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Id
14765789
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2
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2013-02-08T04:40:54.183
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2013-02-08T18:44:45.257
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2013-02-08T18:44:45.257
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973667
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14765485
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2
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2 things: <ol> <li>you are outputting <code>head->next</code> instead of just <code>head</code></li> <li>you are not updating <code>linked</code> to <code>linked->next</code></li> </ol> The head node of a linked list is the first element of a list. head->next, when the linked list is well-formed, is the second element. Either way it shouldn't be changing as the list is made after it is created. If you are not updating <code>linked</code> to be <code>linked->next</code> you are just throwing away the memory you just allocated in <code>linked->next</code> and not creating a list. This works: <pre><code>int d; struct node *linked; struct node *head; linked = malloc(sizeof(struct node)); head = linked; printf ("Location of head %p \n", head);//changed head->next to head d = 1; while (d>0){ printf ("Enter the value of X: "); scanf ("%d", &linked->x); linked->next = malloc (sizeof(struct node)); printf ("Location of linked %p \n", linked->next); printf ("Location of head %p \n", head);// changed head->next to head printf ("To enter further value enter non zero: "); scanf ("%d", &d); linked = linked->next; //this was added if (d==0) linked->next = NULL; } </code></pre> Your code, since it wasn't updating <code>linked</code>, kept <code>head</code> and <code>linked</code> equal. Your list looked like this: Node {x:next} <pre><code> -> {x2:NULL} -> {x3:NULL} {x1:node2} -> {x4:NULL} ^^ ^^ (head and linked) you're outputing this node both times </code></pre> Since you don't have any pointers to node2 or node3 you lose these nodes (a memory leak). You had it almost right. This is what a list generated with this code looks like: <pre><code>{x1:node2} -> {x2:node3} -> {x3:node4} -> {x4:node5} ^^ ^^ ^^ head head->next linked </code></pre> update: I changed the diagram for your code to better reflect what is happening. The <code>ptr->varname</code> operator looks in location pointed to by <code>ptr</code> for the value of <code>varname</code>. So if the pointers are the same value then you are looking in the same space for the variables. In your case, <code>head=linked</code> means that if you set <code>linked->next = ptr</code> then <code>head->next = ptr</code>, and since <code>linked</code> is never updated to be <code>linked->next</code> (and therefore <code>head->next</code>) then you are just assigning and reassigning <code>linked->next</code>
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1. POLinked list maintain the root pointer address to read it later after building it
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1. USRaufio
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1. POLinked list maintain the root pointer address to read it later after building it
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1. COIn my code, I used the head = linked at very beginning and running loop to execute malloc for new linked so all new allocation happened in while loop where as I did head = linked before that so it should not be changing the head. Is it happening because of pointers reference? I am also trying to visualize the move in linked list with linked node, how am I referencing new node with old node next? Here I will have to write the value of new linked node to the linked->next of the previous linked node, right?
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2. CO@devnp, I was a little off on how I represented your code. I added an explination of the `->` operator, which I think might help.
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