I'm a little unclear on this part of C, since it's a bit unlike other languages I've used, but this may just be a dumb question. I'm trying to implement a stack. I have the node struct, it has the information I want to pass:
struct position{
int square[2];
int counter;
struct position *prev;
};
so in main, I declare and initialize the bottom node of the stack, set *prev to NULL, then declare the rest. My question is, what happens when I try to pass it to function pop? I can create a position object that points to this one and return that, but will it be pushed off the stack when the function closes? Or should I return the position and set that equal to a new position object in main? What if I decide to create several of these nodes in a function? Will they remain once the function closes?
Edit: mah reminded me of my followup question which is, if they don't exist outside of the function, should I use malloc to create the space in the memory for them?
The lifetime of your objects depend on where they're created; if you declare for example a structure within a block of code (where a block is everything inside { and its matching }), that structure is no longer valid once execution leaves the block. Pointers to that structure are only valid as long as the structure is valid.
For what you're describing, you want to dynamically allocate your structures, using either malloc() or a similar function. Dynamically allocated data will remain valid (assuming you do not overwrite it) until you free() the memory, or until your program terminates. Pointers to these areas of memory will remain valid for that same period of time.
Consider:
static struct position *topOfStack = NULL;
void push(struct position *node)
{
node->prev = topOfStack;
topOfStack = node;
}
struct position *pop()
{
struct position *popped = topOfStack;
if (topOfStack) topOfStack = topOfStack->pref;
return popped;
}
To use this, you can:
f() {
struct position *node = malloc(sizeof(*node));
/* ... fill in node details ... */
push(node);
}
Notice that I allocated the node dynamically. Had I just declared a struct position node;, I could legally call push(&node); but once my function left scope, the stack would have an invalid item in it (which would likely cause havoc).
what happens when I try to pass it to function pop?
it depends on your pop() function prototype. If the pop's function prototype should be:
struct position* pop(struct position* stack);
I can create a position object that points to this one and return that, but will it be pushed off the stack when the function closes?
your question is quite unclear, and it looks like a big misunderstanding of instance scoping in C. Basically, you have two ways to allocate variables, either on the stack or on the heap. The scoping you're talking about is stack instances scope.
What if I decide to create several of these nodes in a function? Will they remain once the function closes?
basically, if you use the stack, they will live as long as the scope they're declared in. In C, scope is defined by { and }. for example:
int main() {
struct position pos1;
struct position pos2;
struct position pos3;
pos3.prev = pos2;
pos2.prev = pos1;
pos1.prev = NULL;
pop(&pos3);
}
there you declare 3 variables, and associate them, and the pop function just resets the .prev link. But for a stack that kind of architecture is not really useful, because it is quite limited.
There you definitely need to push your instances in the heap, thus using malloc() and free():
// push() pseudocode:
// take stack, iterate over each prev until prev is NULL
// allocate prev with malloc() the same way as for "stack" in main()
// insert values in prev
void push(struct position* stack, int* value);
// pop() pseudocode:
// take stack, iterate over each prev until prev->prev is NULL,
// then keep prev->prev in a temporary variable
// set prev to NULL
// return temporary variable (former prev->prev)
struct position* pop(struct position* stack);
int main() {
int value[2];
struct position* stack = malloc(sizeof(struct position));
// value is what you want to push to the stack
value[0] = 42;
value[1] = 42;
push(stack, value);
value[0] = 2;
value[1] = 20;
push(stack, value);
struct position* pos;
pos = pop(stack);
// do something with pos->value
free(pos);
}
there you create a pointer to a node for which you allocate some memory in the heap. the push() function is allocating some new memory, assigning .prev for that new space to stack's address and populating that memory with the value. pop() should get to the value before the last one, reset its pointer to that value, and return that value.
Of course, I'm just giving concepts and ideas here, I'm leaving you get to the real implementation. One advice though, instead of using square that contains an array, use two separate values in your struct, that will make it simpler for a first implementation.
Related
I have a struct Node and Box given by
typedef struct Node{
Particle p;
Box box;
struct Node *son[4];
}Node
and
typedef struct Box{
double low[3];
double up[3];
}Box
I have two functions insert() and sonumb() where I want to use these structures.
void insert(Particle *p, Node *t){
Box sonbox;
int b=sonumb(&t->box, &sonbox, p);
t->son[b]->box = sonbox; // <--- Produces Segmentation fault (core dumped)
}
int sonumb(Box *box, Box *sonbox, Particle *p){
int b=0;
for(int d=0;d<3;d++){
sonbox->up[d] = 0.5*box->up[d];
sonbox->low[d] = 0.5*box->low[d];
}
b=1; // b=[0,3] just for this example
}
sonum() returns an integer value b. sonbox represents after the call of sonumb() a smaller box inside of t->box. I return the right values for sonbox after the call. So sonbox is not empty. But if I want to copy those values like t->son[b]->box = sonbox I get an segmentatioin fault. What did I miss?
You have nearly certainly missed the allocation of son elements. In order for the expression t->son[b]->box to produce a valid target of an assignment, t->son[b] needs to be assigned a pointer to a valid Node structure. The pointer needs to point to some Node that you have previously allocated.
If child nodes are shared among nodes, this should be a malloc-ed node. This adds quite a bit of complexity, because deleting shared nodes is non-trivial. Two common approaches to working with shared nodes are to (1) allocate all nodes at once in a large array, and use them one-by-one as they become needed, and (2) add reference count to the struct, increment it when taking a pointer, and decrement it when the reference is no longer needed. The second approach is extremely difficult to implement; see if you can avoid it before committing to it.
On the other hand, if child nodes are owned exclusively by their parent, you have a very simple solution: allocate Node with malloc before the assignment of son[b] elements, and free them when you are done with the node:
Box sonbox;
int b=sonumb(&t->box, &sonbox, p);
t->son[b] = calloc(1, sizeof(Node)); // Allocate the node
t->son[b]->box = sonbox;
Use of calloc ensures that the memory of the Node is cleared prior to making other assignments. If this is not necessary, because you assign all members in the rest of your function, replace the call with malloc:
t->son[b] = malloc(sizeof(Node));
adding to #dasblinkenlight's comment.
Box sonbox; ---> This variable is on stack
int b=sonumb(&t->box, &sonbox, p); --> content of t->box is COPIED to sonbox, not by reference but by value.
t->son[b]->box = sonbox; // --> Assigning stack variable is incorrect, because it will vanish once you exit function. OR as #dasblinkenlight suggested pass the value but not the pointer.
My question is an extension of this: Returning pointer to a local structure
I wrote the following code to create an empty list:
struct node* create_empty_list(void)
{
struct node *head = NULL;
return head;
}
I just read that returning pointers to local variables is useless, since the variable will be destroyed when the function exits. I believe the above code is returning a NULL pointer, so I don't think it's a pointer to a local variable.
Where is the memory allocated to the pointer in this case. I didn't allocate any memory on the heap, and it should be on the stack, as an automatic variable. But what happens when the code exits (to the pointer), if I try to use it in the program, by assigning this pointer some pointees / de-referencing and alike?
struct node* create_empty_list(void)
{
struct node *head = NULL;
return head;
}
is equivalent to:
struct node* create_empty_list(void)
{
return NULL;
}
which is perfectly fine.
The problem would happen if you had something like:
struct node head;
return &head; // BAD, returning a pointer to an automatic object
Here, you are returning the value of a local variable, which is OK:
struct node* create_empty_list()
{
struct node* head = NULL;
return head;
}
The value of head, which happens to be NULL (0), is copied into the stack before function create_empty_list returns. The calling function would typically copy this value into some other variable.
For example:
void some_func()
{
struct node* some_var = create_empty_list();
...
}
In each of the examples below, you would be returning the address of a local variable, which is not OK:
struct node* create_empty_list()
{
struct node head = ...;
return &head;
}
struct node** create_empty_list()
{
struct node* head = ...;
return &head;
}
The address of head, which may be a different address every time function create_empty_list is called (depending on the state of the stack at that point), is returned. This address, which is typically a 4-byte value or an 8-byte value (depending on your system's address space), is copied into the stack before the function returns. You may use this value "in any way you like", but you should not rely on the fact that it represents the memory address of a valid variable.
A few basic facts about variables, that are important for you to understand:
Every variable has an address and a value.
The address of a variable is constant (i.e., it cannot change after you declare the variable).
The value of a variable is not constant (unless you explicitly declare it as a const variable).
With the word pointer being used, it is implied that the value of the variable is by itself the address of some other variable. Nonetheless, the pointer still has its own address (which is unrelated to its value).
Please note that the description above does not apply for arrays.
As others have mentioned, you are returning value, what is perfectly fine.
However, if you had changed functions body to:
struct node head;
return &head;
you would return address (pointer to) local variable and that could be potentially dangerous as it is allocated on the stack and freed immediately after leaving function body.
If you changed your code to:
struct node * head = (struct node *) malloc( sizeof( struct node ) );;
return head;
Then you are returning value of local value, that is pointer to heap-allocated memory which will remain valid until you call free on it.
Answering
Where is the memory allocated to the pointer in this case. I didn't
allocate any memory on the heap, and it should be on the stack, as an
automatic variable. But what happens when the code exits (to the
pointer), if I try to use it in the program, by assigning this pointer
some pointees / de-referencing and alike?
There is no memory allocated to the pointer in your case. There is memory allocated to contain the pointer, which is on the stack, but since it is pointing to NULL it doesn't point to any usable memory. Also, you shouldn't worry about that your pointer is on the stack, because returning it would create a copy of the pointer.
(As others mentioned) memory is allocated on the stack implicitly when you declare objects in a function body. As you probably know (judging by your question), memory is allocated on the heap by explicitly requesting so (using malloc in C).
If you try to dereference your pointer you are going to get a segmentation fault. You can assign to it, as this would just overwrite the NULL value. To make sure you don't get a segmentation fault, you need to check that the list that you are using is not the NULL pointer. For example here is an append function:
struct node
{
int elem;
struct node* next;
};
struct node* append(struct node* list, int el) {
// save the head of the list, as we would be modifying the "list" var
struct node* res = list;
// create a single element (could be a separate function)
struct node* nn = (struct node*)malloc(sizeof(struct node));
nn->elem = el;
nn->next = NULL;
// if the given list is not empty
if (NULL != list) {
// find the end of the list
while (NULL != list->next) list = list->next;
// append the new element
list->next = nn;
} else {
// if the given list is empty, just return the new element
res = nn;
}
return res;
}
The crucial part is the if (NULL != list) check. Without it, you would try to dereference list, and thus get a segmentation fault.
I have a linked list struct, i want to pass one node (another struct) pointer to a function (the node is part of the linked list, but i'm passing the node seperately to a deleter function
I want it to copy the next node data into itself (overriding its data), and to delete the next node, thus deleting itself (this part is working)..
I made it check whether the passed node is the last node in the list, and if so, to delete itself.
I don't know how to delete structs from the stack (i know i can malloc() and free() it using the heap memory).
#include <stdlib.h>
#include <stdio.h>
typedef struct {
int data;
struct node * next;
}node;
typedef struct {
struct node * head;
}linked_list;
void print_list(linked_list * list) {
node *current = list->head;
while (current) {
printf("Current node has %d\n",current->data);
current = current->next;
}
}
void delete_node(node * n) {
node * next = n->next;
if (next) {
n->data = next->data;
n->next = next->next;
}
else {
*n = NULL; /*This of course won't compile because assigning void* (null) to node variable
but if i make n point to NULL, nothing will happen because i'm inside a function
and the pointer is duplicated (the passed pointer will still work) */
}
}
void main(){
node first;
node second;
first.data = 1;
first.next = &second;
second.data = 2;
second.next = NULL;
linked_list l;
l.head = &first;
print_list(&l);
delete_node(&second);
print_list(&l);
}
As others have said, you can't.
If you want to be able to store both allocated (by malloc) and non-allocated (static or automatic) storage objects in your list and have a "delete" function that removes objects from the list and frees them, you need to store as part of each list member a flag indicating whether it's in allocated storage or not, and only free the ones which are.
Also note that you'll be in big trouble if the lifetime of the structure with automatic storage ends before you remove it from the list! If dealing with this is confusing at all for you, then you would probably be better-off just using allocated storage (malloc) for all list members.
You can't :)
On most computer architectures, local variables are either allocated directly on a CPU register or on the stack. For local variables allocated on the stack, the top of the stack (the same stack that is used to hold the return addresses of function calls) is manipulated to make space for them when a function enters, and it is restored to "release" the memory when the function exits. All this stack management is handled automatically by the compiler.
You can use the 'free' operator for freeing / deleting a malloc assigned object in the memory. For that, in your code you can write :
free(n);
I have a struct called State:
typedef struct State{
char alphabets[2][6];
struct State *PREV; /*this points to the previous state it came from*/
struct State *NEXT; /*this points to the next state in the linked list*/
int cost; /*Number of moves done to get to this position*/
int zero_index;/*this holds the index to the empty postion*/
} State;
Here's my memAllocator() method:
memAllocator(){
struct State *p = (State*) malloc(sizeof(State));
if (p==NULL){
printf("Malloc for a new position failed");
exit(1);
}
return p;
}
Here's my main method.
main(){
State *start_state_pointer=memAllocator();
State start_state;
start_state.zero_index=15;
start_state.PREV = NULL;
start_state.alphabets[0][0]='C';
start_state.alphabets[0][1]='A';
start_state.alphabets[0][2]='N';
start_state.alphabets[0][3]='A';
start_state.alphabets[0][4]='M';
start_state.alphabets[0][5]='A';
start_state.alphabets[1][0]='P';
start_state.alphabets[1][1]='A';
start_state.alphabets[1][2]='N';
start_state.alphabets[1][3]='A';
start_state.alphabets[1][4]='L';
start_state.alphabets[1][5]='_';
start_state_pointer=&(start_state);
/*start_state=*start_state_pointer;*/
}
I think the statement start_state_pointer=&(start_state); is just assigning the pointer start_state_pointer to to the small amount of temporary space created during State start_state, rather than to the space I allocated.
But when I try the commented out statement start_state=*start_state_pointer to deference the pointer and allocate the space to start state. It gives me a segmentation fault.
I am just starting out in C. Can some one help me with this?
Your memAllocator and main functions don't have explicit return types. This style of code has been deprecated for over 10 years. Functions in C should always have a return type. For main, the return type should be int, and for your memAllocator function, it should be State *.
The second issue is that you allocate space for a State struct, but then fill a different State struct and overwrite the pointer to the previously allocated State struct using start_state_pointer = &(start_state);.
To use the memory that you just allocated, you want to use something like this:
State *start_state = memAllocator();
start_state->zero_index = 15;
start_state->PREV = NULL;
start_state->alphabets[0][0] = 'C';
// etc.
There is no need to create two State structs. When you use State start_start; in your original code, you are creating a struct with something called automatic storage. This means the space for this struct is allocated automatically and is deallocated automatically for you at the end of the scope it is declared in. If you take the address of this struct and pass it around other parts of your program, then you will be passing around a pointer to a deallocated struct, and this could be why your program is crashing.
I'm studying Data Structures, and I'm not getting why stacks and queues need to be declared like:
struct stack *Stack;
(forget about the struct syntax)
I mean, why it is always declared as a pointer?
They are not always declared like that!
In general, declaring a variable as a pointer is useful for later allocating it dynamically. This can be due to a couple of reasons:
The variable is too big for the program stack
You want to return that variable from a function
In your case, let's think of two different implementations of stack:
struct stack
{
void *stuff[10000];
int size;
};
This is a terrible implementation, but assuming you have one like this, then you'd most probably not want to put it on the program stack.
Alternatively, if you have:
struct stack
{
void **stuff;
int size;
int mem_size;
};
You dynamically change the size of stuff anyway, so there is absolutely no harm in declaring a variable of type struct stack on the program stack, i.e. like this:
struct stack stack;
Unless, you'd want to return it from a function. For example:
struct stack *make_stack(int initial_size)
{
struct stack *s;
s = malloc(sizeof(*s));
if (s == NULL)
goto exit_no_mem;
if (initial_size == 0)
initial_size = 1;
s->stuff = malloc(initial_size * sizeof(*s->stuff));
if (s->stuff == NULL)
goto exit_no_stuff_mem;
s->size = 0;
s->mem_size = initial_size;
return s;
exit_no_stuff_mem:
free(s);
exit_no_mem:
return NULL;
}
Personally, though, I would have declared the function like this:
int make_stack(struct stack *s, int initial_size);
and allocate the struct stack on the program stack.
It depends on how your stack structure is defined (not just the layout of the struct, but the operations that manipulate it as well).
It's entirely possible to define a stack as a simple array and index, such as
struct stack_ {
T data[N]; // for some type T and size N
size_t stackptr; // Nobody caught that error, so it never existed, right? ;-)
} stack;
stack.stackptr = N; // stack grows towards 0
// push operation
if (stack.stackptr)
stack.data[--stack.stackptr] = some_data();
else
// overflow
// pop operation
if (stack.stackptr < N)
x = stack.data[stack.stackptr++];
else
// underflow
However, fixed-sized arrays are limiting. One easy method of implementing a stack is to use a list structure:
struct stack_elem {
T data;
struct stack_elem *next;
};
The idea is that the head of the list is the top of the stack. Pushing an item onto the stack adds an element at the head of the list; popping an item removes that element from the head of the list:
int push(struct stack_elem **stack, T data)
{
struct stack_elem *s = malloc(sizeof *s);
if (s)
{
s->data = data; // new element gets data
s->next = *stack; // set new element to point to current stack head
*stack = s; // new element becomes new stack head
}
return s != NULL;
}
int pop(struct stack_elem **stack, T *data)
{
int stackempty = (*stack == NULL);
if (!stackempty)
{
struct stack_elem *s = *stack; // retrieve the current stack head
*stack = (*stack)->next; // set stack head to point to next element
*data = s->data; // get the data
free(s); // deallocate the element
}
return r;
}
int main(void)
{
struct stack_elem *mystack = NULL; // stack is initially empty
T value;
...
if (!push(&mystack, some_data()))
// handle overflow
...
if (!pop(&mystack, &value))
// handle underflow
...
}
Since push and pop need to be able to write new pointer values to mystack, we need to pass a pointer to it, hence the double indirection for stack in push and pop.
No, they don't have to be declared as pointers.
One can as well allocate stacks and queues as global variables:
struct myHash { int key; int next_idx; int data[4]; } mainTable[65536];
struct myHash duplicates[65536*10];
int stack[16384];
myHash also includes a linked list for duplicate entries using indices.
But as stated in the comments, if one has to add more elements to the structures, that was initially planned, then pointers come handy.
An additional reason to declare structures as pointers is that it typically with pointers one can access both the complete structure as a whole, any individual element of the structure or some subset of the elements. That makes the syntax more versatile. Also when the structure is passed as a parameter to some external function, a pointer is inevitable.
There's really no need to implement stacks and queues with pointers - others have already stated this fact clearly. Look at #JohnBode 's answer for how a stack can be perfectly implemented using arrays. The thing is that modelling certain data structures (such as stacks, queues and linked lists) using pointers, allows you to program them in a very efficient way in terms of both execution speed and memory consumption.
Usually an underlying array for holding a data structure is very good implementation choice if your use-cases require frequent random access to an element in the structure, given it's positional index (this is FAST with an array). However growing the structure past its initial capacity can be expensive AND you waste memory with the unused elements in the array. Insertion and deletion operations can also be very expensive since you may need to rearrange elements to either compact the structure or make space for the new elements.
Since a queue and a stack don't have this random-acess requirement and you don't insert or delete elements in the middle of them, it is a better implementation choice to dynamically allocate each individual element "on the fly", requesting memory when a new element is required (this is what malloc does), and freeing it as an element is deleted. This is fast and will consume no more memory than it is actually needed by your data structure.
As aleady pointed out it depends on how big the struct is.
Another reason is encapsulation. The stack implementation might not expose the definition of struct stack in its header file. This hides the implementation detail from the user, with the downside that free store allocation is required.