Question

I was just going through glibc manual for description about posix_memalign function when I encountered this statement:

The address of a block returned by malloc or realloc in the GNU system is always a
multiple of eight (or sixteen on 64-bit systems). If you need a block whose address is a
multiple of a higher power of two than that, use memalign, posix_memalign, or valloc.

If I consider a simple structure containing just an int data member:

struct Mystruct
{
 int member;
};

Then I can see that Mystruct should be 4-byte aligned. But according to libc manual on a 64-bit architecture, dynamically allocating memory for such structure would return memory allocated on an address with 16 byte alignment.

Correct me if I am wrong. To me it seems like compiler uses natural alignment of a structure only for global/static/automatic variables (data, bss, stack). But on the other hand, to allocate the same structure on heap memory, the malloc call uses a predefined alignment(8 on 32-bit architectures and 16 on 64 bit architectures) ?

Answer 1

One thing worth remembering is that malloc() is nothing magical — it’s just a function. Likewise, the heap is just a bunch of memory that the compiler promises not to touch for its stuff. Allocating things statically (on the stack or in data/bss) is something that the compiler does do, so it can align them with whatever alignment is specified. malloc() (and the associated functions) are functions that manage the heap, but they are called at runtime. The compiler doesn’t know anything about malloc() except that it’s a function, and malloc() doesn’t know anything about the memory that you’re asking it to allocate except how much you’re asking for. Because 16-bit aligned is usually fine for most common uses, malloc() ensures this alignment to be safe.

PS: A fun (and eye-opening) exercise is to write your own malloc() implementation.