Why is writing to a 24-bit struct not atomic (when writing to a 32-bit struct appears to be)?

Question

I am a tinkerer—no doubt about that. For this reason (and very little beyond that), I recently did a little experiment to confirm my suspicion that writing to a struct is not an atomic operation, which means that a so-called "immutable" value type which attempts to enforce certain constraints could hypothetically fail at its goal.

I wrote a blog post about this using the following type as an illustration:

struct SolidStruct
{
    public SolidStruct(int value)
    {
        X = Y = Z = value;
    }

    public readonly int X;
    public readonly int Y;
    public readonly int Z;
}

While the above looks like a type for which it could never be true that X != Y or Y != Z, in fact this can happen if a value is "mid-assignment" at the same time it is copied to another location by a separate thread.

OK, big deal. A curiosity and little more. But then I had this hunch: my 64-bit CPU should actually be able to copy 64 bits atomically, right? So what if I got rid of Z and just stuck with X and Y? That's only 64 bits; it should be possible to overwrite those in one step.

Sure enough, it worked. (I realize some of you are probably furrowing your brows right now, thinking, Yeah, duh. How is this even interesting? Humor me.) Granted, I have no idea whether this is guaranteed or not given my system. I know next to nothing about registers, cache misses, etc. (I am literally just regurgitating terms I've heard without understanding their meaning); so this is all a black box to me at the moment.

The next thing I tried—again, just on a hunch—was a struct consisting of 32 bits using 2 short fields. This seemed to exhibit "atomic assignability" as well. But then I tried a 24-bit struct, using 3 byte fields: no go.

Suddenly the struct appeared to be susceptible to "mid-assignment" copies once again.

Down to 16 bits with 2 byte fields: atomic again!

Could someone explain to me why this is? I've heard of "bit packing", "cache line straddling", "alignment", etc.—but again, I don't really know what all that means, nor whether it's even relevant here. But I feel like I see a pattern, without being able to say exactly what it is; clarity would be greatly appreciated.

Answer 1

The pattern you're looking for is the native word size of the CPU.

Historically, the x86 family worked natively with 16-bit values (and before that, 8-bit values). For that reason, your CPU can handle these atomically: it's a single instruction to set these values.

As time progressed, the native element size increased to 32 bits, and later to 64 bits. In every case, an instruction was added to handle this specific amount of bits. However, for backwards compatibility, the old instructions were still kept around, so your 64-bit processor can work with all of the previous native sizes.

Since your struct elements are stored in contiguous memory (without padding, i.e. empty space), the runtime can exploit this knowledge to only execute that single instruction for elements of these sizes. Put simply, that creates the effect you're seeing, because the CPU can only execute one instruction at a time (although I'm not sure if true atomicity can be guaranteed on multi-core systems).

However, the native element size was never 24 bits. Consequently, there is no single instruction to write 24 bits, so multiple instructions are required for that, and you lose the atomicity.

Answer 2

x86 CPU operations take place in 8, 16, 32, or 64 bits; manipulating other sizes requires multiple operations.

Answer 3

The compiler and x86 CPU are going to be careful to move only exactly as many bytes as the structure defines. There are no x86 instructions that can move 24 bits in one operation, but there are single instruction moves for 8, 16, 32, and 64 bit data.

If you add another byte field to your 24 bit struct (making it a 32 bit struct), you should see your atomicity return.

Some compilers allow you to define padding on structs to make them behave like native register sized data. If you pad your 24 bit struct, the compiler will add another byte to "round up" the size to 32 bits so that the whole structure can be moved in one atomic instruction. The downside is your structure will always occupy 30% more space in memory.

Note that alignment of the structure in memory is also critical to atomicity. If a multibyte structure does not begin at an aligned address, it may span multiple cache lines in the CPU cache. Reading or writing this data will require multiple clock cycles and multiple read/writes even though the opcode is a single move instruction. So, even single instruction moves may not be atomic if the data is misaligned. x86 does guarantee atomicity for native sized read/writes on aligned boundaries, even in multicore systems.

It is possible to achieve memory atomicity with multi-step moves using the x86 LOCK prefix. However this should be avoided as it can be very expensive in multicore systems (LOCK not only blocks other cores from accessing memory, it also locks the system bus for the duration of the operation which can impact disk I/O and video operations. LOCK may also force the other cores to purge their local caches)

Answer 4

The C# standard (ISO 23270:2006, ECMA-334) has this to say regarding atomicity:

12.5 Atomicity of variable references Reads and writes of the following data types shall be atomic: bool, char, byte, sbyte, short, ushort, uint, int, float, and reference types. In addition, reads and writes of enum types with an underlying type in the previous list shall also be atomic. Reads and writes of other types, including long, ulong, double, and decimal, as well as user-defined types, need not be atomic. (emphasis mine) Aside from the library functions designed for that purpose, there is no guarantee of atomic read-modify-write, such as in the case of increment or decrement.

Your example X = Y = Z = value is short hand for 3 separate assignment operations, each of which is defined to be atomic by 12.5. The sequence of 3 operations (assign value to Z, assign Z to Y, assign Y to X) is not guaranteed to be atomic.

Since the language specification doesn't mandate atomicity, while X = Y = Z = value; might be an atomic operation, whether it is or not is dependent on a whole bunch of factors:

• the whims of the compiler writers
• what code generation optimizations options, if any, were selected at build time
• the details of the JIT compiler responsible for turning the assembly's IL into machine language. Identical IL run under Mono, say, might exhibit different behaviour than when run under .Net 4.0 (and that might even differ from earlier versions of .Net).
• the particular CPU on which the assembly is running.

One might also note that even a single machine instruction is not necessarily warranted to be an atomic operation—many are interruptable.

Further, visiting the CLI standard (ISO 23217:2006), we find section 12.6.6:

12.6.6 Atomic reads and writes A conforming CLI shall guarantee that read and write access to properly aligned memory locations no larger than the native word size (the size of type native int) is atomic (see §12.6.2) when all the write accesses to a location are the same size. Atomic writes shall alter no bits other than those written. Unless explicit layout control (see Partition II (Controlling Instance Layout)) is used to alter the default behavior, data elements no larger than the natural word size (the size of a native int) shall be properly aligned. Object references shall be treated as though they are stored in the native word size.

[Note: There is no guarantee about atomic update (read-modify-write) of memory, except for methods provided for that purpose as part of the class library (see Partition IV). (emphasis mine) An atomic write of a “small data item” (an item no larger than the native word size) is required to do an atomic read/modify/write on hardware that does not support direct writes to small data items. end note]

[Note: There is no guaranteed atomic access to 8-byte data when the size of a native int is 32 bits even though some implementations might perform atomic operations when the data is aligned on an 8-byte boundary. end note]