segmentation fault for 2D arrays

Question

I want to define a 2D array of very big size. But it is giving me segmentation fault?

  #include 

  int main () {
       int i;
       int te         


        

Answer 1

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);

Answer 2

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.

Answer 3

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]);
}