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Thanks a lot ..it is working perfectly fine now..
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I'm not sure how your "I need to change the Mainframe icon at runtime" requirement is being met. 
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Some people are making such thorough preparation for rainy days that they aren't enjoying today's sunshine." - William Feather
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I am sorry for mentioning as changing the icon at runtime in the question posted earlier, actually I need to change Icon at compile time. For changing it at compile time everyone has helped me to achieve it and it's been achieved.
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Syntax:
#if defined _CHANGEICO
IDR_MAINFRAME ICON "res\\icon1.ico"
#else
IDR_MAINFRAME ICON "res\\icon2.ico"
#endif
And You have to remove IDR_MAINFRAME From the .rc file
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How to show only unique elements from this double linked list using UniqueList function? I made something, but it removes only the first...
<pre>
#include "iostream"
using namespace std;
void Add(int n);
void UniqueList();
struct Elem
{
int key;
Elem *prev;
Elem *next;
} *start;
int main()
{
int num;
while(cin >> num)
{
Add(num);
}
cout << endl;
UniqueList();
return 0;
}
void Add(int n)
{
Elem *p = start;
start = new Elem;
start->key = n;
start->next = p;
start->prev = NULL;
}
void UniqueList()
{
Elem *p = start;
if(NULL != start) cout << "The list is:" << endl;
else cout << "Empty list.";
while(NULL != p)
{ if(p->prev != p->next)
cout << p->key << endl;
p = p->next;
}
}
</pre>
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one problem is that you're never setting the 'prev' members of the added items to anything but NULL. if you have a double-linked list, you need to do something like this:
void Add(int n)
{
Elem *p = start;
start = new Elem;
start->key = n;
start->next = p;
start->prev = NULL;
p->prev=start;
}
but that will probably blow up the first time through, since start itself hasn't been initialized, and *start is never allocated.
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and what about the other main problem with the duplicates?
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i'm going to guess that that's probably a side effect of:
a) not setting prev to anything but NULL and
b) not initializing 'start' the first time around.
and, maybe i don't understand the task, but why are you doing this:
if(p->prev != p->next)
why compare prev and next pointers ? and why compare prev and next and not p itself:
if(p->key != p->next->key)
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I fixed the prev problem. "if(p->prev != p->next)" is a shot in the dark. Any ideas how to show only uniques 
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you're on the right track. but the key thing that's missing is that the list needs to be sorted before you can be guaranteed that duplicates will show up in p and p->next.
duplicates, but p, p->next will never be the same: 1,4,3,5,3
this will work: 1,3,3,4,5
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it's getting messy. How to sort a list??? (Doubly linked)
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You do not need to actually sort the list, since that would mean you first compare a lot of them and move around some nodes, then remove some (the duplicates). That is too much effort for what is needed.
What you need to do is compare a lot and remove, but never move. In pseudo-code (always solve an algorithmic problem in pseudo-code first):
foreach(Node n1) {
foreach (Node n2 before n1) {
if (n2.value==n1.value) {
remove(n1);
break;
}
}
}
PS: and that could be done on a single-linked list as well, as both loops are enumerating from the start and never need to go backward.
Luc Pattyn [Forum Guidelines] [My Articles] Nil Volentibus Arduum
Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.
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i did not understand anything you said. And whats that code C++?
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if you don't want to sort:
for each node, Cur
for each node, Temp
if Temp!=Cur
if Temp.value==Cur.value, duplicate
Temp=Temp.next
Cur=Cur.next
basically, for each node, you compare its value to every other node. it's inefficient, but it will work.
you can also sort your list when you add:
1. walk the list, find the first node with a value greater than the new value. this is 'Cur'
2. change the next pointer of Cur->prev to point to your new node.
3. set your new node's prev to Cur->prev. now you're all linked in the prev direction.
4. change the prev pointer of Cur to point to your new node.
5. set your new node's cur to point to Cur. now you're all linked in the next direction.
that way, the list is always sorted, and you can just walk it, looking for duplicates.
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OK, I decided to sort the list, but can you tell me why is not working well?
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TCPMem wrote: How to sort a list???
Insert the items in their "sorted" position within the list.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Some people are making such thorough preparation for rainy days that they aren't enjoying today's sunshine." - William Feather
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for(int i = 1; i < N; i++)
for(int j = N - 1; j >= i; j--)
if(p->key > p->next->key)
{
t = p->key;
p->key = p->next->key;
p->next->key = t;
}
Here is the sort but it doesn' t work
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first problem: you're never moving p
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p = p->next;
p = p->prev; (forgot you're inserting at the beginning, not at the end of list)
if (p==NULL) bail.
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OK, here is the full code that I' m working on. Can you tell me my mistakes exactly because I can' t understand where to put what. The UniqueList function is important, which now only shows the list and nothing more. The sort does not work...
#include "iostream"
using namespace std;
void Add(int n);
void UniqueList(int N);
struct Elem
{
int key;
Elem *prev;
Elem *next;
} *start;
int main()
{
int num;
int N = 0;
start = NULL;
while(cin >> num)
{
Add(num);
N++;
}
cout << endl;
UniqueList(N);
return 0;
}
void Add(int n)
{
Elem *p = start;
start = new Elem;
start->key = n;
start->next = p;
start->prev = NULL;
if(NULL != p) p->prev = start;
}
void UniqueList(int N)
{
Elem *p = start;
int t;
if(NULL != start) cout << "List:" << endl;
else cout << "Empty list!";
for(int i = 1; i < N; i++)
for(int j = N - 1; j >= i; j--)
{
if(p->key > p->next->key)
{
t = p->key;
p->key = p->next->key;
p->next->key = t;
}
}
while(NULL != p)
{
cout << p->key << endl;
p = p->next;
}
}
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maybe it's best to rethink this from the beginning.
if your goal is to have a list with no duplicates, the easiest thing to do is to simply not add duplicates.
so, every time, you go to add a new Elem, walk through your list and see if you already have an Elem with that key. if you do, don't add anything.
to do that, you need a function that can walk through the list and look for an Elem with a given key:
bool haveKey(Elem *p, int k)
{
while (p!=NULL)
{
if (p->key=k) return true;
p = p->next;
}
return false;
}
just call that at the top of your Add. if it returns false, don't add anything.
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No, the assignment is to add duplicates, but when show the list to show it without them. I think the sort thing will do the job but can' t do it properly.
modified on Monday, April 18, 2011 4:57 PM
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ok, then i would recommend sorting the nodes on input. it is far more efficient than sorting the whole list, and it's actually fairly straightforward. (i outlined it above)
unfortunately, i have to leave for the day. so.. good luck!
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