What is wrong in this??i am tired!

Hello,
i tried running this program but couldn't after having a look at about 30 mins i still couldn't understand what wrong am i doing. I will leave that upto you now. Below is the code..

Code:

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#include <iostream.h>


class linklist
{
        struct node
        {
                int data;
                struct node *next;
        }*first;

        int *d;
                public:
                        friend void display(linklist);

};
void display(linklist q)
{
 struct node *temp=q.first;
 int *t=q.d;
}

int main(void)
{

 linklist q;
 display(q);

 return 0;
}

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why can't q.first's address be assigned to a pointer and why can q.d's address be assigned to pointer?

and if i want to assign the node's address to some other pointer via friend function then how can i do?? :(

probably because they are private

in the future - show your error message

So what??if they are private??i can access all private from member functions or friend functions..

and the error message i get is :-

cannot convert linklist::node* to node*

Now can anybody tell me what wrong am i doing?

node is nested in linklist, so you should write:

Code:

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void display(linklist q)
{
    linklist::node* temp = q.first;
    int* t = q.d;
}

---

Also, <iostream.h> should be <iostream>.

Thanks Laserlight :)
you solved my problem!but still explain me how come it's not recognising the node?please explain me?

Quote:

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but still explain me how come it's not recognising the node?please explain me?

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As I noted, node is nested in linklist.

So if it's nested we have to always use scope resolution??

Quote:

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So if it's nested we have to always use scope resolution??

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Yes, unless you are in the given scope.

Ok that's fine and why is this code not working??even though i have used scope resolution??

Code:

---


#include <iostream.h>


class linklist
{
  {
        int *d;
  }
                public:
                        friend void display(linklist);

};
void display(linklist q)
{
 int linklist::*t=q.d;
}

int main(void)
{
 linklist q;
 display(q);

 return 0;
}

---

i get error declaration terminated incorrectly?and also d is not a member of linklist?but i have declared d in class linklist only?if i have used braces so what?you said if it's nested we have to use scope resolution then why is it not working in this case?

Quote:

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if i have used braces so what?you said if it's nested we have to use scope resolution then why is it not working in this case?

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You have taken my statement out of context. If you have a nested class, then you qualify its name with the scope resolution operator. If you have a member variable, you access it as usual. What are you trying to do?

Ok fine but i m not able to declare only
whenever i provide braces i get error during compilation that "declaration terminated"

i am just trying to do as many errors as i could and then solve it. I feel this is the best way to learn programming language..

Quote:

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Ok fine but i m not able to declare only
whenever i provide braces i get error during compilation that "declaration terminated"

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d exists in an inner scope, but you have no way to refer to this scope. I daresay that d is quite useless like this. Just remove the braces and let it live as a normal member variable of linklist.

Quote:

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i am just trying to do as many errors as i could and then solve it. I feel this is the best way to learn programming language..

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I think you should read a good book first, e.g., Accelerated C++ by Koenig and Moo. If not, you would not be able to reason out the errors, and can only rely on guess and check.

Types, classes, structs, etc, do not belong to any given instance of any object. It is simply a type part of that blueprint, so you have to type the full scope.
Member variables, however, are part of the class instances, so you can't use he scope to access them (unless they are static, but that's another lesson).
I think you got it around your backfoot. You are trying to create a pointer to a class of type linklist. You are not accessing any type inside the class scope, so remove the "::".
Further, it seems that seem t think that "t" is somehow part of linklist? How do you figure this? The first part of the line is completely wrong. If we remove the dependant name, we get:
int* t = q.d;
Which would work.

Check this out, perhaps it might help you:

Code:

---

#include <iostream>

class linklist{
       
        public:
               
                linklist() : head(0) {};
              ~linklist();
               
                void push_back(int d);
                friend void display(linklist& m_list);
       
        private:
       
                struct node{
                       
                        node(int d) : data(d), next(0) {};
                       
                        int data;
                        node* next;
                };
               
                node* head;
};

linklist::~linklist(){
       
        node *curr, *prev;
       
        while(head->next != 0){
               
                curr = head;
               
                while(curr->next != 0){
                       
                        prev = curr;
                        curr = curr->next;
                }
               
                prev->next = 0;
                delete curr;
        }
       
        delete head;
}
       

void linklist::push_back(int d){
       
        if(head != 0){
               
                node *curr = head;
               
                while(curr->next != 0){
                       
                        curr = curr->next;
                }

                curr->next = new node(d);
        }
       
        else{

                head = new node(d);
        }
}

void display(linklist& m_list){
       
        linklist::node *curr = m_list.head;
       
        while(curr != 0){
               
                std::cout << curr->data << "\n";
                curr = curr->next;
        }
}


/*  MAIN:
*/
int main(){
       
        linklist m_list;
       
        for(int i = 0; i < 10; i++)
                m_list.push_back(i + 1);
       
        display(m_list);
       
        return 0;
}

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Quote:

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Originally Posted by dudeomanodude

Check this out, perhaps it might help you:

Code:

---

linklist::~linklist(){
       
        node *curr, *prev;
       
        while(head->next != 0){
               
                curr = head;
               
                while(curr->next != 0){
                       
                        prev = curr;
                        curr = curr->next;
                }
               
                prev->next = 0;
                delete curr;
        }
       
        delete head;
}

---

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Hmm, that's the worst singly-linked-list destructor I've ever seen - It's O(n*n)!
I really hope you don't use that in production code, and strongly suggest you learn how to do it in O(n), preferably without recursion too.

how would you do it?

with only the functions i've provided?

I agree O(2n) is not desirable. Should I just pop the front until it's empty? (I was being lazy...), sorry for the bad example on the destructor.

This is probably more kosher then:

Code:

---

#include <iostream>

class linklist{
       
        public:
               
                linklist() : head(0) {};
              ~linklist();
               
                void push_back(int d);
                void pop_front();
                friend void display(linklist& m_list);
       
        private:
       
                struct node{
                       
                        node(int d) : data(d), next(0) {};
                       
                        int data;
                        node* next;
                };
               
                node* head;
};

linklist::~linklist(){
       
        while(head != 0){
               
                pop_front();
        }
       
        delete head;
}
       

void linklist::push_back(int d){
       
        if(head != 0){
               
                node *curr = head;
               
                while(curr->next != 0){
                       
                        curr = curr->next;
                }

                curr->next = new node(d);
        }
       
        else{

                head = new node(d);
        }
}

void linklist::pop_front(){
       
        if(head != 0){
               
                node *curr = head;
               
                if(head->next == 0){    // Only one element in list

                        delete curr;
                        head = 0;
                }
               
                else{

                        head = curr->next;
                        delete curr;
                }
        }
}
                       
                       
                       

void display(linklist& m_list){
       
        linklist::node *curr = m_list.head;
       
        while(curr != 0){
               
                std::cout << curr->data << "\n";
                curr = curr->next;
        }
}


/*  MAIN:
*/
int main(){
       
        linklist m_list;
       
        for(int i = 0; i < 10; i++)
                m_list.push_back(i + 1);
       
        display(m_list);
       
        return 0;
}

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Now you're talking!
This would do it:

Code:

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linklist::~linklist() {       
        while(head != 0)
                pop_front();
}

---

Thanks dudeomanodude :) it's nice code

There's one thing I don't like about the list implementation, the fact that I need to declare display() as a friend function. What I'd rather do is make an iterator class within list that outsiders can use to traverse the list.

So, we can add the iterator class to the public interface of list:

Code:

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public:
       
        class iterator{
               
            public:
           
                iterator( node* c = 0 ) : curr( c ) {};
               
                int operator*( ) { return curr->data; };
                bool operator!=( const iterator& i ) { return i.curr != curr; };
                bool operator==( const iterator& i ) { return i.curr == curr; };
                iterator& operator++( ) { curr = curr->next; return *this; };

            private:
               
                node* curr;
        };
       
        const iterator begin() const{ return iterator(head); };

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Then instead of needing to hard code friend functions into the header we can write our own functions and use iterators to gain access to the next->

Code:

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void iter_display(linklist& m_list){
       
        for(linklist::iterator it = m_list.begin(); it != 0; ++it){
               
                std::cout << *it << "\n";
        }
}

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