com

CD wrote:
> Is this possible:
>
> class base;
> class derived; //ublic base
>
> vector <base*> bList;
> vector<derived*> dList;
>
> //add some derived class pointer entries to dList;
>
> bList = dList;
> //OR bList = (vector <base*>) dList;
>
> //use the entries as base class pointers

Yes, it is possible. Since 'derived*' is convertible to 'base*',
a simple 'std::copy' should suffice:

std::copy(dList.begin(), dList.end(), std::back_inserter(bList));

Since vector<base*> and vector<derived*> are not the same type, and even
not related, an assignment cannot work.

Victor