about c program

Hi,

I am making a program a Teller class for queue and it is class which has a Teller, customer and a counter .

Code:

---


class Teller
{

public:

Teller();//constructor/

~Teller();//destructor/

int enqueue();//to add customer to queue/

int dequeue();//to remove customer from the queue/


private;



 int *front; //front pointer for the queue/
 
 int *rear; //rear pointer for the queue/
 
 
 Teller *nextptr; //next pointer/

 int node;//node variable for customer/

 int newNode;// node for the new customer/

};




main()


Teller teller; //object for class Teller/
 
 {
 teller.enqueue()
 
 
 newNode->setnext(Null);
 
 rear->setnext(newNode);
 
 rear=newNode;

}

---

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I have just created one object "teller" for this class and this object is basically a counter. I have used the method "enqueue()" to add customers to this counter(teller).

I wanted to know that for the object "teller" ,if we use method "enqueue()" to add customers to the queue, can we define its return type if we pass an argument of "int node" to it?



Thanks

You can make its return type anything you want it to be, unless you have a spec sheet that says otherwise. What do you want it to be?

thanks for the reply,

I want to pass "int node" as an argument to this method "enqueue()" but ,it must also have a return type, but when I write it as:

"int teller.enqueue(int node);"

and when i try to compile it, the compiler gives an error message of syntax error. I am not able to understand how I should define its return type.

You specify the return type and the parameters when you define the function, not when you call it. After all, look at these lines from your code:

Code:

---

newNode->setnext(Null);
 
 rear->setnext(newNode);

---

Do you see any "int" or anything like that in those calls?

Quote:

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Originally Posted by tabstop

You specify the return type and the parameters when you define the function, not when you call it. After all, look at these lines from your code:

Code:

---

newNode->setnext(Null);
 
 rear->setnext(newNode);

---

Do you see any "int" or anything like that in those calls?

---

I am using these nodes as I want to implement "Queues" using linked list.

Quote:

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Originally Posted by student111

I am using these nodes as I want to implement "Queues" using linked list.

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I don't at all want you to focus on why those lines are there. I want you to focus on what they look like -- the syntax you use when you call a function.

Sir, I have declared the "int node" and "int newNode" in this program , I have seen some of the linked list programs which are written this way. I basically want to set the next pointer of newNode as Null and want to join the new node to the queue list.

But can you tell me what exactly is the mistake I am making when using enqueue method?

when I compile this program ,it gives me some errors including the error on line where I have
newNode->setnext(Null);

Also how do I define or declare a null pointer?

Don't modify the first post, it makes it hard to track the conversation.

Some problems:
*main needs a return type, namely int.
*Teller teller is currently declared as if it were a K&R style arguement to main; that's not where it should be.
*A queue is not a node; it doesn't have a pointer to the next queue. A queue may contain nodes, but they should be of a different type then the queue.
*enque should logically take as parameters what you intend to enqueue
*when you access a member of a class, you need to do it throught the instance of the class. In this case the syntax would look like "teller.rear" for instance.
*main only call enque and deque. nodes and other private variables can and should only be modified by class methods.
*nullptr is the null pointer constant in C++11. You can use the literal 0 for earlier versions of C++, but you should learn C++11.

And remember, that this is an academic exercise, and most of the time when you want a queue, you should use std::deque

Quote:

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Originally Posted by King Mir

most of the time when you want a queue, you should use std::deque

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Nah, use std::queue if you want a queue, and std::deque if you want a double ended queue. It so happens that std::queue will adapt a std::deque by default, but the type and hence the interface signals the intention.

Good point. I had forgotten about that container adapter.