Assigning Array Values w/ Pointer

Question

I am practicing basic OOP w/ C++ and while using a pointer I found that I did not need to dereference the pointer to assign a value. Why is this? Menu.h:

class Menu { public:
int MenuCount;
int NumItems;
Drinks *Items;

Menu.cpp

Menu::Menu(Drinks a, Drinks b, Drinks c)  {
std::cout << "How many items would you like on this menu? "; std::cin >> NumItems;
Items = new Drinks[NumItems];
Items[0] = a; Items[1] = b; Items[2] = c; // why is it I can assign values by pointer without dereferencing? 
MenuCount = 3;

To my understanding I would've had to dereference Items with * prior to assigning a new object to the array, nonetheless this code works.

Henry Prickett-Morgan · Accepted Answer · 2018-12-18 21:27:31Z

1

The [] operator implicitly dereferences the pointer. It hides the pointer arithmetic with arrays, and because arrays decay to pointers to the first element, it works just as well on a pointer as an array.

answered Dec 18, 2018 at 21:27

Henry Prickett-Morgan

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1 Comment

Spencer Over a year ago

Thank you for the clarification

CS Pei · Accepted Answer · 2018-12-18 21:27:02Z

1

In your case, Items[0] is actually the same as *(Items + 0) so it is already doing the dereference for you.

answered Dec 18, 2018 at 21:27

CS Pei

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1 Comment

Spencer Over a year ago

Thank you this was quite helpful as well

Joshua Waring · Accepted Answer · 2018-12-18 21:30:40Z

0

Menu::Menu(Drinks a, Drinks b, Drinks c)  {
    std::cout << "How many items would you like on this menu? "; std::cin >> NumItems;
    1#  Items = new Drinks[NumItems];
    2#  Items[0] = a; Items[1] = b; Items[2] = c; // why is it I can assign values by pointer without dereferencing? 
    MenuCount = 3;
}

We are assigning a pointer to a pointer, new Drinks[] will create an array on the heap and return a pointer to it.
Items[0] is the same as (*Items + 0), it'll increment the pointer dereference it at that point.

answered Dec 18, 2018 at 21:30

Joshua Waring

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