1
我想知道是否有可能使用remove_if和lambda表達式表示此表達式。list :: remove_if等效
std::list< gh::Actor* >::iterator astit = actors.begin();
while (astit != actors.end())
{
if((*astit)->state == DELETE_STATE)
{
Actor* reference = *astit;
actors.erase(astit++);
delete reference;
}
else
{
++astit;
}
}
這是行不通的? http://en.cppreference.com/w/cpp/container/list/remove – chris