Thread: another list question....

U wud have nailed this ages back if u had paid attention to what is being posted here..
even now, U r almost there, BUT ,not there yet....

for one last time , read on...

Code:

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

temp->next->next=temp->next->next->next; /* A-C-NULL */   [ correct]


loc->next=temp->next; /*we form a spliced list (loc)B->C->NULL*/   [ wrong !!]

I have forgotten how many times I have written that
BOTH "B->next" and "A->next" point to C

Its not hard to imagine.. We just have two pointer pointing to one location..
AND A is still there in ur list..


go back to the prev page and look at the explanation for the same thing.

Code:

temp->next=loc; /* here we simply just make A->next point to B which forms A->B->C->NULL  */    [ correct]

U have reached the end result correctly , but u must perfectly understand how..