What's correct way to remove a boost::shared_ptr from a list?

Question

I have a std::list of boost::shared_ptr<T> and I want to remove an item from it but I only have a pointer of type T* which matches one of the items in the list.

However I cant use myList.remove( tPtr ) I'm guessing because shared_ptr does not implement == for its template argument type.

My immediate thought was to try myList.remove( shared_ptr<T>(tPtr) ) which is syntactically correct but it will crash from a double delete since the temporary shared_ptr has a separate use_count.

std::list< boost::shared_ptr<T> > myList;

T* tThisPtr = new T(); // This is wrong; only done for example code.
                       // stand-in for actual code in T using 
                       // T's actual "this" pointer from within T
{
   boost::shared_ptr<T> toAdd( tThisPtr ); // typically would be new T()
   myList.push_back( toAdd );
}

{
   //T has pointer to myList so that upon a certain action, 
   // it will remove itself romt the list

   //myList.remove( tThisPtr);                      //doesn't compile
   myList.remove( boost::shared_ptr<T>(tThisPtr) ); // compiles, but causes
                                                    // double delete
}  

The only options I see remaining are to use std::find with a custom compare, or to loop through the list brute force and find it myself, but it seems there should be a better way.

Am I missing something obvious, or is this just too non-standard a use to be doing a remove the clean/normal way?

Answer 1

enable_shared_from_this can help with your problem, but it will require that the types you're using in the list derive from it:

• http://www.boost.org/doc/libs/1_42_0/libs/smart_ptr/enable_shared_from_this.html

If the type enables that functionality, you can get the shared pointer from the object itself by calling shared_from_this().

Answer 2

std::list's remove_if member is what you need:

Define a predicate

template <typename T> struct shared_equals_raw
{
  shared_equals_raw(T* raw)
    :_raw(raw)
    {}
  bool operator()(const boost::shared_ptr<T>& ptr) const
    {
      return (ptr.get()==_raw);
    }
private:
  T*const _raw;
};

then you can call

myList.remove_if(shared_equals_raw(tThisPtr));

to have the list clean up nodes which have shared_ptrs to tThisPtr.

(Untested, so maybe some syntactic stuff needs fixing).

Michael Burr's advice re enable_shared_from_this is good though; it'd be better to avoid having the raw tThisPtr in play at all.

Answer 3

You're correct, we can't directly compare the pointers. But there does exist remove_if, and we can specify our own predicate. The solution:

template <typename T>
struct ptr_contains_predicate
{
    ptr_contains_predicate(T* pPtr) :
    mPtr(pPtr)
    {}

    template <typename P>
    bool operator()(const p& pPtr) const
    {
        return pPtr.get() == mPtr;
    }

    T* mPtr;
};

template <typename T>
ptr_contains_predicate<T> ptr_contains(T* pPtr)
{
    return ptr_contains_predicate<T>(pPtr);
}

Just keep the above predicate in a header somewhere, and you can use it wherever you want.

myList.remove_if(ptr_contains(tThisPtr));

The best solution is to never lose hold of the shared_ptr in the first place, so we can just use remove, but the above is harmless anyway.

Answer 4

Can you use the shared pointer to remove it?

std::list< boost::shared_ptr<T> > myList;

boost::shared_ptr<T> tThisPtr = new T(); 

{
    myList.push_back(tThisPtr);
}

{
    myList.remove(tThisPtr);                     
}