What's correct way to remove a boost::shared_ptr from a list?
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Published on 2010-04-08T00:59:34Z
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2010/04/08
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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?
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