Question

On Linux, with 16 GB of RAM, why would the following segfault:

#include <stdlib.h>

#define N 44000

int main(void) {
    long width = N*2 - 1;
    int * c = (int *) calloc(width*N, sizeof(int));
    c[N/2] = 1;
    return 0;
}

According to GDB the problem is from c[N/2] = 1 , but what is the reason?

Answer 1

You’re allocating around 14-15 GB memory, and for whatever reason the allocator cannot
give you that much at the moment- thus calloc returns NULL and you segfault as you’re dereferencing a NULL pointer.

Check if calloc returns NULL.

That’s assuming you’re compiling a 64-bit program under a 64-bit Linux. If you’re doing something else – you might overflow the calculation to the first argument to calloc if a long is not 64 bits on your system.

For example, try

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

#define N    44000L

int main(void)
{
    size_t width = N * 2 - 1;
    printf("Longs are %lu bytes. About to allocate %lu bytes\n",
           sizeof(long), width * N * sizeof(int));
    int *c = calloc(width * N, sizeof(int));
    if (c == NULL) {
        perror("calloc");
        return 1;
    }
    c[N / 2] = 1;
    return 0;
}