| | |
remove method linked list
Thread Solved |
•
•
Join Date: May 2005
Posts: 42
Reputation:
Solved Threads: 0
I am having a small problem removing ALL items from a linked list. I have written a function that will remove all discharged patients from the linked list.
in this function I check to see if the status of the patient is =5 and if it is I remove it from the linked list.
HOWEVER when I check my linked list after it will always contain 1 patient. even though the status of the patient is ==5. It delelts all but 1.
When I run the function again it will remove the last element.
Why do I have to run the function twice to remove all the elements from the linked list? is there a problem with my remove method in my linked list. or am I specifing the wrong parameters in my for loop function at the top of screen?
any suggestions greatly appreciated.
for(int a=1;a<(S.size()+1);a++)
{
int x=S.retrieve_status(a);
if (x==5)
{
S.remove(a);
}
}in this function I check to see if the status of the patient is =5 and if it is I remove it from the linked list.
C Syntax (Toggle Plain Text)
typedef patient SeqItemType; class ListNode { public: SeqItemType item; ListNode *next; }; class Sequence { public: Sequence(); bool isEmpty(); int size(); void insert(int index, SeqItemType newItem); void remove(int index); void alter_status(int index, int change); int retrieve_status(int index); void display(int index, int a,int b, int c, int d, int e, int f,int g); void move(int from, int to); void alter_priority(int index); int retrieve_ward(int index); void alter_ward(int index, int change); void alter_doctors_name(int index, int y); private: ListNode *find(int index); int length; ListNode *head; }; void Sequence::remove(int index) { ListNode *cur; // remove the first node in the list if (index == 1) { cur = head; head = head->next; } else{ ListNode *prev = find(index-1); cur = prev->next; prev->next = cur->next; } // clean things up! length--; cur->next = NULL; delete cur; }
HOWEVER when I check my linked list after it will always contain 1 patient. even though the status of the patient is ==5. It delelts all but 1.
When I run the function again it will remove the last element.
Why do I have to run the function twice to remove all the elements from the linked list? is there a problem with my remove method in my linked list. or am I specifing the wrong parameters in my for loop function at the top of screen?
any suggestions greatly appreciated.
This is what I am thinking...
Your loop condition is based on a=1 and will increment while a < size()+1.
But my guess is.. your linked list is "zero based".. much like how an array is.. which would account for why all other elements would be deleted.. except for your very first one.. which is, "node zero" or the head node..
try this and see what happens:
Your loop condition is based on a=1 and will increment while a < size()+1.
C Syntax (Toggle Plain Text)
for(int a=1;a<(S.size()+1);a++)
But my guess is.. your linked list is "zero based".. much like how an array is.. which would account for why all other elements would be deleted.. except for your very first one.. which is, "node zero" or the head node..
try this and see what happens:
for(int a=0; a<S.size()+1; a++)•
•
Join Date: May 2005
Posts: 42
Reputation:
Solved Threads: 0
try this and see what happens:
for(int a=0; a<S.size()+1; a++)Clinton thanks for the response. I tryed that but unfortunately it did not work. HOWEVER you got me thinking about the for loop and I came up with a solution.
What I forgot is.... when it comes to removing a node from a linked list the size of that linked list will automatically chage size. So on the question I asked above
C Syntax (Toggle Plain Text)
for(int a=1;a<(S.size()+1);a++) { int x=S.retrieve_status(a); if (x==5) { S.remove(a); } }
If X==5, S.remove(a) would remove the node AND the size of the list S.size() would automatically change. This then caused complications with my for loop as size started to decrease and "a" was still being incremented.
this is the soloution I came up with and works fine.
C Syntax (Toggle Plain Text)
static int wx=0; for(int a=1;a<(S.size()+1+wx);a++) { int x=S.retrieve_status(a-(wx)); if (x==5) { S.remove(a-(wx)); wx++; } } wx=0;
![]() |
Similar Threads
- Single Linked Circular Link List (Java)
- Removing an item from head of linked list (C)
- Linked List Retrieve Method (C)
- recursive linked list (C++)
- Why doesn't this remove the last node in a linked list? (C++)
Other Threads in the C Forum
- Previous Thread: Parsing a String into Tokens Using strcspn, Part 3
- Next Thread: C preprocessor
| Thread Tools | Search this Thread |
adobe api array arrays binarysearch calculate centimeter char cm convert copyanyfile copypdffile cprogramme createcopyoffile createprocess() csyntax directory dynamic feet fflush file floatingpointvalidation fork forloop frequency getlasterror givemetehcodez global graphics gtkgcurlcompiling hacking highest homework i/o inches incrementoperators intmain() iso kernel kilometer km linked linkedlist linux linuxsegmentationfault list locate logical_drives loopinsideloop. match matrix microsoft mqqueue mysql oddnumber odf open opendocumentformat openwebfoundation owf pattern pdf performance pointer posix power probleminc program programming pyramidusingturboccodes read recursion recv recvblocked repetition research scanf scheduling segmentationfault send shape single socketprograming socketprogramming stack standard strchr string suggestions systemcall test unix urboc user variable voidmain() wab whythiscodecausesegmentationfault win32api windows.h





