Re: fails to call destructor in a linked list

From:
Jens Schmidt <Jens.Schmidt-HH@gmx.de>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 2 Mar 2013 04:16:15 -0800 (PST)
Message-ID:
<apda2eFnd53U1@mid.uni-berlin.de>
alexo wrote:

I'm starting to study C++ after a brief panoramic of the C language.


So I'm trying to give some advice beyond the explicitly asked questions.

#include <iostream>

using std::cout;


You are writing everything to cout. Don't do that, the error messages
and tracing belong to cerr or clog.

using std::endl;

// the node class containing data

class Node
{
    public:
        Node(int);
       ~Node();

        Node *next;


next should not be public. The ability of an outsider of Node
to cause internal inconsistency is a clear warning.

        int getValue() const { return itsDatum; }

    private:
        int itsDatum;

};


As others wrote, use the services of the standard library where
applicable. Re-inventing the wheel is redoing the mistakes of
others. For learning algorithms this is O.K., but for learning
the language you might find more appropriate tasks.

// constructor implementation

Node::Node(int itsDatum):
next(0),
itsDatum(itsDatum) {}


Dangerous to have the same name in a parameter and a member. As
soon as you modify the constructor to have some code inside {}
strange things will happen.

// destructor implementation

Node::~Node()
{
    delete next;
    cout << "node deleted." << endl;
}


Think about what will happen on the stack if your list
has not 4, but 4000000 elements.

void print(Node *);
void push(Node **, int);
void pushBack(Node **, int);


Again, these are typical candidates for member functions.

int main()
{

    Node *head = 0; // head is the head of the empty list

    push(&head, 1);
    push(&head, 2);
    push(&head, 3);
    push(&head, 4);

    print(head);

    return 0;
}

// insert an element to the top of the list

void push(Node **phead, int value)
{
    Node *n = new Node(value);

    if(n != 0)
    {

        if(*phead == 0)
        {
            *phead = n;
        }
        else
        {
            n->next = *phead;
            *phead = n;
        }


You save a cheap assignment of null_ptr to next happening
only once for each list, while doing an expensive (on modern
processors) test for every insertion. The else-path works
in all cases.

    }
    else
        cout << "\nmemory allocation error." << endl;


Writing error messages to some output is just like swallowing
exceptions. Let the caller know about the problem and do
something about it.

}

// pushes an element to the end of the list


You wrote this function, but didn't test it.

void pushBack(Node **phead, int value)
{
    Node *temp = new Node(value);

    if(temp != 0)
    {
        if(*phead == 0)
        {
            *phead = temp;
        }
        else
        {
            Node *index = *phead;

            while(index->next != 0)
            {
                index = index->next;
            }
            index->next = temp;
        }


Again an unnecessary distinction. For the inner if-else
I would just write
           while (*phead != 0)
               phead = &(*phead)->next;
           *phead = temp;

    }
    else
    {
        cout << "memory allocation error.";
    }
}

// prints the list checking wheter it is an empy list

void print(Node *index)
{
    if(index != 0)
    {
        while(index != 0)
        {
            cout << index->getValue() << " ";
            index = index->next;
        }
        cout << endl;
    }
    else
    {
        cout << "\athe list is empty." << endl;
    }
}


Other comments:
Some functions do several things at once, while others a replicated
in more than on function. Try to restructure them:
for push and pushBack: allocation of the new node, finding the insertion
point, doing the insertion
for print: walking the list, formattion the output
--
Greetings,
   Jens Schmidt

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