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Home/ Questions/Q 6682313
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T04:41:29+00:00 2026-05-26T04:41:29+00:00

I’m doing a restaurant management program. Right now I’m trying to output private vector

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I’m doing a restaurant management program. Right now I’m trying to output private vector data and am getting stuck.

So I have a Menu.h

private:
vector<Category> categories;
vector<Menu_Item> menu_items;
vector<Recipe> recipes;
vector<Ingredient> ingredients;
vector<Order> orders;
vector<Order_Item> order_items;

And Menu.cpp

Menu.read()
Menu.show()

The read function reads from a file like this

1010    Appetizers                                      
1901    Entrees                                     
1576    Desserts                                        
1320    Drinks  

And stores those values into the appropriate vector, for example this one would be vector categories.

I also have a .h file for all the different types of things like, Menu_Item.h, Recipe.h, etc. And I store values into the vector like such:

menu_items.push_back(Menu_Item(meniID, catID, rID....

However in Menu_Item.h the values are

private:
int menu_item_id;
int cat_id;
int recipe_id;
string menu_item_name;
double price;   

The show() function queries the user what he/she wants to see. Let’s say the user wants to see a specific menu item like Onion rings. What I can’t do is

if(menu_items[0].menu_item_name == "Onion Rings")

because it says that menu_item_name value is private within Menu_Item.h. How can I access the private data?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T04:41:30+00:00Added an answer on May 26, 2026 at 4:41 am

    You have to make menu_items public or make a public getter function like the following.

    public: 
       vector<Menu_Item> get_menu_items(){ return menu_items;}
    

    Then say if you had a Menu object of this type called Menu you can do this:

    if(Menu.get_menu_items()[0].menu_item_name == "Onion Rings")
    

    The other possible option is that you make a friend class if another specific class needs access, though usually this won’t be the best design decision.

    In response to the comment you could do this:

    for(size_t n=0, n<menu_items.size()-1, ++n){    
         if(Menu.get_menu_items()[n].menu_item_name == "Onion rings")
              cout << "something";
    }
    
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