Question

I was working on an assignment (details in another question). As part of it, I was increasing the size of the array. And found that when I try to initialize an array:

int arr[2097152]; // 8MB

I got segmentation fault … I think its because I am trying to declare an array that too large? Then I found the way to get around this is to use malloc. But being new to C (mainly use JavaScript/Python/Java …). I get very confused with pointers and stuff …

I’ve declared an array 8MB:

int *arr = malloc (MBs * 1024 * 1024 / sizeof(int)); // MBs = 8

But now … how do I access it, or write to it? When I use it like arr am I getting the address, and if I use *arr I get the 1st element?

Answer 1

Use it like arr[index], just as if it were declared as an array. In C, the notation x[y] is exactly equivalent to *(x + y). This works in the case of an array because the array name is converted to a pointer to its first element.

int *arr = malloc (MBs * 1024 * 1024 / sizeof(int));

This is not a good approach (and doesn’t make it the size you want), because you don’t have the number of elements handy. You should declare it based on the number of elements, e.g.,

#define ARR_LENGTH 2097152
int *arr = malloc (ARR_LENGTH * sizeof *arr);

You need to multiply by the element size because malloc’s argument is a byte count.