c - gcc memory alignment using malloc

Question

I've the following struct:

#define M 3

#pragma pack(push)
#pragma pack(1)
struct my_btree_node {
    struct my_btree_node *pointers[M];
    unsigned char *keys[M - 1];
    int data[M - 1];
    unsigned char number_of_keys;
};
#pragma pack(pop)

The sizeof(struct my_btree_node) function returns a value of 49 byte for this struct. Does allocating memory for this struct using malloc return a 64 byte block because on 64 bit systems pointers are 16-byte-aligned or will it indeed be 49 bytes?

Is there a way to align memory with a smaller power of two than 16 and is it possible to get the true size of the allocated memory inside the application?

I want to reduce the number of padding bytes in order to save memory. My applications allocates millions of those structs and I do not want to waste memory.

Answer 1

malloc(3) is defined to

The malloc() and calloc() functions return a pointer to the allocated memory, which is suitably aligned for any built-in type. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.

So a conforming implementation has to return a pointer aligned to the largest possible machine alignment (with GCC, it is the macro __BIGGEST_ALIGNMENT__)

If you want less, implement your own allocation routine. You could for example allocate a large array of char and do your allocation inside it. That would be painful, perhaps slower (processors dislike unaligned data, e.g. because of CPU cache constraints), and probably not worthwhile (current computers have several gigabytes of RAM, so a few millions of hundred-byte sized data chunks is not a big deal).

BTW, malloc is practically implemented in the C standard library (but -on Linux at least- the compiler knows about it, thanks to __attribute__-s in GNU glibc headers; so some internal optimizations inside GCC know and take care of calls to malloc).