another list question....

**Axel** · 10-28-2006

A is spliced out.but ,U still have "loc" pointing to A

i think it makes sense. So you are saying after this line:

curr->next = (curr->next)->next

A and B become split (independant of each other) and then then you just insert B after a

like this:

loc->next (A) = curr(B);

**kris.c** · 10-28-2006

ah! better...

**Axel** · 10-28-2006

i have another bit of the code which i dont understand. It handles the case where there are 3 nodes. If i have a list containing ACB and it pass it 'C' it swaps B and C around to form ABC.

Code:

 while(temp->next->next != NULL)
        {
            if(temp->next->value=='C') // check to see if we have the correct value to swap
            {
                loc = temp->next->next; // save B
                temp->next->next = temp->next->next->next; // we then splice B to form B -> NULL and AC -> NULL
                loc->next = temp->next; // assign C after B to form BC
                temp->next = loc; // assign B after A to form AB
                break;
            }
            temp = temp->next; 
        }
    }

i'm not sure how it all comes together.

**Axel** · 10-28-2006

we start off with

Code:

loc = temp->next->next;

A | C | B
loc = b;

Code:

 temp->next->next = temp->next->next->next;

A | C | NULL

Code:

loc->next = temp->next;

A | C | NULL | C

Code:

 temp->next = loc;

A | B | NULL | C

i just don't get why i have NULL as the final result

**Axel** · 10-28-2006

or is this close to what it does?

http://img120.imageshack.us/img120/7555/drawing1fu5.jpg

**kris.c** · 10-28-2006

THe point that u r missing here is , NULL is not a node. U cant have

NULL --> C . It simply makes no sense. NULL is the end of ur list, there is nothing beyond that point.

Let me explain thru what u have written here itself :

we start off with

loc = temp->next->next;

A | C | B
loc = b;

temp->next->next = temp->next->next->next;

A | C | NULL

" all's fine upto this point "

and loc is pointing to B ..
temp is pointing to A.

loc->next = temp->next;

answer these questions :

what is temp->next?
what is loc->next?

u r equating loc->next to what is stored in temp-> next, that is C right?

U r having both temp->next and loc->next poiting to the same location now..

temp-- > C -- > NULL
^
loc -------|

NOTE : I am not able to get it to appear correctly here.. loc - > next points to C

temp->next = loc;

this makes " loc " as the next node for "temp" . The previous value stored is now lost..
the next node for "loc" is already set to be C right??

so, we now have :

temp --> loc -- >C -- > NULL

or

A -- > B -- > C --> NULL

I hope this does the job..

**Axel** · 10-28-2006

Originally Posted by kris.c

u r equating loc->next to what is stored in temp-> next, that is C right?

loc is NULL because of the previous lline yes?

temp->next->next = temp->next->next->next;

as with loc->next it is also null.

so i read the following line

Code:

 loc->next = temp->next;

as: make loc->next ( NULL ) point to temp->next (C)

what is temp->next?

C

what is loc->next?

NULL

**kris.c** · 10-28-2006

>Loc is NULL because of the previous lline yes?

tell me , how can "loc" be NULL here?

loc -> val = B
loc -> next = NULL
and (temp->next)->next = NULL

after B has been spliced out.

**Axel** · 10-28-2006

sorry i meant loc->next

so when you do:

Code:

 loc->next = temp->next;

it's saying:

NULL = C;

**kris.c** · 10-28-2006

well, this is C and not math.

when u do,
loc->next = temp->next;

U r storing the value in temp->next in loc->next... and not any other way that can be imagined.

so, loc->next is NOT null, but the value that was stored in temp->next .. that is C..

I can only suggest 2 things ..

1) re-read my explanation once again slowly .. write down at each step.
if that still does not help, parhaps U have been reading the wrong material
2) look up for stuff on the net which display waht is happening pictorially..

**Axel** · 10-28-2006

but after loc->next = temp->next;
is run do they *BOTH* contain C? or just loc->next ?

**kris.c** · 10-28-2006

yes, both WILL point to C

after this , u r doing
temp->next = loc ;

this stores the address of loc into temp->next
so, temp->next is pointing to loc... fine?

which means, u have updated the value in temp->next to loc. temp->next does not point to C anymore.
but u havent altered loc->next .
so,loc -> next is still pointing to C

so, the new list is :
temp--> loc -- > C -- > NULL

**Axel** · 10-28-2006

ACB

loc=temp->next->next; /* store B somewhere */

temp->next->next=temp->next->next->next; /* A-C-NULL */

loc->next=temp->next; /* ACC */

temp->next=loc; /* A->B->C->NULL */

right ?

**vart** · 10-28-2006

Do not use something like that

Code:

temp->next->next->next

in a real program - you cannot suppose without checks that for example temp->next->next is not null. Check it first.

**kris.c** · 10-28-2006

its getting a little clearer now... except for this :
>loc->next=temp->next; /* ACC */

U have this instead :

A- C- NULL
|
B---|

as I said earlier both A->next and B->next point to C

next u change A->next to B.so , u get

A C - -> NULL
| |
| |
-- B

U better get this right now...