Does vector::erase() on a vector of object pointers destroy the object itself?
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 obj开发者_开发技巧ect 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.
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;
}
- If you have a
vector<MyObject>
thenMyObject::~MyObject
will be called. - If you have a
vector<MyObject*>
thendelete <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();
}
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.
Yes. vector::erase
destroys the removed object, which involves calling its destructor.
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.
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