parse 2d dynamic int array to Shared Memory - c

Im trying to build a program that parsing a 2d dynamic array to other program by using shared memory.I search a lot but im a bit confused because im not familiar at this one.
My code so far:
int main (int argc, char* argv []){
int rows,columns;
if( argc < 3 ){
printf("Need The size of the 2d array\n");
return 0;
}
rows = atoi(argv[1]);
columns = atoi(argv[2]);
time_t t;
srand((unsigned) time(&t));
key_t key = ftok(".",'a');
size_t size = sizeof(key_t) + (rows * columns + 2 + rows) * sizeof(int);
int shmid = shmget(key,size,IPC_CREAT|IPC_EXCL|S_IRWXU);
int *memory = shmat(shmid, NULL, 0);
printf("Shared Memory Key: %d\n", key);
int *argsflag = memory;
int *resflag= memory + 1;
int *res = memory + 2;
int **array = (int **) memory + (rows*columns);
for(int i = 0; i < rows ; i++) {
for(int j = 0; j < columns; j++) {
array[i][j] = rand() % 100;
}
}
for(int i = 0; i < rows ; i++) {
for(int j = 0; j < columns; j++) {
printf("%d ",array[i][j]);
}
printf("\n");
}
shmctl(shmid,IPC_RMID,NULL);
shmdt(memory);
return(0);
}
Im getting a Segmentation fault (core dumped) and i dont know why.Also by searching i find a solution with struct but i dint get how i can build that.

You cannot have a int** point at a 2D array. It can only point to the first element in a 1D array of int*.
Furthermore, what's the logic of memory + (rows*columns)? You end up setting the pointer to the last item of the array, rather than the first.
Try this instead:
void* memory = shmat( ...
...
int (*array)[columns] = memory;
...
array[i][j] = ... ;
Where int (*array)[columns] is an array pointer, which ends up point at the first array in the 2D array.
For details, see Correctly allocating multi-dimensional arrays.

Related

Why do I get a segmentation fault by declaring a 2d array in c?

I am new to threads and I have a program that uses threads to find the minimum number out of a 2d array and later on, it finds the distance that the other elements of the array have from the minimum number and stores them in another array.
The user should enter the size of the array and the number of threads he wants to use.
I tried the program below for 1d array and it worked just fine. When I converted it to work for a 2d array it started crashing and throwing a segmentation fault. I, however, cannot find which part of the 2d declaration is wrong.
Any help is really appreciated.
Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>
#include <string.h>
#include <pthread.h>
struct Parameters
{
// input
int s,p; //n is size of array, p is number of threads
int** array; //array with elements
int start;
int end;
// output
int smallest;
int pos; //position if minimum
int** B; //array that holds the distances
};
void* min(void* args)
{
struct Parameters* p = (struct Parameters*)args;
int **array = p->array;
int **B1 = p->B;
int start = p->start;
int end = p->end;
int smallest = array[start][start];
int pos = p->pos;
int distance;
//find the smallest
for (int i = start; i < end; i++)
{
for(int j = start; j < end; j++)
{
if (array[i][j] < smallest)
{
smallest = array[i][j];
pos = i;
}
}
}
//find the distances
for(int i = 0; i < ((struct Parameters*)args) -> s; i++)
{
for(int j = 0; j < ((struct Parameters*)args) -> s; j++)
{
distance = abs(pos - i);
B1[i][j] = distance;
}
}
params->smallest = smallest;
params->B = B1;
return NULL;
}
int main()
{
int smallest,pos;
int s,p;
struct Parameters *ptr = (struct Parameters *)malloc(sizeof(struct Parameters));
if(ptr == NULL)
{
printf("Not enough. Try again \n");
exit(0);
}
printf("Type s\n");
scanf("%d",&(ptr->s));
printf("Type p\n");
scanf("%d", &(ptr->p));
// declare an array of threads and associated parameter instances
pthread_t threads[(ptr->p)];
struct Parameters thread_parameters[(ptr->p)] ;
int arr[ptr->s][ptr->s];
int B2[ptr->s][ptr->s];
// intialize the array
for(int i=0; i< ptr->s; i++)
{
for(int j=0; j< ptr->s; j++)
{
printf("Type a \n");
scanf("%d",&arr[i][j]);
}
}
// smallest needs to be set to something
smallest = arr[0][0];
// start all the threads
for (int i = 0; i < ptr->p; i++)
{
memcpy(arr, thread_parameters[i].array, sizeof(arr));
thread_parameters[i].s = ptr->s;
memcpy(Bb, thread_parameters[i].B, sizeof(B2));
thread_parameters[i].start = i * (ptr->s / ptr->p);
thread_parameters[i].end = (i+1) * (ptr->s / ptr->p);
pthread_create(&threads[i], NULL, min, &thread_parameters[i]);
}
// wait for all the threads to complete
for (int i = 0; i < ptr->p; i++)
{
pthread_join(threads[i], NULL);
}
// Now aggregate the "smallest" and "largest" results from all thread runs
for (int i = 0; i < ptr->p; i++)
{
if (thread_parameters[i].smallest < smallest)
{
smallest = thread_parameters[i].smallest;
}
}
printf("Smallest is %d\n", smallest);
thread_parameters[ptr->p].B[ptr->s][ptr->s];
for (int i = 0; i < 1; i++)
{
for(int j = 0; j < ptr->s;j++)
{
for(int k = 0; k < ptr->s; k++)
{
printf("Element %d is %d away from min\n",j,thread_parameters[i].B[j][k]);
}
}
}
return 0;
}
Thank you!!
The issue with your code might also come from :
memcpy(arr, thread_parameters[i].array, sizeof(arr));
...
memcpy(Bb, thread_parameters[i].B, sizeof(B2));
as thread_parameters[i].array and thread_parameters[i].B are not allocated, if you are only reading the array it might b fine to only pass them by address
thread_parameters[i].array = arr
but for thread_parameters[i].B you would need to allocate the arrays and perform a deep copy (memcpy would not work)
The below text does not answer the question but does provide some insight on VLA usage
One reason for causing the segmentation with a declaration of a Variable Length Array is that the value is to large to allocate the array on the stack (some compiler choose this option, this choice might have performance reason).
The is not much option to recover cleanly from failure to allocate memory on the stack as there is little way to clean up stack memory during runtime within the same stack context.
You can mitigate the issue by allocating your 2D arrays on the heap instead, some of the strategies are available here(thanks #Lundin) and here.
int** alloc_2d_int_array(size_t rows, size_t cols) {
int **result = malloc(rows * sizeof(int *));
if(result == NULL) {
// could not allocate more memory
return NULL;
}
size_t row_size = cols * sizeof(int);
for(int i=0; i < rows; ++i) {
result[i] = malloc(row_size);
if(result[i] == NULL) {
// could not allocate more memory
// cleanup
return NULL;
}
}
return result;
}
the above implementation have not been tested, but does compile, there are still risk of integer overflow.
Then use the above define function as following:
int **arr = alloc_2d_int_array(ptr->s, ptr->s);
int **B2 = alloc_2d_int_array(ptr->s, ptr->s);
easier implementation (see here(thanks #Lundin))
int **arr = malloc(sizeof(int[ptr->s][ptr->s]);
int **B2 = malloc(sizeof(int[ptr->s][ptr->s]);

Dynamically allocating space for a 2D array

I am a novice C programmer trying to write a function that dynamically allocates space for a 2D array. I am getting a segmentation fault when running this code & i'm not sure why.
#include <stdio.h>
#include <stdlib.h>
int allocate_space_2D_array(int **arr, int r, int c) {
int i,j;
arr = malloc(sizeof(int *) * r);
for (i = 0; i < r; i++)
arr[i] = malloc(sizeof(int *) * c);
for (i = 0; i < r; i++) {
for (j = 0; j < c; j++) {
printf("%p", arr[r][c]);
}
printf("\n");
}
return arr;
}
I expected to be able to print out and see the contiguous memory locations of each spot in the array, but I am never reaching that point in my code, because when I run it, i get a segmentation fault. Would appreciate any help.
Seeing your program i see 3 errors one while you allocate memory for 2D-array,one while you're printing and another one is how you declare the function.
First malloc is ok,the second one is wrong cause you already allocated memory for r(size of row) pointers so it's just like if you have * arr[r],so to allocate memory correctly now you should allocate memory just for int and not for int*.
Second error while printing you put as index for row and column the values r and c,but r and c are the size of matrix , as we know the size of an array or 2D-array goes from 0 to size-1,in your case goes from 0 to r-1 and from 0 to c-1.
Third error you should declare the function not as int but as int** cause you want to return a matrix so the return type is not int but int**.
I change your code to make it work correctly,it should be work.
int** allocate_space_2D_array(int **arr, int r, int c) {
int i,j;
arr = malloc(sizeof(int *) * r);
for (i = 0; i < r; i++)
arr[i] = malloc(sizeof(int ) * c);
for (i = 0; i < r; i++) {
for (j = 0; j < c; j++) {
printf("%p", arr[i][j]);
}
printf("\n");
}
return arr;
}

Allocating contiguous memory for a 3D array in C

I need to allocate contiguous space for a 3D array. (EDIT:) I GUESS I SHOULD HAVE MADE THIS CLEAR IN THE FIRST PLACE but in the actual production code, I will not know the dimensions of the array until run time. I provided them as constants in my toy code below just to keep things simple. I know the potential problems of insisting on contiguous space, but I just have to have it. I have seen how to do this for a 2D array, but apparently I don't understand how to extend the pattern to 3D. When I call the function to free up the memory, free_3d_arr, I get an error:
lowest lvl
mid lvl
a.out(2248,0x7fff72d37000) malloc: *** error for object 0x7fab1a403310: pointer being freed was not allocated
Would appreciate it if anyone could tell me what the fix is. Code is here:
#include <stdio.h>
#include <stdlib.h>
int ***calloc_3d_arr(int sizes[3]){
int ***a;
int i,j;
a = calloc(sizes[0],sizeof(int**));
a[0] = calloc(sizes[0]*sizes[1],sizeof(int*));
a[0][0] = calloc(sizes[0]*sizes[1]*sizes[2],sizeof(int));
for (j=0; j<sizes[0]; j++) {
a[j] = (int**)(a[0][0]+sizes[1]*sizes[2]*j);
for (i=0; i<sizes[1]; i++) {
a[j][i] = (int*)(a[j]) + sizes[2]*i;
}
}
return a;
}
void free_3d_arr(int ***arr) {
printf("lowest lvl\n");
free(arr[0][0]);
printf("mid lvl\n");
free(arr[0]); // <--- This is a problem line, apparently.
printf("highest lvl\n");
free(arr);
}
int main() {
int ***a;
int sz[] = {5,4,3};
int i,j,k;
a = calloc_3d_arr(sz);
// do stuff with a
free_3d_arr(a);
}
Since you are using C, I would suggest that you use real multidimensional arrays:
int (*a)[sz[1]][sz[2]] = calloc(sz[0], sizeof(*a));
This allocates contiguous storage for your 3D array. Note that the sizes can be dynamic since C99. You access this array exactly as you would with your pointer arrays:
for(int i = 0; i < sz[0]; i++) {
for(int j = 0; j < sz[1]; j++) {
for(int k = 0; k < sz[2]; k++) {
a[i][j][k] = 42;
}
}
}
However, there are no pointer arrays under the hood, the indexing is done by the magic of pointer arithmetic and array-pointer-decay. And since a single calloc() was used to allocate the thing, a single free() suffices to get rid of it:
free(a); //that's it.
You can do something like this:
int ***allocateLinearMemory(int x, int y, int z)
{
int *p = (int*) malloc(x * y * z * sizeof(int));
int ***q = (int***) malloc(x * sizeof(int**));
for (int i = 0; i < x; i++)
{
q[i] = (int**) malloc(y * sizeof(int*));
for (int j = 0; j < y; j++)
{
int idx = x*j + x*y*i;
q[i][j] = &p[idx];
}
}
return q;
}
void deallocateLinearMemory(int x, int ***q)
{
free(q[0][0]);
for(int i = 0; i < x; i++)
{
free(q[i]);
}
free(q);
}
I use it and works fine.

Segmentation fault in int matrix

I was experimenting some basic C code that defines an int matrix with pointers.
typedef int **Matrix;
Matrix createMatrix(int lines, int columns) {
int i, j;
Matrix m = (Matrix) malloc(sizeof(int) * lines * columns);
for (i = 0; i < lines; ++i) {
for (j = 0; j < columns; ++j) {
m[i][j] = 0;
}
}
return m;
}
int main(int argc, char**argv) {
Matrix m = createMatrix(5, 10);
// ...
if (m[2][3] == 20) {
// ...
}
return 0;
}
However, these m[i][j] accesses are throwing segmentation faults. What's wrong here? Too many asterisks?
I was convinced that a pointer to a pointer to an int was effectively the same as a matrix.
Your allocation of the Matrix data item assumes you're accessing it linearly with a single index. If you want to access it with two indices, e.g., m[1][1] you need to allocate each dimension:
Matrix m = malloc(sizeof(int *) * lines);
for ( int i = 0; i < lines; i++ )
m[i] = malloc(sizeof(int) * columns);
Note also that you should not type cast malloc.

segmentation fault for 2D arrays

I want to define a 2D array of very big size. But it is giving me segmentation fault?
#include <stdio.h>
int main () {
int i;
int temp[4000][5000];
for (i = 0; i < 5; i++)
{
printf ("Hello World\n");
}
}
Can anyone please suggest me some other way? Is there some problem with memory initialization? Thanks in advance
You can allocate the whole table in only one array but you won't be able to access array data with indices using two square brackets:
int * temp = malloc(4000*5000*sizeof(int));
to access the element (i,j) where previously you wrote temp[i][j], now you should now compute the index the following way:
temp[i*5000+j];
and do not forget to free the memory allocated for your table afterward:
free(temp);
int temp[4000][5000];
That's a VERY BIG array, way bigger than the normal size of stack, you get a segmentation fault because of stack overflow. Consider using dynamic allocation instead.
You need to use dynamic allocated arrays for such big arrays.
Try:
int* temp[4000];
for(i = 0; i < 4000; ++i) temp[i] = malloc(5000 * sizeof(int));
...
for(i = 0; i < 4000; ++i) free(temp[i]).
Whole program with error checking:
int main () {
int i, j;
int* temp[4000];
for (i = 0; i < 4000; ++i)
{
temp[i] = malloc(5000 * sizeof(int));
if (temp[i] == NULL)
{
for (j = 0; j < i; ++j) free(temp[i]);
exit(1);
}
}
for (i = 0; i < 5; i++)
{
printf ("Hello World\n");
}
for (i = 0; i < 4000; ++i) free(temp[i]);
}
Here you can find function which would use single malloc call to allocate two dimension array.
And simpler version of my own:
int main () {
int i, j;
int* temp[4000];
int* array = malloc(4000 * 5000 * sizeof(int));
if (malloc_tmp == NULL) exit(1);
for (i = 0; i < 4000; ++i)
{
temp[i] = array + (i * 5000);
}
for (i = 0; i < 5; i++)
{
printf ("Hello World\n");
}
free(temp[0]);
}

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