Question

have this code:

void set(list<Person*>* listP){
    Person timmy = Person(10);
    listP->push_back(&timmy);
}
int main()
{
    list<Person*> listP;
    set(&listP);
    Person* timmy  = listP.back();
}

If i understand correct (please correct me) timmy is allocated on the stack , so i cannot count on the values of timmy when i use them in the main. Am i correct? do i need to create timmy like this:

Person* timmy = new Person(10);

in order to create it on the heap and not on the stack ,so it will not be destroyed after method return?

Thank you

Answer 1

void set(list<Person*>* listP){
    Person timmy = Person(10); // create timmy on automatic storage (stack)
    listP->push_back(&timmy); //push timmy's address
} //timmy is destroyed. pushed address points to deallocated memory

Yes, you need to use Person* timmy = new Person(10); to allocate on heap.

void set(list<Person*>* listP){
    Person *timmy = new Person(10); // timmy is a pointer now
    listP->push_back(timmy); //push timmy's copy (copy of pointer)
} //timmy (pointer) is destroyed, but not the memory it points to

Also prefere to use smart_pointers such as std::shared_ptr, std::unique_ptr or boost smart pointers. It will simplify memory management and writing exception-safe code