Does vector::erase() on a vector of object pointers destroy the object itself?

Question

I have a vector of pointers to objects. I need to remove an element from the vector and place that element in another list.

I read that erase can be used to remove the object from the vector, but I also read that it calls the objects destructor before doing so.

I need to know whether or not erasing the object will destroy it as well.

Answer 1

vector::erase
Removes from the vector container and calls its destructor but If the contained object is a pointer it doesnt take ownership of destroying it.

You will have to explicitly call delete on each contained pointer to delete the content it is pointing to, for example:

void clearVectorContents( std::vector <YourClass*> & a ) 
{    
    for ( int i = 0; i < a.size(); i++ ) 
    {       
        delete a[i];    
    }    
    a.clear(); 
} 

Storing raw pointers in standard containers is not a good idea. If you really need to store resources that have to be allocated by new, then you should use boost::shared_ptr. Check out the Boost documentation.

An more generic & elegant solution:
This solution makes use of for_each & templates as @Billy pointed out in comments:

// Functor for deleting pointers in vector.
template<class T> class DeleteVector
{
    public:
    // Overloaded () operator.
    // This will be called by for_each() function.
    bool operator()(T x) const
    {
        // Delete pointer.
        delete x;
        return true;
    }
};

And this can be called as:

for_each( myclassVector.begin(),myclassVector.end(),
          DeleteVector<myclass*>());

where, myclassVector is your vector containing pointers to myclass class objects.

Usage Example:

#include "functional"
#include "vector"
#include "algorithm"
#include "iostream"

//Your class
class myclass
{
    public:
        int i;
        myclass():i(10){}
};


// Functor for deleting pointers in vector.
template<class T> class DeleteVector
{
    public:
    // Overloaded () operator.
    // This will be called by for_each() function.
    bool operator()(T x) const
    {
        // Delete pointer.
        delete x;
        return true;
    }
};


int main()
{
    // Add 10 objects to the vector.
    std::vector<myclass*> myclassVector;

    for( int Index = 0; Index < 10; ++Index )
    {
        myclassVector.push_back( new myclass);
    }

    for (int i=0; i<myclassVector.size(); i++) 
    {
        std::cout << " " << (myclassVector[i])->i;
    }

    // Now delete the vector contents in a single  line.
    for_each( myclassVector.begin(),
              myclassVector.end(),
              DeleteVector<myclass*>());

    //Clear the vector 
    myclassVector.clear();

    std::cout<<"\n"<<myclassVector.size();

    return 0;
}

Answer 2

• If you have a vector<MyObject> then MyObject::~MyObject will be called.
• If you have a vector<MyObject*> then delete <MyObject instance> will not be called.

To move MyObject between two vectors, you need to first insert it into the destination vector, then erase the original. Note that this will create a new copy of the object. When moving pointers, you can just keep the pointer in a temporary variable, erase it from the vector, then insert wherever you need.

And here's another simple way to delete and then remove all the items in a vector:

template<class T> void purge( std::vector<T> & v ) {
    for ( auto item : v ) delete item;
    v.clear();
}

Answer 3

Yes, erase destroys the element. However, if you're placing the element in another container you probably made a copy putting it into that other container. The only way you'd run into issues is if you copied a pointer or something like that into the other container.

Answer 4

Yes, of course it will. If the object doesn't exist in the vector, where else would it exist?

Edit: This will not delete anything pointed to by a pointer. You should use automatic life-time managing pointers such as shared_ptr to manage object lifetimes.

Answer 5

Yes. vector::erase destroys the removed object, which involves calling its destructor.