What is aligned memory allocation?

Alignment requirements specify what address offsets can be assigned to what types. This is completely implementation-dependent, but is generally based on word size. For instance, some 32-bit architectures require all int variables start on a multiple of four. On some architectures, alignment requirements are absolute. On others (e.g. x86) flouting them only comes with a performance penalty.

malloc is required to return an address suitable for any alignment requirement. In other words, the returned address can be assigned to a pointer of any type. From C99 §7.20.3 (Memory management functions):

The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated).

Suppose that you have the structure.

struct S {
    short a;
    int b;
    char c, d;
};

Without alignment, it would be laid out in memory like this (assuming a 32-bit architecture):

 0 1 2 3 4 5 6 7
|a|a|b|b|b|b|c|d|  bytes
|       |       |  words

The problem is that on some CPU architectures, the instruction to load a 4-byte integer from memory only works on word boundaries. So your program would have to fetch each half of b with separate instructions.

But if the memory was laid out as:

 0 1 2 3 4 5 6 7 8 9 A B
|a|a| | |b|b|b|b|c|d| | |
|       |       |       |

Then access to b becomes straightforward. (The disadvantage is that more memory is required, because of the padding bytes.)

Different data types have different alignment requirements. It's common for char to be 1-byte aligned, short to be 2-byte aligned, and 4-byte types (int, float, and pointers on 32-bit systems) to be 4-byte aligned.

malloc is required by the C standard to return a pointer that's properly aligned for any data type.

glibc malloc on x86-64 returns 16-byte-aligned pointers.