It was, actually, the display function after all. Sorry if I wasted your time (I didn't actually put the code of the display function here, my bad. However, I did learn some new things, so thank you all!
I have created a structure which involves an array that represents a set of integers.
When I first create a structure, I use malloc() to allocate space for the structure and thus, the array.
The problem resides, however, in the fact that when I create the structure, the first element of the array is being assigned a number. I want, however, to just create a structure where the array in it is uninitialised.
typedef struct set{
int numberOfValues; /* The number of values in the set */
int value[40]; /* the set */
} *Set; /* Set is thus a pointer to struct set */
Set aOfPStructs[4];
void create()
{
static int setnum = 0;
Set S; /* S is a pointer to struct set */
if (setnum <= 4)
{
S = (struct set *) malloc(sizeof(struct set));
S->numberOfValues = 0;
aOfPStructs[setnum] = S;
printf("The set's number is %d\n", setnum);
setnum++;
}
else
printf("Error. No more sets can be created.\n");
}
The above is only a snippet from the whole program, but any other code is irrelevant to the question tbh.
So with that code, if I were to display the contents of the array in the structure, I would get 1 value. Thing is, I have not added any elements to the array, so I should get no values at this point. Can I fix this?
Any help would be appreciated! Thanks!
Edit:
I tried memset(), but it's not exactly what I want. When I use the create(), I want 0 elements in the array value[], but the code is generating 1 random element once I use the create function.
The function. to display the element seems to work just fine though. Funny thing is, when I run a function to delete one element from the array, that extra element is removed, only to be added again if I run the add() function.
Edit 2:
Added a screenshot for further explanation of my problem. I appreciate every comment thus far, and future ones too.
http://imgur.com/cza3ip6
Edit 3:
Ok, I get the use of memset() now and the concept of initialisation. Thing is, if I wanted to check the number of elements of the set, just right after I create it, I should get 0 because I have not added any elements, but an element is being generated with create...
I think you may not understand what "initialized" and "uninitialized" mean.
"Uninitialized" means "holding whatever bit pattern happens to be left over from the last computation that used this memory or from bootstrapping the machine."
"Initialized" just means you have assigned a set value.
The malloc call gives you uninitialized storage according to this definition. So what you are seeing is the best you can get. You are printing the integer formed from the arbitrary bit pattern in the first array slot.
For what it's worth, the simplest way to get a pre-zeroed memory block of n ints (initialized to zero) is to say.
int *blk = calloc(n, sizeof(int));
Use memset to set to 0 all the data in the structure. By default when you use malloc the memory you get is not initiliazed:
S = (struct set *) malloc(sizeof(struct set));
memset(S, 0, sizeof(struct set));
Try this. It uses a little different approach to create and initialize array of structs, but the relevant point is that it uses memset to init array of ints for each element of struct array:
#include<stdio.h>
typedef struct {
int numberOfValues; /* The number of values in the set */
int value[40]; /* the set */
}SET;
SET set[4], *pSet; //initialize SET *, pSet to the first position of set
main()
{
int i;
pSet= &set[0];
for(i=0;i<4;i++)
{
memset(pSet[i].value, 0, sizeof(int)*40); //set all elements of member value to 0
}
}
You haven't posted the function displaying your set. I suppose that the problem may be there. Maybe you use something like for(i=0; i <= S->numberOfValues; i++) instead of for(i=0; i < S->numberOfValues; i++), so you print the first element even if not there.
Related
I have had a look around but have not been able to find an answer to this question already. I am trying to create a hash table of which each element is a struct. In each struct is a variable to let the program know if the cell has been occupied, for this to work I need to set all of them to zero. The thing is it worked fine but now and then (seemingly randomly) I'd get an access violation. I thought I fixed it but when I come to grow my array the error creeps up again, leading me to believe that I have made an error. My pointer knowledge is not that good at all, so any help would be appreciated. This is what the function looks like:
HashTableCell *initialiseTable(HashTableCell *hashTable, int *tableSizePtr)
{
int i = 0;
int totalSize = *tableSizePtr * sizeof(HashTableCell);
HashTableCell *tempStartingcell;
tempStartingcell = (HashTableCell*)malloc(sizeof(HashTableCell));
*tempStartingcell = *hashTable;
while (i <= *tableSizePtr)
{
/*we keep moving forward, need to use the first entry*/
*hashTable = *(tempStartingcell + (i * sizeof(HashTableCell)));
hashTable->isOccupied = 0;
i++;
}
free(tempStartingcell);
return hashTable;
}
And before I malloced some space for the table and passed it in another function like so:
HashTableCell *hashTable;
hashTable = (HashTableCell*)malloc((sizeof(HashTableCell)*tableSize));
hashTable = initialiseTable(hashTable, tableSizePtr);
The idea is to start at the beginning and move along the correct number of spaces along per iteration of the while loop. When I come to resize I merely make a new array with double the malloced space and pass it to the initialise function but this throws up an access violation error at seemingly random indexes.
I am using VS2015 if that helps anything.
Thank you for your help.
The problem is in this line:
*hashTable = *(tempStartingcell + (i * sizeof(HashTableCell)));
When you are adding an integer to a pointer, C and C++ already take into account the size of the array elements, so you should not multiply with sizeof(HashTableCell), but rather do:
*hashTable = *(tempStartingcell + i);
Otherwise, your extra multiplication will cause an access outside of the tempStartingCell array. It makes even more sense to write it like this:
*hashTable = tempStartingcell[i];
But there is more wrong with your code; if you just want to set isOccupied to zero for each element in hashTable, just do:
void initialiseTable(HashTableCell *hashTable, int tableSize)
{
for (int i = 0; i < tableSize; i++)
hashTable[i].isOccupied = 0;
}
I am trying to delete an array of initialized structs e.g. reset the array
My struct:
struct entry{
char name[NAME_SIZE];
int mark;
};
typedef struct entry Acct;
Acct dism2A03[MAX_ENTRY];
Acct clear[0]; << temp struct to set original struct to null
My attempt:
entry_total keeps track of how many structs in the struct array dism2A03[x] have values set in them.
I tried to create an empty array of the same struct clear[0]. Looped through initialized arrays in dism2A03[x] and set them to clear[0]
for(m=0;m<entry_total;m++){
dism2A03[m]=clear[0];
}
break;
However, it is setting them to 0, i want them to become uninitialized e.g. no values in them
You cannot have memory with no value in it. It's physically impossible. It's due to the laws of physics of our universe :-)
Also, this:
Acct clear[0];
is wrong. You cannot have an array with zero elements. Some compilers will allow this as an extension, but it's not valid C. And for the compilers that allow this, it doesn't do what you think it does.
It would seem to me that what you want instead is to resize the array. To do that, you would need to copy the elements you want to keep into a new array, and then free() the old one. To do that, you need to create dism2A03 using dynamic memory:
Acct *dism2A03 = malloc(sizeof(Acct) * MAX_ENTRY);
if (dism2A03 == NULL) {
// Error: We're out of memory.
}
(malloc() returns NULL if there's no more free memory, and the code checks that. Usually all you can do if this happens is terminate the program.)
When you want a new array with some elements removed, then you should back up the starting address of the current one:
Acct* oldArray = dism2A03;
then create a new one with the new size you want:
dism2A03 = malloc(sizeof(Acct) * NEW_SIZE);
if (dism2A03 == NULL) {
// Error: We're out of memory.
}
copy the elements you want from the old array (oldArray) to the new one (dism2A03) - which is up to you, I don't know which ones you want to keep - and after than you must free the old array:
free(oldArray);
As a final note, you might actually not want to create a new array at all. Instead, you could keep having your original, statically allocated array ("statically allocated" means you're not using malloc()):
Acct dism2A03[MAX_ENTRY];
and have a index variable where you keep track of how many useful elements are actually in that array. At first, there are 0:
size_t dism2A03_size = 0;
As you add elements to that array, you do that at the position given by dism2A03_size:
dism2A03[dism2A03_size] = <something>
++dism2A03_size; // Now there's one more in there, so remember that.
While doing so, you need to make sure that dism2A03_size does not grow larger than the maximum capacity of the array, which is MAX_ENTRY in your case. So the above would become:
if (dism2A03_size < MAX_SIZE) {
dism2A03[dism2A03_size] = <something>
++dism2A03_size; // Now there's one more in there, so remember that.
} else {
// Error: the array is full.
}
As you can see, adding something to the end of the array is rather easy. Removing something from the end of the array is just as easy; you just decrement dism2A03_size by one. However, "removing" something from the middle of the array means copying all following elements by one position to the left:
for (size_t i = elem_to_remove + 1; i < dism2A03_size; ++i) {
dism2A03[i - 1] = dism2A03[i];
}
--dism2A03_size; // Remember the new size, since we removed one.
Note that you should not attempt to remove an element if the array is empty (meaning when dism2A03_size == 0.)
There's also the case of adding a new elements in the middle of the array rather than at the end. But I hope that now you can figure that out on your own, since it basically a reversed version of the element removal case.
Also note that instead of copying elements manually one by one in a for loop, you can use the memcpy() function instead, which will do the copying faster. But I went with the loop here so that the logic of it all is more obvious (hopefully.)
when you declare an array in this way Acct dism2A03[MAX_ENTRY]; the array is allocated in the stack, therefore it will be removed when the function will perform the return statement.
What you can do is to allocate the structure in the heap via malloc/calloc, and then you can free that memory area via the free function.
For example :
typedef struct entry Acct;
Acct * dism2A03 = calloc(MAX_ENTRY, sizeof( struct entry));
// ....
free(dism2A03);
I have a recursive struct which is:
typedef struct dict dict;
struct dict {
dict *children[M];
list *words[M];
};
Initialized this way:
dict *d = malloc(sizeof(dict));
bzero(d, sizeof(dict));
I would like to know what bzero() exactly does here, and how can I malloc() recursively for children.
Edit: This is how I would like to be able to malloc() the children and words:
void dict_insert(dict *d, char *signature, unsigned int current_letter, char *w) {
int occur;
occur = (int) signature[current_letter];
if (current_letter == LAST_LETTER) {
printf("word found : %s!\n",w);
list_print(d->words[occur]);
char *new;
new = malloc(strlen(w) + 1);
strcpy(new, w);
list_append(d->words[occur],new);
list_print(d->words[occur]);
}
else {
d = d->children[occur];
dict_insert(d,signature,current_letter+1,w);
}
}
bzero(3) initializes the memory to zero. It's equivalent to calling memset(3) with a second parameter of 0. In this case, it initializes all of the member variables to null pointers. bzero is considered deprecated, so you should replace uses of it with memset; alternatively, you can just call calloc(3) instead of malloc, which automatically zeroes out the returned memory for you upon success.
You should not use either of the two casts you have written—in C, a void* pointer can be implicitly cast to any other pointer type, and any pointer type can be implicitly cast to void*. malloc returns a void*, so you can just assign it to your dict *d variable without a cast. Similarly, the first parameter of bzero is a void*, so you can just pass it your d variable directly without a cast.
To understand recursion, you must first understand recursion. Make sure you have an appropriate base case if you want to avoid allocating memory infinitely.
In general, when you are unsure what the compiler is generating for you, it is a good idea to use a printf to report the size of the struct. In this case, the size of dict should be 2 * M * the size of a pointer. In this case, bzero will fill a dict with zeros. In other words, all M elements of the children and words arrays will be zero.
To initialize the structure, I recommend creating a function that takes a pointer to a dict and mallocs each child and then calls itself to initialize it:
void init_dict(dict* d)
{
int i;
for (i = 0; i < M; i++)
{
d->children[i] = malloc(sizeof(dict));
init_dict(d->children[i]);
/* initialize the words elements, too */
}
}
+1 to you if you can see why this code won't work as is. (Hint: it has an infinite recursion bug and needs a rule that tells it how deep the children tree needs to be so it can stop recursing.)
bzero just zeros the memory. bzero(addr, size) is essentially equivalent to memset(addr, 0, size). As to why you'd use it, from what I've seen around half the time it's used, it's just because somebody though zeroing the memory seemed like a good idea, even though it didn't really accomplish anything. In this case, it looks like the effect would be to set some pointers to NULL (though it's not entirely portable for that purpose).
To allocate recursively, you'd basically just keep track of a current depth, and allocate child nodes until you reached the desired depth. Code something on this order would do the job:
void alloc_tree(dict **root, size_t depth) {
int i;
if (depth == 0) {
(*root) = NULL;
return;
}
(*root) = malloc(sizeof(**root));
for (i=0; i<M; i++)
alloc_tree((*root)->children+i, depth-1);
}
I should add that I can't quite imagine doing recursive allocation like this though. In a typical case, you insert data, and allocate new nodes as needed to hold the data. The exact details of that will vary depending on whether (and if so how) you're keeping the tree balanced. For a multi-way tree like this, it's fairly common to use some B-tree variant, in which case the code I've given above won't normally apply at all -- with a B-tree, you fill a node, and when it's reached its limit, you split it in half and promote the middle item to the parent node. You allocate a new node when this reaches the top of the tree, and the root node is already full.
EDIT: Thank you very much for your responses. I understand this properly now!
I am trying to learn more on C pointers. Tinkering around, I am questioning the difference between two actions I am using.
This code seems to work at first glance, but I am not sure of what's the difference, and if any of these two approaches is wrong in some way.
I'd like to know what's the difference between the two pieces of code, when I should I pass the adress, and when a pointer to an array?
Is any of the pieces wrong? If so, what would be the proper way?
having a simple struct grid pretty much like struct grid { int val; } (for demonstration purposes)
First piece of code. Passing address of the pointer to the array.
void set (mygrid *grid, int foo){
grid->bar = foo; //should this be '*grid->bar?' But this seems to work properly.
}
void main(){
int i;
int* array;
int max = 24;
array = malloc(sizeof(grid) * max);
for(i = 0; i < max; i++){
set(&array[i], 0);
}
}
Second piece of code. I am not entirely sure why this works, but the compiler doesn't output any warning.
I am supposed to be passing the pointer to the start of the array like this?
void set(mygrid *grid, int foo){
int i; int max = 24; //so this example code compiles :P
for(i = 0; i < max; i++){
grid[i].bar = foo;
}
}
void main(){
int* array;
int max = 24;
array = malloc(sizeof(grid) * max);
set(array, 0); //Why not &array?
}
Passing an array decays into a pointer that points to the first member of the array, just like &array[0].
In your second example, array is just a pointer, and the return value from malloc is just the address of the start of the block of memory you get.
It doesn't have to be used for an array; it could be used for storage of an arbitrary sizeof(int) * max bytes of data. An array (in C) is really just a nice way of thinking about & working with a solid block of memory divided up into equal size portions.
Secondly, you should understand how my_array[i] works. All it does is take the address of where your block of array data starts (which is the actual value of my_array), and then look at what value is stored at a particular offset from there. Specifically, if my_array is of a (made up) type of WhatEver, then it will access the data from my_array + i*sizeof(WhatEver) to my_array + (i+1)*sizeof(WhatEver).
On a related note (since you're learning C), it's highly recommended to check that the return from malloc is not NULL before doing anything with it.
I'm no C guru but am also trying to improve my understanding so if this is incorrect, please leave a comment or edit my answer so I can learn from my mistakes :)
In your first piece of code
grid->bar is same as (*grid).bar
. and using name of an array refers to its base address. so writing array is equivalent &array[0]
&array[i] is equivalent to array+i
array[i] is equivalent to *(array +i)
In you second piece of code i dont understand why there is no error because in your function set you do not declare max and i dont see a global max variable too.
also in your second piece of code you use
set(array,0) because array is already an integer pointer(see the declaration int * array).As far as i understand the mygrid is not a struct but is an array of structs in the second example
In C, an array is pretty much the same as a pointer. For me this isn't so amazing, since it is one of the earlier programming languages I learned, but if you're coming from a high level language where an array is a different type of object, then it might come across as strange.
If I create a struct in C and want to add them to an array that is not set to a fixed size, how is the array created?
Can one create a tempStruct which is used on every iteration while getting user input and store this in an array, always using the same tempStruct struct in the loop?
How is an array created if the size is unknown as it depends on user input, and how are structs added to this array?
When the size is unknown at compile time, you'll need to allocate the memory on the heap, rather than in the data segment (where global variables are stored) or on the stack (where function parameters and local variables are stored). In C, you can do this by calling functions like malloc.
MyStructType *myArray = (MyStructType *)malloc(numElements * sizeof(MyStructType)
... do something ...
free(myArray)
If you're actully using C++, it's generally better to use new[] and delete[], e.g.
MyStructType *myArray = new MyStructType[numElements]
... do something ...
delete [] myArray
Note that new[] must be paired with delete[]. If you're allocating a single instance, use new and delete (without "[]"). delete[] and delete are not equivalent.
Also, if you're using C++, it's generally easier and safer to use an STL vector.
the C array must be with fixed size this is what we have learned years ago
but memory allocation functions may help you to find a solution
in c++ you can use the operator new
int num=0;
cout<<"please enter the number"<<endl;
cin>>num;
int *x=new int[num];
for(int i=0;i<num;i++)
{
cout<<"enter element "<<(i+1)<<endl;
cin>>x[i];
}
//do your work
and as
Mr Fooz
mentioned delete[] is used to free the memory allocated by new[]
and this is a general example
If you are using the older C89 standard compiler, you cannot use variable length arrays. If you use C99 then you can create variable length array. For clarification: variable-lenght doesnt mean that the array lenght can change during execution. It just means that you can set it during execution instead of fixing a value during compile time.
For eg:
CreateArray(const int numberOfElements)
{
struct MyStruct arrayOfStructs[numberOfElements];
}
This is valid in C99 but not in C89. Check your compiler documentaion.
Yes, you can use a tempStruct during input which you add later to the array.
If the size of the array is unknown, then you are in trouble. You must keep track of the array's size somewhere. Just have an integer variable that you change every time you change your array to keep track of your array's size.
If the size of the struct is not known at compile time it is even more complicated. You either just store Pointers in the array which point to your actual struct elements in memory, or you have to keep track of the sizes of every struct in the array. In the later case you would have to do addressing in the array completely manually calculating a lot. While this is very memory efficient, it is also error prone and extremely hard to debug.
OK. sample to create an array that hold your struct using pointers:
struct MyStruct
{
/* code */
}
main()
{
counter = 0;
struct MyStruct** array = malloc(counter);
// Create new Element
struct MyStruct myElement;
myElement.someData = someValue;
// Add element to array:
array = realloc(array, sizeof(struct MyStruct*) * (counter + 1);
array[counter] = &myElement;
counter++;
// Create yet another new Element
struct MyStruct myElement;
myElement.someData = someOtherValue;
array = realloc(array, sizeof(struct MyStruct*) * (counter + 1);
array[counter] = &myElement;
counter++;
// Now remove the last element
free(array[counter -1]); // may have something more complicated than that, depending on your struct
array = realloc(array, sizeof(struct MyStruct*) * (counter - 1);
counter--;
}
this code is not tested!