Question

I am dynamically allocating memory as follows:

char* heap_start1 = (char*) malloc(1);
char* heap_start2 = (char*) malloc(1);

When I do printf as follows surprisingly the addresses are not consecutives.

printf("%p, %p \n",heap_start1,heap_start2);

Result:

   0x8246008, 0x8246018

As you can see there is a 15 bytes of extra memory that are left defragmented. It’s definitely not because of word alignment. Any idea behind this peculiar alignment?

Thanks in advance!

I am using gcc in linux if that matters.

Answer 1

glibc’s malloc, for small memory allocations less than 16 bytes, simply allocates the memory as 16 bytes. This is to prevent external fragmentation upon the freeing of this memory, where blocks of free memory are too small to be used in the general case to fulfill new malloc operations.

A block allocated by malloc must also be large enough to store the data required to track it in the data structure which stores free blocks.

This behaviour, while increasing internal fragmentation, decreases overall fragmentation throughout the system.

Source:
http://repo.or.cz/w/glibc.git/blob/HEAD:/malloc/malloc.c
(Read line 108 in particular)

/*
...
Minimum allocated size: 4-byte ptrs:  16 bytes    (including 4 overhead)
...
*/

Furthermore, all addresses returned by the malloc call in glibc are aligned to: 2 * sizeof(size_t) bytes. Which is 64 bits for 32-bit systems (such as yours) and 128 bits for 64-bit systems.