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Returning a struct pointer
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returning a local variable from function in C [duplicate]
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Closed 8 years ago.
I'm creating a structure inside a function and then returning its pointer. For some reason I keep getting the same memory address each time.
typedef struct list_type
{
node_t *head;
node_t *tail;
} list_t;
list_t newList() {
list_t list = {NULL, NULL};
list_t *listptr = &list;
printf("newList: %p\n", listptr);
return listptr;
}
Outputs:
newList: 0x7fffb42c8ae0
newList: 0x7fffb42c8ae0
newList: 0x7fffb42c8ae0
What am I doing?...
You are allocating list on the stack, but you're trying to return a reference to it from a function. This is a big no-no in C, since the struct's memory is automatically freed by the compiler when it goes out of scope. If you want memory to persist beyond the scope it is allocated in, you need to use malloc.
Furthermore, list_t and a pointer to it (list_t*) are two different types, and you need to indicate that as such in your function definition.
list_t* newList() {
list_t *listptr = malloc(sizeof(list_t));
printf("newList: %p\n", listptr);
return listptr;
}
However, be careful when using malloc. Since you're allocating memory manually, you'll now need to ensure that you call free on the pointer when you're done with it, or it will leak, since the compiler cannot free the memory for you.
list_t list = {NULL, NULL};
allocates memory on stack. Memory allocated on stack is automatically freed when the variable, list in your case, goes out of scope.
That is why you are getting the same address again and again as memory allocated to list on stack gets freed when program comes out of function newList() and is available for allocation. And, when you call the newList() again, same memory is alloacted again.
You should read about different memory allocations and pay more attention to the compiler warnings.
You're returning a pointer to an object allocated on the stack. If you use the value that the pointer points to in the function that calls newList(), you'll cause undefined behavior to occur.
You should read about malloc() and free().
Your function returns a list_t, a structure type, which is perfectly fine by-value. I see no evidence you're trying to dynamically allocate a list_t, and no evidence you even need to do so.
Lose the pointer stuff for this specific structure type entirely and just do this:
list_t newList()
{
list_t list = {NULL, NULL};
return list;
}
void freeList(list_t lst)
{
// TODO: delete lst nodes by walking lst.head through lst.tail
}
int main()
{
list_t lst = newList();
...
freeList(lst);
}
Don't step into the arena of managing dynamic memory because you can; do it when you need to do so, and in this case, you don't. You obviously need to in order to manage the actual list nodes, but thats a different issue than this one.
Related
I have the following struct which I use to implement a priority queue
struct q_element
{
//Declaration of struct members
int element;
int priority;
struct q_element *next_element;
};
and I use a pointer to a struct 'cur' to start from the first item in the queue
struct q_element* cur = start_element;
and keep moving until I find the one I want to delete from list.
while (cur->priority!=max_priority)
cur = cur->next_element;
Does the following line of code actually free the struct? Because 'cur' is a pointer to the struct I wasn't completely sure.
free(cur);
You need to pass a pointer to free, so free(cur) is the way to go, assuming that the struct itself has been allocated using malloc/calloc/realloc. Specifically, if you allocated your curr in the automatic memory (i.e. on the stack) you are not supposed to call free on it.
It looks like q_element is part of a linked list. Freeing the struct itself will not free other structs pointed to by it, so if you'd like to free the struct along with its tail, you need to write a loop.
Finally, when you free memory pointed to by some pointer in your program, it is a very good idea to assign NULL to the pointer that you freed to avoid accidental double-freeing and undefined behavior on accessing freed memory.
I have a struct which contains 2 integers and a pointer to another struct. I allocate memory for struct first and then for the pointer. When I free the memory I free up the pointer first and then I free up the struct.
When I run my program and call the function that frees memory it crashes when the call is made. When I don't call the function that frees memory it works fine, but then I'm not freeing up the memory.
I tried removing the line that frees the memory allocated to the pointer and the program doesn't crash, but I don't think thats right since a "free" is needed for every "malloc/calloc" right? Anyone see anything wrong with the freeing function?
//Define a struct data type
struct q_element
{
//Declaration of struct members
int element;
int priority;
struct q_element *next_element;
};
//Method to allocate memory
struct q_element* allocateStruct()
{
//Declaration of a variable
struct q_element *e;
//Allocate memory for one queue element
e = malloc(sizeof(struct q_element));
//Allocate memory for one pointer to a queue element
e->next_element = calloc(1,sizeof(struct q_element*));
//Initialize integer members of queue element
e->element = 0;
e->priority = 0;
return e;
}
//Method to free memory allocated
void freeStruct(struct q_element* e)
{
//Free up pointer member
free(e->next_element);
//Free up struct
free(e);
}
You don't need to allocate memory for the next_element pointer. The pointer is already there, just like int element for example.
So if you want to allocate just one element, you can set the next_element pointer to NULL and everything is fine.
You are not allocating enough memory for e->next_element in the line:
e->next_element = calloc(1,sizeof(struct q_element*));
// ^^^ remove the *
That should be:
e->next_element = calloc(1,sizeof(struct q_element));
If you used e->next_element as though it were a valid pointer, you most likely ended up accessing memory that you did not allocate. That clobbered some of the bookkeeping information created by calloc, which lead to problems when you called free.
In
//Allocate memory for one pointer to a queue element
e->next_element = calloc(1,sizeof(struct q_element*));
you allocate space for a pointer to a q_element structure, rather than a q_element structure. Do you attempt to write to this structure, because if so, that's probably where it goes wrong.
As a side note you might be better off just doing
e->next_element = 0
inside allocate_struct and then doing e->next_element = allocate_struct() outside the function later.
In addition to what everyone else is mentioning about allocation, you also need a sentinel to check if the next_element was already freed. You may be attempting a double free.
Try the following code:
void freeStruct(struct q_element* e)
{
//Free up pointer member
if(e->next_element != 0){
free(e->next_element);
e->next_element = 0;
}
//Free up struct
free(e);
}
I wanted to create a generic Linked List in C. Following is the structure of the node:
typedef struct node {
void *value;
int size; // n bytes
ind index; // index of the node
struct node *next;
} Node;
And my delete_node function is as following. The search function sends a pointer to the Node I want to delete.
Node *search_list(Node *list, void *data, int n_bytes);
int delete_node(Node *list, Node *to_be_deleted); // returns 1 on success
Inside the delete_node function I want to free up the memory pointed by void *value and then free up the memory allocated for the Node itself.
free(to_be_deleted->value); // Would this work??
free(to_be_deleted);
Since it is void pointer we don't know that how many bytes the object it is pointing to has occupied. How can we free up the memory for that?
Sorry if it is a stupid questions?
The memory allocator keeps track of how large memory allocations are on its own -- there's no need to tell free() how much memory to free.
As such, you should be able to just get rid of size and n_bits.
free(to_be_deleted->value); // Would this work??
Straight forward answer , Yes this will work.
simple thing :
see the definitions of free() and malloc()
void free(void *) // free takes void* as argument so it will work
void* malloc(sizeof(type))
In mallocwe have to pass thesize that how many bytes we want to allocate.
but in free just pass the pointer and whatever bytes allocated to that pointer on heap storage it will be freed
Yes, what you wrote should work. The reason is that malloc (which is a library call) creates metadata that is used to determine which parts of memory are free and which ones are taken. When you call free(), you are actually only modifying this metadata such that subsequent calls to malloc know that this memory can be re-used (note that most implementations will not zero the existing data).
I have a generic linked-list that holds data of type void* I am trying to populate my list with type struct employee, eventually I would like to destruct the object struct employee as well.
Consider this generic linked-list header file (i have tested it with type char*):
struct accListNode //the nodes of a linked-list for any data type
{
void *data; //generic pointer to any data type
struct accListNode *next; //the next node in the list
};
struct accList //a linked-list consisting of accListNodes
{
struct accListNode *head;
struct accListNode *tail;
int size;
};
void accList_allocate(struct accList *theList); //allocate the accList and set to NULL
void appendToEnd(void *data, struct accList *theList); //append data to the end of the accList
void removeData(void *data, struct accList *theList); //removes data from accList
--------------------------------------------------------------------------------------
Consider the employee structure
struct employee
{
char name[20];
float wageRate;
}
Now consider this sample testcase that will be called from main():
void test2()
{
struct accList secondList;
struct employee *emp = Malloc(sizeof(struct employee));
emp->name = "Dan";
emp->wageRate =.5;
struct employee *emp2 = Malloc(sizeof(struct employee));
emp2->name = "Stan";
emp2->wageRate = .3;
accList_allocate(&secondList);
appendToEnd(emp, &secondList);
appendToEnd(emp2, &secondList);
printf("Employee: %s\n", ((struct employee*)secondList.head->data)->name); //cast to type struct employee
printf("Employee2: %s\n", ((struct employee*)secondList.tail->data)->name);
}
Why does the answer that I posted below solve my problem? I believe it has something to do with pointers and memory allocation. The function Malloc() that i use is a custom malloc that checks for NULL being returned.
Here is a link to my entire generic linked list implementation: https://codereview.stackexchange.com/questions/13007/c-linked-list-implementation
The problem is this accList_allocate() and your use of it.
struct accList secondList;
accList_allocate(&secondList);
In the original test2() secondList is memory on the stack. &secondList is a pointer to that memory. When you call accList_allocate() a copy of the pointer is passed in pointing at the stack memory. Malloc() then returns a chunk of memory and assigns it to the copy of the pointer, not the original secondList.
Coming back out, secondList is still pointing at uninitialised memory on the stack so the call to appendToEnd() fails.
The same happens with the answer except secondList just happens to be free of junk. Possibly by chance, possibly by design of the compiler. Either way it is not something you should rely on.
Either:
struct accList *secondList = NULL;
accList_allocate(&secondList);
And change accList_allocate()
accList_allocate(struct accList **theList) {
*theList = Malloc(sizeof(struct accList));
(*theList)->head = NULL;
(*theList)->tail = NULL;
(*theList)->size = 0;
}
OR
struct accList secondList;
accList_initialise(secondList);
With accList_allocate() changed to accList_initialise() because it does not allocate
accList_initialise(struct accList *theList) {
theList->head = NULL;
theList->tail = NULL;
theList->size = 0;
}
I think that your problem is this:
You've allocated secondList on the stack in your original test2 function.
The stack memory is probably dirty, so secondList requires initialization
Your accList_allocate function takes a pointer to the list, but then overwrites it with the Malloc call. This means that the pointer you passed in is never initialized.
When test2 tries to run, it hits a bad pointer (because the memory isn't initialized).
The reason that it works when you allocate it in main is that your C compiler probably zeros the stack when the program starts. When main allocates a variable on the stack, that allocation is persistent (until the program ends), so secondList is actually, and accidentally, properly initialized when you allocate it in main.
Your current accList_allocate doesn't actually initialize the pointer that's been passed in, and the rest of your code will never see the pointer that it allocates with Malloc. To solve your problem, I would create a new function: accList_initialize whose only job is to initialize the list:
void accList_initialize(struct accList* theList)
{
// NO malloc
theList->head = NULL;
theList->tail = NULL;
theList->size = 0;
}
Use this, instead of accList_allocate in your original test2 function. If you really want to allocate the list on the heap, then you should do so (and not mix it with a struct allocated on the stack). Have accList_allocate return a pointer to the allocated structure:
struct accList* accList_allocate(void)
{
struct accList* theList = Malloc( sizeof(struct accList) );
accList_initialize(theList);
return theList;
}
Two things I see wrong here based on the original code, in the above question,
What you've seen is undefined behaviour and arose from that is the bus error message as you were assigning a string literal to the variable, when in fact you should have been using the strcpy function, you've edited your original code accordinly so.. something to keep in mind in the future :)
The usage of the word Malloc is going to cause confusion, especially in peer-review, the reviewers are going to have a brain fart and say "whoa, what's this, should that not be malloc?" and very likely raise it up. (Basically, do not call custom functions that have similar sounding names as the C standard library functions)
You're not checking for the NULL, what if your souped up version of Malloc failed then emp is going to be NULL! Always check it no matter how trivial or your thinking is "Ah sher the platform has heaps of memory on it, 4GB RAM no problem, will not bother to check for NULL"
Have a look at this question posted elsewhere to explain what is a bus error.
Edit: Using linked list structures, in how the parameters in the function is called is crucial to the understanding of it. Notice the usage of &, meaning take the address of the variable that points to the linked list structure, and passing it by reference, not passing by value which is a copy of the variable. This same rule applies to usage of pointers also in general :)
You've got the parameters slightly out of place in the first code in your question, if you were using double-pointers in the parameter list then yes, using &secondList would have worked.
It may depend on how your Employee structure is designed, but you should note that
strcpy(emp->name, "Dan");
and
emp->name = "Dan";
function differently. In particular, the latter is a likely source of bus errors because you generally cannot write to string literals in this way. Especially if your code has something like
name = "NONE"
or the like.
EDIT: Okay, so with the design of the employee struct, the problem is this:
You can't assign to arrays. The C Standard includes a list of modifiable lvalues and arrays are not one of them.
char name[20];
name = "JAMES" //illegal
strcpy is fine - it just goes to the memory address dereferenced by name[0] and copies "JAMES\0" into the memory there, one byte at a time.
I'm just reading about malloc() in C.
The Wikipedia article provides an example, however it justs allocate enough memory for an array of 10 ints in comparison with int array[10]. Not very useful.
When would you decided to use malloc() over C handling the memory for you?
Dynamic data structures (lists, trees, etc.) use malloc to allocate their nodes on the heap. For example:
/* A singly-linked list node, holding data and pointer to next node */
struct slnode_t
{
struct slnode_t* next;
int data;
};
typedef struct slnode_t slnode;
/* Allocate a new node with the given data and next pointer */
slnode* sl_new_node(int data, slnode* next)
{
slnode* node = malloc(sizeof *node);
node->data = data;
node->next = next;
return node;
}
/* Insert the given data at the front of the list specified by a
** pointer to the head node
*/
void sl_insert_front(slnode** head, int data)
{
slnode* node = sl_new_node(data, *head);
*head = node;
}
Consider how new data is added to the list with sl_insert_front. You need to create a node that will hold the data and the pointer to the next node in the list. Where are you going to create it?
Maybe on the stack! - NO - where will that stack space be allocated? In which function? What happens to it when the function exits?
Maybe in static memory! - NO - you'll then have to know in advance how many list nodes you have because static memory is pre-allocated when the program loads.
On the heap? YES - because there you have all the required flexibility.
malloc is used in C to allocate stuff on the heap - memory space that can grow and shrink dynamically at runtime, and the ownership of which is completely under the programmer's control. There are many more examples where this is useful, but the one I'm showing here is a representative one. Eventually, in complex C programs you'll find that most of the program's data is on the heap, accessible through pointers. A correct program always knows which pointer "owns" the data and will carefully clean-up the allocated memory when it's no longer needed.
What if you don't know the size of the array when you write your program ?
As an example, we could imagine you want to load an image. At first you don't know its size, so you will have to read the size from the file, allocate a buffer with this size and then read the file in that buffer. Obviously you could not have use a static size array.
EDIT:
Another point is: When you use dynamic allocation, memory is allocated on the heap while arrays are allocated on the stack. This is quite important when you are programming on embedded device as stack can have a limited size compared to heap.
I recommend that you google Stack and Heap.
int* heapArray = (int*)malloc(10 * sizeof(int));
int stackArray[10];
Both are very similar in the way you access the data. They are very different in the way that the data is stored behind the scenes. The heapArray is allocated on the heap and is only deallocted when the application dies, or when free(heapArray) is called. The stackArray is allocated on the stack and is deallocated when the stack unwinds.
In the example you described int array[10] goes away when you leave your stack frame. If you would like the used memory to persist beyond local scope you have to use malloc();
Although you can do variable length arrays as of C99, there's still no decent substitute for the more dynamic data structures. A classic example is the linked list. To get an arbitrary size, you use malloc to allocate each node so that you can insert and delete without massive memory copying, as would be the case with a variable length array.
For example, an arbitrarily sized stack using a simple linked list:
#include <stdio.h>
#include <stdlib.h>
typedef struct sNode {
int payLoad;
struct sNode *next;
} tNode;
void stkPush (tNode **stk, int val) {
tNode *newNode = malloc (sizeof (tNode));
if (newNode == NULL) return;
newNode->payLoad = val;
newNode->next = *stk;
*stk = newNode;
}
int stkPop (tNode **stk) {
tNode *oldNode;
int val;
if (*stk == NULL) return 0;
oldNode = *stk;
*stk = oldNode->next;
val = oldNode->payLoad;
free (oldNode);
return val;
}
int main (void) {
tNode *top = NULL;
stkPush (&top, 42);
printf ("%d\n", stkPop (&top));
return 0;
}
Now, it's possible to do this with variable length arrays but, like writing an operating system in COBOL, there are better ways to do it.
malloc() is used whenever:
You need dynamic memory allocation
If you need to create array of size n, where n is calculated during your program execution, the only way you can do it is using malloc().
You need to allocate memory in heap
Variables defined in some functions live only till the end of this function. So, if some "callstack-independent" data is needed, it must be either passed/returned as function parameter (which is not always suitable), or stored in heap. The only way to store data in heap is to use malloc(). There are variable-size arrays, but they are allocated on stack.