I have to print a list of a set in c using linked lists (hence pointers). However,when I delete the first element of the list and try to print the list, it just displays a lot of addresses under each other. Any suggestions of what the problem might be? Thanks!
Delete function:
int delete(set_element* src, int elem){
if (src==NULL) {
fputs("The list is empty.\n", stderr);
}
set_element* currElement;
set_element* prevElement=NULL;
for (currElement=src; currElement!=NULL; prevElement=currElement, currElement=currElement->next) {
if(currElement->value==elem) {
if(prevElement==NULL){
printf("Head is deleted\n");
if(currElement->next!=NULL){
*src = *currElement->next;
} else {
destroy(currElement);
}
} else {
prevElement->next = currElement->next;
}
// free(currElement);
break;
}
}
return 1;
}
void print(set_element* start)
{
set_element *pt = start;
while(pt != NULL)
{
printf("%d, ",pt->value);
pt = pt->next;
}
}
If the list pointer is the same as a pointer to the first element then the list pointer is no longer valid when you free the first element.
There are two solutions to this problem:
Let all your list methods take a pointer to the list so they can update it when neccesary. The problem with this approach is that if you have a copy of the pointer in another variable then that pointer gets invalidated too.
Don't let your list pointer point to the first element. Let it point to a pointer to the first element.
Sample Code:'
typedef struct node_struct {
node_struct *next;
void *data;
} Node;
typedef struct {
Node *first;
} List;
This normally happens when you delete a pointer (actually a piece of the memory) which doesn't belong to you. Double-check your function to make sure you're not freeing the same pointer you already freed, or freeing a pointer you didn't create with "malloc".
Warning: This answer contains inferred code.
A typical linked list in C looks a little something like this:
typedef struct _list List;
typedef struct _list_node ListNode;
struct _list {
ListNode *head;
}
struct _list_node {
void *payload;
ListNode *next;
}
In order to correctly delete the first element from the list, the following sequence needs to take place:
List *aList; // contains a list
if (aList->head)
ListNode *newHead = aList->head->next;
delete_payload(aList->head->payload); // Depending on what the payload actually is
free(aList->head);
aList->head = newHead;
The order of operations here is significant! Attempting to move the head without first freeing the old value will lead to a memory leak; and freeing the old head without first obtaining the correct value for the new head creates undefined behaviour.
Addendum: Occasionally, the _list portion of the above code will be omitted altogether, leaving lists and list nodes as the same thing; but from the symptoms you are describing, I'm guessing this is probably not the case here.
In such a situation, however, the steps remain, essentially, the same, but without the aList-> bit.
Edit:
Now that I see your code, I can give you a more complete answer.
One of the key problems in your code is that it's all over the place. There is, however, one line in here that's particularly bad:
*src = *currElement->next;
This does not work, and is what is causing your crash.
In your case, the solution is either to wrap the linked list in some manner of container, like the struct _list construct above; or to rework your existing code to accept a pointer to a pointer to a set element, so that you can pass pointers to set elements (which is what you want to do) back.
In terms of performance, the two solutions are likely as close so as makes no odds, but using a wrapping list structure helps communicate intent. It also helps prevent other pointers to the list from becoming garbled as a result of head deletions, so there's that.
Related
First time asking a question but I did look around Google and stackoverflow to see if someone has asked something similar before. In malloc, recasting and free, it looked like the OP asked something similar for example. But it was more complicated.
I was wondering whether it's possible to create a generic function for a list structure in C that traverses the list given that you know that the different types of structures will always have a "next" field.
For example, given these two list-type structures:
typedef struct _list1 {
int value;
list1 *next;
} list1;
typedef struct _list2 {
int value;
char *string;
list2 *next;
} list2;
Is it possible to create a generic void freeList((void *) list) function or something which looks something like the below? I am aware it's a simple thing to write both free functions for each individual list separately.
void freeList((void *) list) {
// Included this because the structs would have different sizes
// so I thought it would be possible to cast it in order to properly dereference the field.
if (sizeof *list == sizeof list1)
*list = (list1) list;
else if (sizeof *list == sizeof list2)
*list = (list2) list;
if (!list) return;
else {
free(list->next);
free(list);
}
}
So far, my experiments with the code shown above didn't fare well given that gcc would complain about dereferencing a void * pointer.
Making a heterogeneous list can be achieved by the use of a tagged union, or just a tag and casting:
struct list_item {
struct list_item *next;
enum datatype type;
void *contents;
};
or
struct list_item {
struct list_item *next;
enum datatype type;
union {
int some_int;
char some_char;
} contents;
};
Then while traversing the list you just have to verify the type stored in type before using the contents of the element.
This check:
if (sizeof *list == sizeof list1)
*list = (list1) list;
else if (sizeof *list == sizeof list2)
*list = (list2) list;
doesn't work because sizeof is a static construct: its value is defined at compilation time. You're just asking for the sizeof void.
is it possible to create a generic function for a list structure in C that traverses the list given that you know that the different types of structures will always have a "next" field.
Yes, as mentioned before; you must be careful that every structure starts with the "next" field; the two structures in your post should therefore be reordered like this:
typedef struct _list1 {
list1 *next;
int value;
} list1;
typedef struct _list2 {
list2 *next;
int value;
char *string;
} list2;
It is not clean code, because the compiler could reorder (and pad) the fields of the structure, but in general it should work.
Is it possible to create a generic void freeList((void) *list) function or something which looks something like...
This is possible if your structs do not refer malloced memory; or they do, but in a uniform (and known) way (note the first case is a sub-case of this last).
If the structs contain pointers pointing to memory that has to be freed, in fact, while freeing the struct the freeList() function should also free the referenced memory. A few solutions come to my mind:
1 - If all the different structs contain the same "pointers" layout, the routine can free those pointers in a uniform manner, knowing in advance what to do. In such scenario, one can also use pointer fields that are not used by all the structs, but only some.
2 - Every single instance of a struct could contain some helper field describing the pointer's layout. For example, just after the "next" field, another "mempntcnt" field could tell how many pointers (to be freed) follow the "next" field. Or, this "mempntcnt" could be passed as a parameter to freeList().
3 - This problem could be managed by a totally separated mechanism, outside the scope of freeList(). Much depends on the final usage: I mean, for a given (kind of) linked list, first call a routine that frees all the memory referenced by the list itself, then free the list by calling the common freeList(). After all, if different structs are needed, then different routines are used on them...
I hope I've been clear enough...
If you ensure that the next pointer is the first member of the struct then this is possible.
typedef struct list1 {
// next pointer must be first
struct list1 *next;
int value;
} list1;
typedef struct list2 {
// next pointer must be first
struct list2 *next;
int value;
char *string;
} list2;
void freeList(void *list) {
if (list) {
freeList(*(void**)list);
free(list);
}
}
Seemingly simple C code is seemingly not allowing me to remove the first element from a linked list. I can, however, successfully remove any other individual element and can successfully delete the whole linked list.
typedef struct list{
int data;
struct list * next;
} list;
void remove_element(list * node, unsigned int index){
if (node == NULL)
exit(-1);
list *currElem = node;
if (index == 0) {
node = node->next;
currElem->next = NULL;
free(currElem);
return;
}
Produces the follwing:
"free(): invalid pointer: 0xbfabb964"
I've followed the same format for all of my other manipulation functions with no issues. Similar threads on forums don't seem to be dealing with this particular problem.
You can read the explanation in this pdf on the Push function which explains it:
http://cslibrary.stanford.edu/103/
This is where c gets funky pschologically. You instinctively want to label a pointer as a pointer, which it is. But it is a pointer value, not a pointer reference. It's like the holy spirit of the C divinty. The triumvirate. C passed arguments to functions by value, not by address/reference. So, what do you do to pass a variable by reference? Remember, the solution is so obvious, it really didn't make sense to me for a week, I swear to god.
I have a function that adds one item to a list that I created. If it's the first time and the list points to NULL, it allocates the list and completes it, returning the address. If it's not the first time, it adds another item and again returns the first item (by now I could disregard this return). The list and the function WORKS fine, here is the prototype:
typedef struct structTAppConsoleList {
char *text;
void (*cbFunction)(int);
int number; // This is the item number
struct structTAppConsoleList *previous;
struct structTAppConsoleList *next;
} TList;
TList *AppConsoleListAddItem(TList *p_list, const char *p_string, void (*p_funcPtr)(int));
So, somewhere in my code I have to create a lot of them and I'm trying to make it as the code below shows. Thing is, I can't make it work... I want to create something to group the lists I want to create and then use it in the function. The code below is an idea of what I'm trying to do. Consider only the part where I try to allocate the 3 lists, the rest is not important for this example.
TList *list1;
TList *list2;
TList *list3;
int main(void)
{
int i,j;
TList **groupMyLists;
TList *temp;
groupMyLists=malloc(sizeof(TList)*3);
*groupMyLists =(TList*)&list1;
*(groupMyLists+1)=(TList*)&list2;
*(groupMyLists+2)=(TList*)&list3;
for(j=0;j<3;j++) {
temp=NULL;
for(i=0;i<10;i++) {
temp=AppConsoleListAddItem(temp,"some text",someFunc);
}
**groupMyLists=temp; // my make won't let me do this
groupMyLists++;
}
}
I'm pretty sure that this would do it, but I can't compile it.
In my head, (*groupMyLists) would be the same as (&list1), (&list2), (&list3), the same way that (**groupMyLists) would be the same as (list1), (list2) and (list3). So why I can't do (**groupMyLists=temp)? Anyone?
I hope I made myself clear!! I's not easy to explain this madness I'm trying to do...
Change this line, you are using the wrong indirection.
*groupMyLists=temp;
In addition to the above two answers about the incorrect indirection of **groupMyLists you probably also want to assign the list1,list2,list3 pointers correct pointer values instead of writing garbage values into the allocated memory in groupMyLists i.e.
TList * groupMyList = malloc(sizeof(TList)*3);
list1 = &groupMyList[0];
list2 = &groupMyList[1];
list3 = &groupMyList[2];
but, this does not really match the rest of your code as it seems that AppConsoleAddListItem allocates the temp list so in that case your malloc would be incorrect as it should allocate the space for the pointers instead of space for the lists as in:
TList ** groupMyList = (TList **)malloc(sizeof(TList *)*3);
TList * temp;
if (!groupMyList) {
/* Print allocation error warning or handle in some proper fashion */
exit(1);
}
for(j=0;j<3;j++) {
temp=NULL;
for(i=0;i<10;i++) {
temp=AppConsoleListAddItem(temp,"some text",someFunc);
}
groupMyLists[j]=temp; // Here you now assign the pointer in temp to the memory for // pointers that you allocated above
}
list1 = groupMyList[0]; // Here we need to assign the list1,2,3 after the AppConsole calls
list2 = groupMyList[1]; // as these calls will changes the pointer addresses written into
list3 = groupMyList[2]; // groupMyList
Although I can not be sure exactly what you are trying to do there are several inconsistencies of pointers and indirections in your original code and the above two examples can hopefully be of some guidance
This would do the job:
**groupMyLists = *temp;
of copying one struct referenced by temp to another struct referenced by *groupMyLists.
But only if *groupMyLists would reference any valid memory, which is does not do - at least not from the source you posted.
I'm trying to destroy my deque but somehow I fail with pointers. I have written the following code (deque is a pointer to a pointer, which points to the first element of the deque). The DequeItem's are structs with fields next (pointer to next element) and data (void *).
void deque_destroy(DequeItem **deque) {
DequeItem *temp;
DequeItem *item;
for (item = *deque; item != NULL; item = temp) {
printf("%d", *((int*)((item)->data)));
temp = item->next;
free(item);
}
}
The struct declaration is:
struct DequeItem {
void *data; // Data stored in the deque item
struct DequeItem *previous; // Pointer to the previous DequeItem in the ring
struct DequeItem *next; // Pointer to the next DequeItem in the ring
};
typedef struct DequeItem DequeItem;
Looks correct, well done to read out temp before you call free() on the item, that's a common beginner mistake you've avoided.
I think you need to provide more information about what goes wrong, and perhaps also the struct declarations.
Is the data member also dynamically allocated memory? If so, you might need a free(item->data); call, too depending on how it was allocated when the item was created.
As commenters have pointed out, your data pointer might be NULL so you should check it before printing:
if(item->data != NULL)
printf("%d\n", *(int *) item->data);
Note:
Simplification of the casting expression makes it easier to read.
Include line feed ('\n') in the printf() string to avoid buffering confusion and separating the values visually.
The problem was, that the next-pointer of the last element (back element) in the deque will point to the first element (front element) even though the first element has been destroyed. I fixed this by setting
(*deque)->previous->next = NULL
before the for-loop above. Thanks for help!
I've seen quite a few example C implementations of linked lists on this site, and most of them place the next pointer at the end of each node like so...
struct intNode1 {
int data;
intNode1 *next;
};
Why do they implement them like that instead of like this?
struct node {
struct node *next;
};
struct intNode2 {
struct node node;
int data;
};
The latter way of implementing linked lists allows your insertion and deletion code work on any kind of node as well as allowing you to create a generic list type while the former way forces you to implement each kind of list from scratch.
For example, here is an (incomplete) implementation of a singly linked list using both kinds of nodes:
struct intList {
struct intNode1 *head;
};
struct list {
struct node *head;
};
Now, obviously any operation on a generic list that needs to compare it's nodes will need a function pointer to a comparison function, but that can often be hidden away in the implementation of a less generic interface for a list. For instance:
/* Returns zero if successful or nonzero otherwise */
int list-insertInt(struct list *list, int n) {
struct intNode2 * newNode;
if(!(newNode = malloc(sizeof *newNode)) {
return -1;
}
newNode->data = n;
return list-insertNode(list, (struct node *)newNode);
}
/* Assumes that the list contains only int nodes. */
int list-containsInt(struct list *list, int n) {
struct intNode2 *current = (intNode2 *)list->head;
while (current) {
if(current->data == n) {
return true;
}
current = current->next;
}
return false;
}
You can of course free a list without knowing what kinds of nodes it has:
void list-free(struct list *list) {
struct node *current = list->head;
struct node *next;
while(current) {
next = current->next;
free(current);
current = next;
}
}
PS. It's a bit late (i.e. it's early in the morning but I haven't slept yet) as I write this. so feel free to edit this question to be more clear.
Because textbooks on datastructures are mostly meant to teach concepts to beginners. That kind of 'optimization' just adds a lot of noise to the beginner's ear. It is what you do with your knowledge after school, that separates you from the rest...
I don't know about anyone else but I do it simply so, when I want to write the data portion to a file, I just write the bit sans the pointers at the end (including prev pointer if it's a doubly linked list).
Very rarely do I have a linked list where the types of each node can be different and almost certainly never when teaching the concepts of lists and other abstract data types to beginners.
The advantage of putting the pointer at the end is that the node and the payload have the same address. This may not seem much of an advantage now, but think back to before ANSI C. Yes I'm talking about a time when the compiler didn't even try to check the data type of pointers.
When you wanted to pass the payload to a function you could just pass the pointer, saving several bytes of typing (and valuable disk space!).
Some linked lists put the Next pointer at the end of the structure, some don't. I don't really see this as an issue.
To be honest, I'm not sure I ever ran across a case in C where a linked list maintained different node types. But doing something like you describe could be used if that's what you needed to do.
Note that most C programmers today use C++, which would allow you to use inheritance to accomplish the same thing. Using inheritance, it wouldn't matter where the Next member was placed within the class.