I'm writing an application and am having problems with passing a dynamically created array through pointers to the calling function.
I create a pointer in the main to contain the dynamically generated array, and an int to contain the length of that array. I pass those to the readNDEF() function. There it allocates memory based on the read of the needed memory, and reads bytes into the generated array.
Followed a lot of the answers givin here on similar questions but none seem to fix it, of give other errors (eg stack smashing)
int main(void) {
uint8_t *recordPTR; //creating pointer
uint8_t length=0; //creating length variable
readNDEF(recordPTR, &length);
int i;
for (i=0;i<1;i++){
printf("%x ",*(recordPTR+i)); //segmentation fault happens here
}
}
bool readNDEF(uint8_t *messagePTR, uint8_t *messageLength){
int NDEFlength;
if(!(NDEFlength=getNDEFmessageLength())<0){ //get length
closeSession();
return false;
}
uint8_t tempLength=0x00|NDEFlength;
messagePTR = malloc(tempLength*sizeof(uint8_t)+5); //+5 overhead for the rest of the frame
if(messagePTR == NULL){ //check if mallok ok
return false;
}
if(!ReadBinary(0x0002, (uint8_t)0x00|NDEFlength, messagePTR)){ //read NDEF memory
closeSession();
return false;
}
messagePTR++; //skip first byte in the array
closeSession();
*messageLength = tempLength;
//print the array (Works, data correct)
int i;
for (i=0;i<tempLength;i++){
printf("%02x ",*(messagePTR+i));
}
return true;
}
The length returns like it should, but the array itself when enumerating it in the for loop gives a segmentation fault. Using an other way I could enumerate it without the fault, but the data was not correct (random data) probably because it was out of scope after returnign from the function.
Your readNDEF method allocates memory for an object inside that method (because argument of the type pointer, as any other argument in C, is passed by value). Hence the pointer outside has not been changed, and the memory allocated within that function was lost (a memory leak). You need to pass a pointer-to-pointer in order to achieve what you want:
bool readNDEF(uint8_t **messagePTR, uint8_t *messageLength){
///
*messagePTR = malloc(tempLength*sizeof(uint8_t)+5);
}
and call it accordingly:
readNDEF(&recordPTR, &length);
Related
I try something like below but all the time I have a segmentation fault.
I don't really want to use (e.g.) #define N 1000 and then declare int buffer[N].
Just in case..I'm not allowed to use any headers except stdio.h as well as dynamic memory.
void input (int *buffer, int *length);
int main()
{
int length, *buffer = NULL, *numbers = NULL;
input(buffer, &length);
}
void input(int *buffer, int *length) {
scanf("%d", length);
if (*length < 0) {
error = 1;
return;
}
for (int i = 0; i < *length; i++) {
scanf("%d", *buffer[i]);
}
}
How to pass an array with unknown 1-d dimension into function
In C, arrays cannot exist until their size is known.
There are other approaches though.
In C, code cannot pass an array to a function. some_function(some_array) converts the array some_array to the address of the first element of the array: &some_array[0]. That is what the function receives, a pointer, not an array. The original size information of the array is not passed, thus also pass the length to the function.
Sample:
Read desired length.
{
int length = 0;
scanf("%d", &length);
Form a variable length array, length >= 1.
if (length <= 0) {
return NULL;
}
int buffer[length];
Now call a function, passing the length and the address of the first element of the array.
// Do stuff with length and buf, like read data
foo1(length, buffer);
// foo1() receives the length & address of the first element of the array as an int *
// Do more stuff with length and buf, like write data
foo2(length, buffer);
}
At the end of the block }, buffer no longer available.
In C, you can't create an array if you can't know its size at compile time (or at least not in certain implementations and standards), so doing something like buffer[length] won't work (again at least not in certain implementations/standards).
What you need to do to make sure this works everywhere is to use a pointer (as I see you're trying to use here). However, what you're doing wrong here that causes your segfault with the pointers is you assign them the value of NULL. This also won't work due to how when you assign a pointer an arbitrary value, there is no memory allocated for the pointer (This applies for everything other than addresses of "regular" variables using the & operator and assigning other pointers that are checked to be OK). Your pointers are just pointing to address 0 and can't be used for anything.
What you need to do here to fix the pointers is to use dynamic memory allocation, so you can have a truly variable-sized array. Specifically, you need to use a function like malloc or calloc to allocate memory for the pointers so they are usable. In your case, using calloc and reading its documentation, we see that it takes 2 parameters: The number of elements it should allocate memory for and the size of each element. We also know that it returns a pointer to the starting address of the allocated memory and that in case of failure (which can only happen if you're out of memory), it returns NULL. Using this, we understand that in your case the call to calloc would be like this:
int *buffer = (int *) calloc(length, sizeof(int));
The sizeof() function returns the size of a data type in bytes. Here you allocated enough memory for the pointer to hold length integers (since you'll use it as an array you need enough memory for all the integers, you're not just pointing to 1 integer but storing all of them), and calloc is also noted to initialize every allocated element to 0, so you have an array of integers that are all initialized to 0 (Also note that type casting has been used to make sure the allocated memory block is appropriate for use with an integer array, you can read more about type casting in this small article from Tutorialspoint if you'd like). Then, after this has been allocated, you can start reading your integers into the array. The complete code looks like this:
void input (int *buffer, int *length);
int main() {
// NOTE: I don't see the numbers pointer used here, maybe remove it?
int length, *buffer, *numbers;
input(buffer, &length);
}
void input(int *buffer, int *length) {
scanf("%d", length);
if (*length < 0) {
// Consider printing the exact error here
error = 1;
return;
}
buffer = (int *) calloc(length, sizeof(int));
if (buffer == NULL) {
printf("Couldn't allocate memory for buffer\n");
error = 1;
return;
}
// Accessing the elements of an array doesn't need * and in fact * here can (and probably will) cause terrible things
for (int i = 0; i < *length; i++) {
scanf("%d", buffer[i]);
}
}
Also don't forget to call free() on the pointer after you're done using it to avoid memory leaks (in your case that'd be after the call to input()).
Hope this helped, good luck!
You cannot use arrays because their memory size must be known to the compiler at compile time. Also you can't use Variable Length Arrays because they are allocated at the point of declaration and deallocated when the block scope containing the declaration exits.
The solution to your problem might be to use malloc
I'm a beginner of C and now I'm learning pointer and dynamic memory allocation. I want to write a simple program to create empty arrays and check for the existence of a given number. Here's my code:
/* create an empty array pointer */
int* createArray(){
int *a = (int*) malloc(sizeof(int));
return a;
}
void findArrayElement(int *list, int element){
int i;
int len = (sizeof(list) / sizeof(int));
if (sizeof(list) == 0) {
printf("NO\n");
return;
}
for (i=0; i<len; i++) {
if (list[i] == element) {
printf("YES\n");
return;
}
}
printf("NO\n");
}
int main(int argc, const char * argv[]) {
int *p;
p = createArray();
printf("size of int is: %lu\n", sizeof(int));
printf("size of p is: %lu\n", sizeof(p));
printf("LENGTH of p is: %lu\n", ARRLENGTH(p));
findArrayElement(p, 2);
findArrayElement(p, 0);
return 0;
}
But when I run the program, I always get 'YES' when I looking for 0, so
Is there a way to differentiate integer 0 and a complete empty array?
Also I'm not sure whether my function createArray() is a correct way to create an empty array.
Thanks guys.
Is there a way to differentiate integer 0 and a complete empty array?
How do you define an empty array? Once you allocate a memory chunk and assign it to a pointer, it already has some value (which is undefined in case of alloc). The most used way to mark a pointer as not used or not allocated os to assign NULL to it.
Also I'm not sure whether my function createArray() is a correct way to create an empty array.
sizeof returns the number of bytes which the given object (or type) occupies in the memory. In your case sizeof(list) returns 8 as it is a pointer.
In oder to allocate an array, the function has to receive its size. Currently it always allocates size for one integer only.
Edit: Adding example.
/* create an empty array pointer */
int* createArray(size_t size)
{
return (size ? (int*) malloc(sizeof(int)*size) : NULL);
}
So now the returned pointer should be 'coupled' with the size of the array. Which means that each function that receives an array as a parameter should receive also its size.
sizeof returns the memory size of the array pointer, regardless of contents.
edit: if it exists in memory, it will be nonzero.
edit 3: removed inaccurate information, see the comments about creating a variable to record the length. Also from comments, note that your createArray function is creating an array for exactly 1 integer. In C, arrays are of fixed length. So this Array will always be the same size (whether you stored something in it or not). sizeof(pointer) will always return the memory allocated for the pointer, not the memory allocated for the array at which it is pointing.
I'm having a problem with my program. I need my program to read from a text file, the first consists of the dimensions of the 2d array the rest is the contents of the array. I have coded the readWord function which reads from textfiles and it works but when i do anything on the main function with the array it crashes. Please could you help.
int main()
{
int num_it, cols, rows;
char** myworld;
num_it = readWorld(myworld,&cols, &rows);
myworld[1][2]='x';/*it crashes when i make this statement*/
}
int readWorld(char** world, int* width,int* height)
{
int result=0,i,cols=0,rows=0;
char buff[25];
FILE* file = fopen ("world.txt", "r");
fscanf(file, "%d %d %d\n", width, height, &result);
/*Dynamic allocation*/
world = (char**)malloc(*(height)* sizeof(char*));
for(i=0;i<*height;i++)
{
world[i] = (char*)malloc(*width*sizeof(char));
}
/*store data in array*/
while(fgets(buff, sizeof buff, file) != NULL)
{
if (strlen(buff) >1){
for(cols=0; cols<=(strlen(buff)); ++cols)
{
world[rows][cols] = buff[cols];
}
++rows;
}
}
fclose(file);
return result;
}
You need to allocate the memory for myworld in the actual caller!
What's happening here is that you are passing the pointer by value to the function.
The pointer value is changed by the function but that's not going to adjust the one in the caller.
Two options: use a triple indirection (ie pass a pointer to the the pointer) or allocate in the main sub. I prefer the latter, mainly because you can control the memory deallocation in a more symmetrical manner; even if you fixed this problem your code still has the memory leak.
What you're experiencing is undefined behaviour as you are attempting to access memory your program does not own.
Your myworld variable in main is never initialized and points to junk, so when you try to access it bad things happen. Think about why: you are passing a copy of the variable to readWorld. You correctly allocate memory inside there, and make the copy point to it, but the original pointer (in main) still points to whatever random location it pointed to before.
If you want the memory for it to be allocated inside the readWorld function and made accessible via the myworld variable in main then you must pass a pointer to myworld to readWorld; in other words, you must pass a triple pointer.
Try this:
int readWorld(char*** world, int* width,int* height)
{
char **tempworld = malloc(...);
// do whatever
*world = tempworld;
return result;
}
int main()
{
char **myworld = NULL;
readWorld(&myworld, ...);
return 0;
}
I have a function that I pass an array into and an int into from my main function. I am doing operations to the array inside this new function, let's call it foo. In foo, I initialize another array with 52 cells all with 0. I do operations on the array that I passed from main, and transfer that data to the newly initialized array. I want to return the new array back to the main function. But of course, I can't return data structures like arrays. So I instead return an int pointer that points to this array. Inside the int main, I pass the pointer to have it point to various cells in the array. When I print the results of what the pointer is pointing to, it should either be pointing to 0 or an integer greater than 0. But instead, I get inconsistent results. For some reason, some of the values that SHOULD be 0, prints out garbage data. I've been trying to spot the bug for some time, but I just wanted a second hand look at it. Here is just the GENERAL idea for the code for this portion anyways...
int main(){
int *retPtr;
char input[] = "abaecedg";
retPtr = foo(input, size);
for(i=0; i<52; i++){
// error displayed here
printf("%d\n", *(retPr + i));
}
}
int foo(char input[], int size)
{
int arr[52] = {0}; // should initialize all 52 cells with 0.
int i=0, value; // looking for non-zero results in the end.
int *ptr = &arr[0];
for(i=0; i<size; i++){
if(arr[i] > 64 && arr[i] < 91){
value = input[i] - 65;
arr[value]++;
}
}
return ptr;
}
Hopefully this makes sense of what I'm trying to do. In the foo function, I am trying to find the frequency of certain alphabets. I know this might be a bit cryptic, but the code is quite long with comments and everything so I wanted to make it as succinct as possible. Is there any possible reason why I'm getting correct values for some (numbers > 0, 0) and garbage values in the other?
The reason you get garbage back is that the array created in foo is allocated in foos stack frame, and you then return a pointer into that frame. That frame is discarded when foo returns.
You should allocate the array on the heap (using malloc and friends) if you want it to remain after foo returns. Don't forget to free() it when you're done with the array.
int main(){
char input[] = "abaecedg";
int retPtr[] = foo(input, size); //An array and a pointer is the same thing
...
free(retPtr);
}
int *foo(char input[], int size)
{
int arr[] = calloc(52*sizeof(int); // should initialize all 52 cells with 0.
...
arr[value]++;
...
return arr;
}
Another way is to let foo take an array as a parameter and work with that, in this way:
int main(){
int ret[52] = {0};
...
foo(input, size, ret);
...
}
void foo(char input[], int size, int *arr)
{
...
arr[value]++;
...
return; //Don't return anything, you have changed the array in-place
}
The reason this works is because an array is the exact same thing as a pointer, so you are really passing the array by reference into foo. arr will be pointing to the same place as ret, into the stack frame of main.
In function foo the array arr is a local array, that is, allocated on the stack. You must not return any pointer of data allocated on the stack, since the stack is rewinded after you return from the function, and its content is no more guaratneed.
If you want to return an array you should allocate it on the heap using malloc, for example, and return the pointer malloc returned. But you will then have to free that memory somewhere in your program. If you fail to free it you will have what's called a "memory leak", which may or may not crash/disturb this program from running again, depending on your environment. A not clean situation, that's for sure.
That's why I consider C not so good for functional programing idioms, such as returning things from function (unless they are primitive types). I would achieve what you tried to do by passing another array to foo - an output array, companioned by a size variable, and fill that array.
Alternately, you could wrap the array within a struct and return that struct. Structs can be returned by value, in which case they are copied via the stack to the caller function's returned value.
Can't find what is wrong with this code, it works as expected when inputting exactly 4 values, but on the fifth call (before it even asks for scanf) it always gives me this error:
* glibc detected ./a2: double free or corruption (fasttop): 0x0916e018 **
Here's some code of my program:
typedef struct {
int i;
char str[25];
} typeX;
int main(){
int dSize = 0;
int *dSizePtr = &dSize;
dPointer = (typeX **)malloc(sizeof(typeX *)); // makes an array of pointers
int i;
for (i = 0; i < 100; i++)
makeElement(dPointer, dSizePtr); // Puts values into those pointers
free(dPointer);
return 0;
}
void makeElement(dPointer **, int *dSizePtr){
dPointer = (typeX **)realloc(dPointer, sizeof(typeX *)*(*dSizePtr+1)); // grow the array by one
if (typeX == NULL)
return; // some kind of quit statement, just return for now
dPointer[*dSizePtr] = (typeX *)malloc(sizeof(typeX)); // make a new pointer in the array
scanf("%s", dPointer[*dSizePtr]->str); // input the values of the struct (have to use scanf)
char input[20];
scanf("%s", input);
dPointer[*dSizePtr]->int = atoi(input);
++(*dSizePtr);
}
I know I don't have to make a dSizePtr and I can just pass in &dSize, but the way my program is currently set up (this isn't exactly the same, just compressed for readability), that's the way I have to pass it.
I honestly have no idea why this error is coming up. Been looking at my code for hours and reading online and haven't found a solution. Any help will be greatly appreciated!
The problem is that your function makeElement get the value of dPointer, not its reference. When you realloc the data, the originally allocated chunk is freed. But the dPointer outside of the makeElement scope is not changed;
The runtime error is delayed as the actual memory allocation is performed in quantities bigger than sizeof(typeX*)
This line is causing the double free.
dPointer = (typeX **)realloc(dPointer, sizeof(typeX *)*(*dSizePtr+1)); // grow the array by one
For the first few iterations of the loop in the caller the block of memory is large enough that realloc() doesn't have to do anything, and thus it returns the same pointer passed to it. But at some point the block of memory is too small and realloc() has to allocate a new block of memory and returns a pointer to it. That returned pointer is assigned to dPointer in makeElement() but it does not change the value of dPointer in the caller. So the caller continues to pass the old dPointer value into makeElement(), which passes it to realloc(), which notices that this pointer has been freed (by the call to realloc() that expanded the size of the array).