I was implementing stacks using dynamically allocated arrays. Once the array is full I need to reallocate the array and make it twice as big as the initial array.
My code:
typedef int Item;
typedef struct stackImp *Stack;
struct stackImp{
Item * items;
int top;
int maxSize;
Stack createStack (void){
Stack s = malloc(sizeof(struct stackImp));
assert(s != NULL);
s->items = malloc(DEFAULT_SIZE * sizeof(Item));
assert(s->items != NULL);
s->top = 0;
s->maxSize = DEFAULT_SIZE;
return s;
void push (Stack stack, Item item)
.
.
.
if (stack->top < stack->maxSize) {
//Over here I'm checking if the top index is less than the
//maximum Items the array can store. If it's less then it pushes the
//item to the top of the array.
stack->items[stack->top] = item;
stack->top++;
}
else {
//If the index is greater than or equal to the maximum size then
//I realloc a new array which is twice the size of the initial array.
temp = realloc(stack->items, 2*(stack->maxSize) * sizeof(Item));
assert (temp != NULL);
stack->items = temp;
.
.
.
}
When I push items onto the stack it works perfectly fine but when I push on more than the initial maxSize it gives me this error:
It is clear I'm doing something wrong with my realloc function but I am unable to find what.
This is the valgrind output:
Your code is kind of confusing, since stack->top does not actually points to the index of the actual stack top. This might be causing some confusion on other parts of your code, since you have to use stack->items[stack->top-1] and not stack->items[stack->top] to access the top of your stack.
Please double check things like that. This error is probably caused by memory corruption. Learn how to use memory debuggers.
And I recommend that you try to implement stacks using lists instead of arrays.
Since the error is coming from realloc and the memory arena seems to be already corrupted, it's almost certainly something you're doing before this point.
Your best bet is to run it with a memory debugger like valgrind and check if you're overflowing buffers.
It may be the initial allocation of the array of Item objects (or lack thereof), it may be incorrect initialisation of the other fields like top or maxSize. Without the full code, it's a little hard for us to tell so, if you want a definitive answer, you should post all the relevant code.
But a memory debugger should help out greatly, should you not wish to do that.
Related
for learning purpose I'm implementing a stack with it's functions in c.
I added some small additional functionality to use malloc the first time and try to understand it properly.
I wrote a function which is initially creating my stack struct. The return value of the function is a new struct with an already allocate memory. What is the best way to handle a malloc exception in a function which return value should be a struct? Maybe should I design the function different? I'm aware that the printf is not doing it's job ;)
My Stack struct:
typedef struct Stack
{
int count;
int capacity;
int *data;
} Stack;
Creating a Stack instance:
Stack create_stack(int initialcapacity)
{
Stack new_stack;
new_stack.count = 0;
new_stack.capacity = initialcapacity;
if (!(new_stack.data = malloc(initialcapacity * sizeof(int))))
printf("not enough memory!");
return new_stack;
}
The function is called with the initial capacity of the stack:
Stack stack = create_stack(10);
A second question came up while I was writing a function to delete the Stack instance.
int delete_stack(Stack *stack)
{
stack->count = 0;
stack->capacity = 0;
free(stack->data);
stack->data = NULL;
return 0;
}
Am I able to remove also the struct instance itself? It feels not complete to just set the values back to 0 and direct int* to NULL.
Last but not least, I have a question to my push function. Also here I added some functionality which allows me to push something on the stack while it is already full:
void push(int value, Stack *stack)
{
if (stack->count == stack->capacity)
{
int *temp = malloc(stack->capacity * sizeof(int));
int i;
for (i = 0; i < stack->count; i++)
temp[i] = stack->data[i];
free(stack->data);
stack->data = NULL;
stack->data = malloc(stack->capacity * 2 * sizeof(int));
for (i; i > -1; i--)
stack->data[i] = temp[i];
free(temp);
temp = NULL;
stack->data[stack->count] = value;
stack->count++;
stack->capacity = stack->capacity * 2;
}
else
{
stack->data[stack->count] = value;
stack->count++;
}
}
Is it necessary to "free" the smaller array and put the pointer to NULL before I allocate a new array double the size?
If there is anything from my code which is unnecessary or not written properly, please let me know, I'm grateful for any hint which makes me better.
Cheeers,
me
I would do it with pointers. That is, your create_stack() would allocate a new Stack struct using malloc, then set the values to the struct and usee malloc again to allocate space for the Stack->data. Like this:
Stack* create_stack(int initialcapacity) {
Stack* new_stack = malloc(sizeof(Stack));
if (new_stack == NULL)
return NULL; // return null to tell the caller that we failed
new_stack->count = 0;
new_stack->capacity = initialcapacity;
new_stack->data = malloc(initialcapacity * sizeof(int))
if (new_stack->data == NULL)
{
free(new_stack);
return NULL;
}
return new_stack;
}
With this, we "handle" a malloc error by returning NULL, so the caller knows we failed.
Now that we have used malloc to allocate the Stack struct, you can (read: MUST) free the space taken by it using free(stack); in delete_stack().
In push(), the temporary array is not needed, that is, you could just right away allocate a bigger array, copy the contents to it from the original stack->data, free stack->data and set it to the newly malloc'd array:
int *temp = malloc(stack->capacity * 2 * sizeof(int));
// TODO: what if malloc fails?
int i;
for (i = 0; i < stack->count; i++)
temp[i] = stack->data[i];
free(stack->data);
stack->data = temp;
stack->data[stack->count] = value;
stack->count++;
stack->capacity = stack->capacity * 2;
Q. What is the best way to handle a malloc exception in a function which return value should be a struct?
There are at least three ways:
1) Instead of returning structure itself, return a pointer to it. This means two mallocs: one is for structure itself and another one is for data field. Returning NULL pointer means that something went wrong during construction.
struct Stack* create_stack(int initialcapacity) {
struct Stack* stack = malloc(sizeof(struct Stack));
...
return stack;
}
2) More flexible way is to pass pointer to already allocated structure. Flexibility comes from idea that calling code controls where to allocate structure: on stack or in dynamic memory. Return value of function may be used solely to notify calling code about errors:
bool create_stack(int initialcapacity, struct Stack* stack) {
...
}
// if calling code wants structure on stack (yeah, "stack" on stack)
struct Stack stack;
if (!create_stack(50, &stack)) {
die();
}
// if calling code wants it in dynamic memory
struct Stack* stack = malloc(sizeof(struct Stack));
if (!stack) {
die();
}
if (!create_stack(50, stack)) {
die();
}
3) If your program is not a 10,000+ LOC production code, easiest way may be to simply print error message and abort program immediately if allocation fails. Usually allocation errors are fatal: you can't recover in any meaningful way if there is not enough memory. You may even create a wrapper function over malloc to automatically catch such errors and exit:
void* my_malloc(size_t count) {
void* ptr = malloc(count);
if (ptr == NULL) {
fprintf(stderr, "Allocation failed");
exit(EXIT_FAILURE);
}
return ptr;
}
Q. Am I able to remove also the struct instance itself?
No, you can't. Because it is allocated on stack (the structure itself, not the data). If you want to delete structure itself, you need to use approach #1 from above.
And, by the way, there is no need to set zeroes and NULLs to fields. It doesn't delete anything. Such approach is used rarely and with only purpose to catch bugs (when calling code first deletes some structure and then tries to use it afterwards).
Q. Is it necessary to "free" the smaller array and put the pointer to NULL before I allocate a new array double the size?
Once again, you don't need to NULLify anything -- it doesn't delete anything. Instead of two mallocs and manual copying use realloc, which will do most of the work for you.
Generally, you should be able to declare a structure, then have an array of say 64 of them, with an integer to say which entry is on the top. Very simple, and no dynamic allocation. But 64 is pretty low, That's because stacks, recursion, and levels of nesting are intimately linked. Usually it should be possible to see that 64 is an insane level of nesting, and no legitimate input will ever even approach it. You then might need a guard to protect from malicious or corrupted input, but that just terminates the program or sub-routine.
If you can't establish a low sanity bound on a stack, it might that you still need one. Either it's a rare case where nesting goes very deep, or it's that you haven't approached the problem in the best way, but a sub-optimal program that still works is better than no program.
So you use the same structure, but the stack is set up with a call to
malloc() and, if it grows out of bounds, regrow with a call to realloc().
You might want to still sanity check it, but now sanity checks are
much higher, a million or so as opposed to 64. You also have to check that
realloc does not fail.
typedef struct
{
int x;
char astring[32];
} ENTRY;
static ENTRY *stack = 0;;
static int top = -1;
static int N = 0;
void push(const ENTRY *e)
{
/* grow logic like this */
ENTRY *temp = realloc(stack, newsize * sizeof(ENTRY));
if(temp)
stack = temp;
else
{
/* reallocation has failed but stack still valid */
free(stack);
N = 0;
top = -1;
/* for the sake of argument do this. We need temp to avoid
a memory leak */
fprintf(stderr, "out of memory\n");
}
/* push here, trivial */
}
int pop(ENTRY *e)
{
/* e is a non-const pointer. Fill and reset stack top */
}
You might want the stack global as in the example or you might want to
wrap it in a structure you pass about. Usually you'll want either pointers
or structures on the stack, but occasionally you might need a stack
of integers or floating point values.
There's no good way of handling memory allocation errors in C, especially
ones which can't happen (a computer with several GB of memory installed
is more likely to develop an electrical fault than to run out
of memory when asked for a couple of kilobytes). The usual rule is to
shunt up. But that makes the push call difficult, because every push
could potentially run the computer out of memory (but it can't really,
it's just your encapsulation allows the function to fail).
Recently, I'm learning C. I found a question on the internet. The question is:
What is the problem with this function in terms of memory allocation?
What is a good solution? You may assume that a struct item type has
been declared. The purpose of this function is to allocate an array of
struct item, which you may assume has been declared prior to this
function.
BOOLEAN allocate_items(struct item * items, size_t howmany)
{
size_t count;
items = malloc(sizeof(struct item) * howmany);
if(!items) {
perror("failed to allocate memory");
return FALSE;
}
return TRUE;
}
So, I think that the 4th line is wrong. It should be like:
items = malloc(sizeof(struct item));
And also the 6th line is wrong. It should be like:
if(items == NULL){
Is it correct?
First of all, both line 4 and 6, as you mentioned seems to be OK.
That said, the basic problem with this function is, you're allocating memory to a local scope of variable. This way
as you don't return the pointer to allocated memory, after the function returns, there would be no way to access the allocated memory.
by not freeing up the allocated memory, you will face memory leak.
If you have to allocate memory to a pointer, you need to pass the address of that pointer to the function and allocate memory. You can also return the pointer but then you need to change the function signature.
Finally, arrays are not pointers and vice-versa. They may appear or beahave similar sometimes, but they are not the same.
The 4th line is not wrong as they are trying to declare an array of the structs.
You should add a line inside the function that declares a new pointer, temp, to hold the current value of items, then after allocating the memory,
the 6th line should be
if(items == temp)
to check whether the value has changed(because that is the closest we can get to checking whether malloc worked)
this is because the ! operator is used to check if a condition is true or not(at least at a basic level in most languages) and as a pointer isn't a condition or an int that can be used as true or false, the operator won't work.
Here a fixed version, as it would probably be written in the "industry".
bool allocate_items(struct item ** pitems, size_t howmany)
{
// argument validation
assert(NULL != pitems); // some also add release version checks...
if(NULL == pitems ) return false;
// We can also spot memory leak sources here.
// If *pItems != NULL - does that mean we have to free first to prevent
// a leak? What if it is just some random value and not something we can
// free? So contract usually is: *pitems has to be NULL...
assert(NULL == *pitems);
if(NULL != *pitems) return false;
// implementation
*pitems = malloc(sizeof(struct item) * howmany);
if(NULL == *pitems) {
perror("failed to allocate memory");
}
return NULL != *pitems;
}
While the bool defined in stdbool.h sometimes causes trouble with C++ interop (same symbols on both sides, but sometimes sizeof(bool) differs), it is still the better option compared to inventing yet another bool type.
The pitems is a pointer to the location where the pointer to the new chunk of memory shall be written to. A caller of this function might have written:
int main(int argc, const char*[] argv) {
struct item *myBunchOfStuff = NULL;
if(false != allocate_items( &myBunchOfStuff, 20) ) {
// ...
free(myBunchOfStuff);
myBunchOfStuff = NULL;
}
return 0;
}
Defensive programming states: Your function cannot claim "Heh - my function only crashed because I was given a bad value!". Instead, it has to verify itself. It is responsible not to crash. The pointer could still be != NULL but otherwise bad. That is impossible for the function to catch, usually.
In C, everyone is proud of not requiring the cast of malloc()'s return value. You can be proud of that until you compile your code with a C++ compiler. Then you have to change your code and fix that. Well, I guess it is a matter of preference...
While parameter checking is often seen as a separate part of the functions implementation, after that, you should try to stick to "single point of exit". Main reason for that is maintainability. With multiple exit points, if the function gets bigger later on, it gets harder to spot if some early exit forgets to free some memory or cleanup other forms of state.
I want to create a program that creates link list which contains arrays.
However, I want each link to have a smaller array than previous one. at this moment the program works, but I want to know whether logically the program is actually creating smaller arrays, so no free space is wasted.
int main()
{
int c=3;
int d=0;
typedef struct mylist {
struct mylist *link;
int info[c-d];
}Node;
Node *a = (Node*) malloc (sizeof (Node));
a -> link = NULL;
Node *b = a;
int i=0,j=0;
while (i!=4){
while ((j)<=(2-d)){
printf("link%d array%d ",i,j);
scanf("%d",&a->info[j]);
j++;
}
j=0;
if (i !=3 ){
a -> link = (Node*) malloc (sizeof (Node));
a = a -> link;
}
d++;
i++;
}
d=0;
a = b;
while (a != NULL){
while ((j)<=(2-d)){
printf("%d ->", a->info[j]);
j++;
}
j=0;
d++;
a = a -> link;
}
a=b;
printf(" the end");
return 0;
}
First of all, you're not free()ing memory which you allocated, make sure you take care of that.
Now, you're asking whether are you not using any additional memory - reading your code is pretty hard for us, as you can see in the comments. It is always a good idea to try it out yourself - find out the size of the memory occupied during the initialization of the structure, either with the sizeof() operator or with checking your values.
However, I don't think your code will work properly, due to the fact that you are statically initializing the arrays size. As the values are determined compile-time, this is completely fine. However, the problem is that you have already decided the size of the int array at the start.
If you want to use dynamic allocation of memory, you need to use the malloc() function. When creating a new node, you will have to allocate memory for an array in it as well (and calculate how much size you want to give it). This will also solve your problem of not knowing the size of the array or not being sure about it very well - as you are explicitly declaring how much memory you're allocating for the array.
Just a small addition though. When allocating for example 20 bytes of memory, the OS won't give you specifically 20 bytes - you'll get an entire page, which will often be more memory than necessary. So unless you're working with a device where controlling memory properly is really important or you're working with huge sizes of memory, you should not need to overthink the difference of the size of an array of 4 ints and 3 ints.
I have a problem with this code, I tried to understand what's going on, but I cannot understand why it just crushes. the functions works like I expect for the node->left.
But when the last call of the function ends in the case node->left; the function just crushes, I think because of the reallocated array, I don't know if I'm trying to access a forbidden memory.
I'll explain a little more where I think the problem comes from:we are in the last call of helper(node->left,table,len) before printing the array : consider len = N and node->left!=NULL ==> reallocating table len = N+1, assigning node->data to table[len] and everything is fine, node->left == NULL ==> printing the table and we are done with the helper(node->left,table,N) case; now we are in the helper(node->right,table,N) case; in this case the program just crushes ; it's supposed to reallocate table; and assign node->data to table[N]; and so one ...
By the way : this function tries to print all the Root-leaf paths in a binary tree.
struct node {
int data;
struct node* left;
struct node* right;
};
void helper(struct node* node,int *table,int len)
{
if(node == NULL){
return;
}
else{
table = realloc(table,(len+1)*sizeof(int));
table[len]=node->data;
len = len +1;
if(node->left == NULL && node->right == NULL)
{
int cmt=0;
for(cmt = 0; cmt < len ; cmt++ ){
printf("%d ->",table[cmt]);
}
}
else{
helper(node->left,table,len);
helper(node->right,table,len);
}
}
}
void print_rtl(struct node* node) {
if(NULL == node) return;
else{
int *t=malloc(sizeof(int));
t[0]=node->data;
int lenght = 1;
helper(node->left,t,1);
helper(node->right,t,1);
}
}
Here is the thing about realloc: it has the freedom to change not just the size of the allocated memory block, but also its location. In general, realloc can break any pointers you have that are pointing to objects in the block of memory that was realloced. When you try to use those pointers later, you will get undefined behavior.
You should either allocate enough memory at the beginning of your program that you won't need to call realloc, or you should redesign your program so you don't need to allocate that memory block in the first place, or you should make your table pointer be a global variable so that when you change its value, it gets updated for every function that is using the table.
Right now the problem with your code is that when you call helper, it might call realloc and change the location of your table, but it doesn't communicate that change to the caller in any way, so the caller doesn't know where the table is located.
There might be other problems with your code. If you continue to have trouble, I strongly urge you to produce an MCVE and post that in your next question. Doing that makes it much easier for someone else to reproduce your problem and find a solution to it that actually works.
It's hard to tell exactly what going on because it quite a big mess, but generally... The first thing helper() does (after validating node!=NULL) is reallocate table. That means when you get to the recursion point, helper(node->left.. will reallocate, and then immediately after helper(node->right.. will also try to do so, but on an invalid pointer. From there on, it's a wild ride to exceptions...
I'm getting a SIGTRAP signal when trying to free an dynamically created array, and have no idea on why.
I'm allocating the array like this:
int* visited = (int*) malloc( l.nodeCount * sizeof(int));
(l.nodeCount is an integer. In the instance of the program I get this error, it is set to 12.)
And when I try to free(visited), I get the SIGTRAP signal in the debugger.
The whole function is this one:
int Graph_GetSmallestPathCount(AdjacencyList l, int destination){
//One path if destination is root
if(destination == 0) return 1;
if(l.nodeCount == 0)
return 0;
Queue reading = Queue_NewQueue();
Queue storing = Queue_NewQueue();
/*Allocates visited array*/
int* visited = (int*) calloc( l.nodeCount, sizeof(int));
/*Visited array initialization*/
int i;
for(i = 0; i < l.nodeCount; i++)
visited[i] = 0;
/*Marks root node and enqueues it*/
visited[0] = 1;
Queue_Enqueue(&reading, 0);
//While there are nodes to read
while(!Queue_IsEmpty(reading))
{
//Dequeues a node
int v = Queue_Dequeue(&reading);
//Gets it's adjacency list
List* currentList = AdjacencyList_GetAdjacentNodes(l, v);
listCell* auxCell = currentList->head->next;
//While there are nodes in it's adjacency list
while(auxCell != NULL){
//Enqueues it if it has not been visited
if(visited[auxCell->data] == 0){
Queue_Enqueue(&storing, auxCell->data);
}
//Adds to the paths to that node
visited[auxCell->data] += visited[v];
auxCell = auxCell->next;
}
//When the queue ends
if(Queue_IsEmpty(reading)){
//If the destination has been reached, return
if(visited[destination] > 0){
Queue_Destroy(&reading);
Queue_Destroy(&storing);
return visited[destination];
}
else{
//Switch queues
Queue_Destroy(&reading);
reading = storing;
storing = Queue_NewQueue();
}
}
}
//Destination has not been reached before end of algorithms. Deallocate everything and return 0
free(visited);
Queue_Destroy(&reading);
Queue_Destroy(&storing);
return 0;
}
Sorry for the lack of comments, I did this on a run and didn't put any in. Also sorry for the printf overload, I put them there while trying to pinpoint the problem.
EDIT: I cleaned it up a little.
The weird thing is that the program works for certain inputs and doesn't for others.
Hope someone can help me out =D
I can't tell you why you get a SIGTRAP as you haven't published a minimal example.
However, I can tell you how to find out out yourself:
Make your program readable. Use one instruction per line. The indent tool is your friend. Sure, that won't fix the bug, but it will make it easier for you to find it.
Don't malloc like that. There is no need to cast the return value of malloc, and using calloc(l.nodeCount, sizeof (int)); or similar is more readable anyway.
What SIGTRAP actually means is you've hit a breakpoint instruction. No doubt what's actually happening is that you've jumped to something which is not your code, and might not be code at all, but contains the binary code for a breakpoint. Why did that happen? The normal cause would be memory corruption, particularly stack corruption. I'm guessing free() is corrupting its own stack. And I'd guess the reason for that is because you are (somewhere) writing to memory outside the memory you've allocated. To test this, run your program with the malloc()/calloc() immediately followed by the free() and an exit(0). If that works, you know the issue is something you are doing between.
We can't tell what you are doing between because you haven't (thankfully) posted the full program, but try running it under valgrind. When you get an out-of-range write, valgrind will normally pick it up. Fix every valgrind warning. That doesn't guarantee a solution, but will find one 95% of the time in my experience.
Also note that return visited[destination]; appears to exit the function without free()-ing visited, which is thus a memory leak.
First, don't call malloc() like that. l.nodeCount * sizeof(int) could potentially exceed INT_MAX, and you'll either have a security hole or if you're lucky, a crash.
Instead, use calloc(l.nodeCount, sizeof(int)).
You should also check the return value of malloc or calloc for NULL in the event your program runs out of memory to allocate.
Found the answer. There was indeed a chance that on specific cases the array would be created with one element less than it should. My bad.
Thanks to all who helped =D