Why are address are not consecutive when allocating single bytes?

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.

At least three possible reasons:

• malloc needs to produce memory that is suitably-aligned for all primitive types. Data for SSE instructions needs to be 128-bit aligned. (There may also be other 128-bit primitive types that your platform supports that don't occur to me at the moment.)

• A typical implementation of malloc involves "over-allocation" in order to store bookkeeping information for a speedy free. Not sure if GCC on Linux does this.

• It may be allocating guard bytes in order to allow detection of buffer overflows and so on.

malloc guarantees that returned memory is properly aligned for any basic type. Moreover, memory block could be padded with some guard bytes to check for memory corruption, it depends on settings.