Segmentation fault in allocating a large array in C

Question

I intend to allocate a large array of size 1073741824 (equivalent to 1GB eg) and then read from it randomly but i get segmentation fault (core dumped) for only defining the array as i checked which i do like:

unsigned int size = 1073741824;
short arr = malloc(size * sizeof(short));

i also tried casting it as follows but still the same issue:

unsigned int size = 1073741824;
short *arr = (short*) malloc(size * sizeof(short));

also the ulimit command returns unlimited

so what am i doing wrong?

Answer 1

The likely cause of your problem is malloc() fails to allocate close to 2GB of memory on your system. You must check the return value: if malloc() fails, it returns a NULL pointer which cannot be dereferenced.

Furthermore, your first code fragment defines arr as a short instead of a short * .

Here is the typical code pattern you should be using:

    unsigned int size = 1073741824;
    short *arr = malloc(size * sizeof(short));
    if (arr == NULL) {
        fprintf("cannot allocate memory for %u shorts\n", size);
        exit(1);
    }

Answer 2

I think your problem is that the function malloc returns NULL due to the large size of the array. Using malloc directly is quite error prone. Therefor it's better to use a macro function like this:

#include <stdio.h>
#include <stdlib.h>

#define NEW_ARRAY(ptr, length) \
{ \
        (ptr) = malloc((size_t) (length) * sizeof (ptr)[0]); \
        if ((ptr) == NULL) { \
                fputs("error: Memory exhausted\n", stderr); \
                exit(EXIT_FAILURE); \
        } \
}


int main(void)
{
    unsigned int arrLen = 1073741824;
    short *arr;

    NEW_ARRAY(arr, arrLen);
    /*...*/
    free(arr);
    return 0;
}