Align the start of an array in dynamic memory in C

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Aligned memory management?

I have an array which I am declaring like this

int * myarray;
int num_of_element;

myarry = (int*) calloc(num_of_elements, sizeof(int));

The size of an int is 4 bytes however I want to ensure that my array starts on an 8 byte boundary - so I can efficiently load two values each time. Is there a different way or something else I can do?

score 2 · Answer 1 · edited May 23 '17 at 11:44

2

There are several ways.

Over-allocate, and round up the returned address to the desired alignment.
Define a union with a type whose alignment is the desired one, generally double. This requires low-level knowledge but is not optimal when you want to allocate an array.

Also, you shouldn't cast the return value of malloc() in C.

edited May 23 '17 at 11:44

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answered Sep 18 '12 at 14:05

unwind

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Doesn't calloc already ensure the pointer returned is aligned with every possible type, including that double? – Piotr Kalinowski Sep 18 '12 at 14:35
Yes, at least on BSD and Linux. @PiotrKalinowski – alk Sep 18 '12 at 16:42

score 2 · Answer 2 · answered Sep 18 '12 at 14:06

Did you tried posix_memalign ?

The function posix_memalign() allocates size bytes and places the address of the allocated memory in *memptr. The address of the allocated memory will be a multiple of alignment, which must be a power of two and a multiple of sizeof(void *).

Example:

if (posix_memalign(&myarray, 8, num_of_elements*sizeof(int)) != 0) {
    // failed
}

See http://pubs.opengroup.org/onlinepubs/009696699/functions/posix_memalign.html

Linux posix_memalign manpage also documents aligned_alloc() (c11), memalign() (obsolete) :

https://www.kernel.org/doc/man-pages/online/pages/man3/posix_memalign.3.html

Align the start of an array in dynamic memory in C

2 Answers2