Solved

declaring an object with a user defined name

Posted on 2000-04-11
7
235 Views
Last Modified: 2010-04-02
how do I get the program (in c++) to declare an object of a class when the user enters the name of that object?

example:
the user wants to create an object of the class Restaurant and he enters the name Wendy's for it.
0
Comment
Question by:datavirus
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
  • 2
7 Comments
 
LVL 5

Expert Comment

by:pitonyak
ID: 2706462

I would assume that the user would not really know that they wanted to create a particular class..... I would assume that the user would simply say that they wanted to create a Restuarant and then you would create a restuarant and then fill in the values....


How you would do this would depend upon what you really mean to do.

Consider the following:

cout << "User, what do you want to create?" << endl;
cin >> item_type;

if (item_type == "restuarant") {
  item = new restuarant;
  item->ask_user_for_values();  // assume that this is a virtual function
} else if (item_type == "grocery") {
  item = new grocery;
  item->ask_user_for_values();  // assume that this is a virtual function
}


Another option might be:

cout << "User, what do you want to create?" << endl;
cin >> item_name;

item_type = figure_out_item_type(item_name);
if (item_type == "restuarant") {
  item = new restuarant;
  item->get_values();  // assume that this is a virtual function
} else if (item_type == "grocery") {
  item = new grocery;
  item->get_values();  // assume that this is a virtual function
}

0
 

Author Comment

by:datavirus
ID: 2706563
The only option the user has is to create a restaurant. The code I'm making is for editing a database of restaurants. So what the program would do is receive the name of the restaurant and declare it as a member of that class.
0
 

Author Comment

by:datavirus
ID: 2706575
The only option the user has is to create a restaurant. The code I'm making is for editing a database of restaurants. So what the program would do is receive the name of the restaurant and declare it as a member of that class.
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 2

Accepted Solution

by:
AndrewRodionov earned 50 total points
ID: 2706652
Hi datavirus,

You can declare your class Restaurant that contains the instance name, e.g.

class Restaurant {
private:
  char szName[32];
public:
  Restaurant();
  Restaurant( char* pszName );
  ...
  const char* GetName() const;
};

When the user enters restaurant name you instantiate your class, e.g.

char szName[32];
cout << endl << "Enter a restaurant name: ";
cin >> szName;
Restaurant rstnSample( szName );

So you have the named class instance, i.e. the named object, and can check its name calling rstnSample.GetName().

Hope that helps.
Andrew.
0
 

Author Comment

by:datavirus
ID: 2707871
Hi Andrew,
In this line of code:

Restaurant rstnSample( szName );

Is rstnSample(szName) a method of the class Restaurant that returns the name of the restaurant entered by the user?....
0
 
LVL 5

Expert Comment

by:pitonyak
ID: 2709980

Note that I did not write that particular piece of code, but....

Here is his code again with comments...

// Declare an array of characters to hold the name
//
char szName[32];
//
// Tell the user that you desire the name.
// Note that this may not display because the
// stream has not been flushed. You can do this with
// cout << endl;
// or explicitly as
// cout.flush();
//
cout << endl << "Enter a restaurant name: ";
//
// Read the name
//
cin >> szName;
//
// Create a variable of type Restaurant with the name
// rstnSample. Note that he created a variable of type char*
// called szName above.
// When this variable, rstnSample, is created, it calls the constructor
// which takes a char* (I would recommend a const char*)
// He will probably then copy this into the internal data member of
// the variable rstnSample which is of type Restaurant.
//
Restaurant rstnSample( szName );


Enough for now...
Andy
0
 
LVL 2

Expert Comment

by:AndrewRodionov
ID: 2710501
Hi guys,

Andy, your comments for my code are detailed and accurate. Thanks. And I agree with a const char*...

datavirus, feel free to ask more if there are some difficulties.

Andrew.
0

Featured Post

Announcing the Most Valuable Experts of 2016

MVEs are more concerned with the satisfaction of those they help than with the considerable points they can earn. They are the types of people you feel privileged to call colleagues. Join us in honoring this amazing group of Experts.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

When writing generic code, using template meta-programming techniques, it is sometimes useful to know if a type is convertible to another type. A good example of when this might be is if you are writing diagnostic instrumentation for code to generat…
C++ Properties One feature missing from standard C++ that you will find in many other Object Oriented Programming languages is something called a Property (http://www.experts-exchange.com/Programming/Languages/CPP/A_3912-Object-Properties-in-C.ht…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.
The viewer will learn how to clear a vector as well as how to detect empty vectors in C++.

624 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question