everything is in the title. Im new in C++ and im not sure that I undertand shared_ptr properly...
I've this method:
const std::set<Something::Ptr, b> & getSome() const;
that I use to get a set of somethings :
auto s = u->second.getSome();
After that i want to iterate on it with :
for(auto i = s.begin();i != s; s.end();i++)
//(so i= std::shared_ptr<Something> referring to the first element in s )
My question is how can i access i methods?
I tried to understand what I was working with by debugging and cout some things :
auto whatIsThat1 = *i;
cout << "hello" << whatIsThat1; //>> hello0x13fd500
auto whatIsThat2 = i->get();
cout << "hello" << whatIsThat2; >> hello0x21c2500
I think you are confused because arretCourant is not a std::shared_ptr. It is a std::set iterator referring to an element of the std::set, which is a std::shared_ptr.
So in order to call a method on the object that the std::shared_ptr points to, you need to first dereference the iterator to get a reference to the std::shared_ptr and then dereference again to get a reference to the object that the std::shared_ptr is pointing to. So to call a method on that object use:
(*arretCourant)->methodName()
std::shared_ptr overloads the -> operator to make it call the method on the object pointed to, rather than the std::shared_ptr itself. It also overloads the indirection operator * to return a reference to the object it is pointing to, so
(**arretCourant).methodName()
also works.
If you use arretCourant->methodName() you are dereferencing the iterator, but not the std::shared_ptr, so you are calling methodName() on the std::shared_ptr itself.