Printing a linked list

This is a discussion on Printing a linked list within the C++ Programming forums, part of the General Programming Boards category; i have a linked list in my program that is circular . i have to print from a specific number ...

  1. #1
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    Printing a linked list

    i have a linked list in my program that is circular . i have to print from a specific number in the list and when i do im getting an error. now this function is a constant and must stay a constant, the line with the stars is giving me this error
    (C:\Windows\Desktop\Project5\Functions.cpp(117) : error C2166: l-value specifies const object)

    Code:
    void ClosedList::PrintFrom(int item) const
    
    {         
    
        NodeType* currPtr = head->backlink;
    
        NodeType* newNodePtr = new NodeType;
    
        newNodePtr->component = item;
    
        newNodePtr->link = head;
    
        head = newNodePtr;*******************************
    
        newNodePtr = NULL;
    
        if(!IsEmpty())
    
        {
    
    	do
    
    	{
    
    	     cout << currPtr->component << "    ";
    
    	     currPtr = currPtr->backlink;
    
    	}while (currPtr != head);
    
    	   cout << currPtr->component << "    ";
    
    	}
    
    	cout <<endl;
    
    bool ClosedList::IsEmpty() const
    {
    	return (head == NULL);
    }

  2. #2
    Registered User matheo917's Avatar
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    i presume tha the "head" pointer is a GLOBAL pointer that is located in the global namespace, and by including "const" after the function's name you signify that the function will not change anything beyond the scope of this function

    and on the line

    head = newNodePtr;*******************************
    you are changing the head pointer....

    if you make the function not "const" i think the error should go away....




    matheo917

  3. #3
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    Global head

    Head is a private member of a class
    Code:
    struct  NodeType;
    struct NodeType
    {
    int component;
    NodeType* link;
    NodeType* backlink;
    };
    
    class ClosedList
    {
    public:
    	bool IsEmpty() const;
    	void Append(int item);
    	int TotalCount();
    	void Print() const;
    	void PrintReverse() const;
    	void PrintFrom(int item)const;
    	void Delete(int item);
    	ClosedList();
    	~ClosedList();
    	ClosedList(const ClosedList& otherList);
    private:
    	NodeType* head;

  4. #4
    Registered User matheo917's Avatar
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    still....

    still.............that function is designed to not alter anything besides what was created and what dies within that function...

    i don't have time to follow the logic step by step in your entire program but once again i don't realy believe that this function is supposed to give "head" a new address to point to....
    considering the fact that all it supposed to do is print the list, which in result all it does is retrieves your data...(without altering it)


    go over your logic again, perhaps you want to create a "cur" poiner and assign value of "head" to it and then manipulate your program by using "cur" not head


    good luck

  5. #5
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    error fixed but...

    i fixed the error but now i can print in the right order. the number i have to print from is the third number in the list to the end then it has to rap back around and print the first two numbers, any suggestions

    The list in the order of appending
    92 82 62 72 99

    Code:
    void ClosedList::PrintFrom(int item) const
    {
    
    	NodeType* currPtr = head->backlink;
    	NodeType* newNodePtr = new NodeType;
    
    	newNodePtr->component = item;
    	newNodePtr->backlink = currPtr;
    	currPtr = newNodePtr;
    	newNodePtr = NULL;
    	
    	if(!IsEmpty())
    	{
    		do
    		{
    			cout << currPtr->component << "    ";
    			currPtr = currPtr->backlink;
    			
    		}while (currPtr != head);
    		
    		cout << currPtr->component << "    ";
    	}
    	cout <<endl;
    }

  6. #6
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    designate a second pointer to hold the position where you start in the list and keep searching for that pointer every time before you print. If current pointer start pointer then stop printing. I suspect you will need to write a special case for when the list is just one node long, but what the hey.

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