Aligned and unaligned memory accesses?

c++ c memory memory-alignment omap
Can Bal picture Can Bal · Jun 30, 2009 · Viewed 28.2k times · Source

What is the difference between aligned and unaligned memory access?

I work on an TMS320C64x DSP, and I want to use the intrinsic functions (C functions for assembly instructions) and it has 

ushort & _amem2(void *ptr);
ushort & _mem2(void *ptr);

where _amem2 does an aligned access of 2 bytes and _mem2 does unaligned access.

When should I use which?

Answer

Doug picture Doug · Jun 30, 2009

An aligned memory access means that the pointer (as an integer) is a multiple of a type-specific value called the alignment. The alignment is the natural address multiple where the type must be, or should be stored (e.g. for performance reasons) on a CPU. For example, a CPU might require that all two-byte loads or stores are done through addresses that are multiples of two. For small primitive types (under 4 bytes), the alignment is almost always the size of the type. For structs, the alignment is usually the maximum alignment of any member.

The C compiler always puts variables that you declare at addresses which satisfy the "correct" alignment. So if ptr points to e.g. a uint16_t variable, it will be aligned and you can use _amem2. You need to use _mem2 only if you are accessing e.g. a packed byte array received via I/O, or bytes in the middle of a string.

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