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Home/ Questions/Q 7508549
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T22:41:32+00:00 2026-05-29T22:41:32+00:00

Recently I’ve peeked into the Linux kernel implementation of an atomic read and write

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Recently I’ve peeked into the Linux kernel implementation of an atomic read and write and a few questions came up.

First the relevant code from the ia64 architecture:

typedef struct {
    int counter;
} atomic_t;

#define atomic_read(v)      (*(volatile int *)&(v)->counter)
#define atomic64_read(v)    (*(volatile long *)&(v)->counter)

#define atomic_set(v,i)     (((v)->counter) = (i))
#define atomic64_set(v,i)   (((v)->counter) = (i))
  1. For both read and write operations, it seems that the direct approach was taken to read from or write to the variable. Unless there is another trick somewhere, I do not understand what guarantees exist that this operation will be atomic in the assembly domain. I guess an obvious answer will be that such an operation translates to one assembly opcode, but even so, how is that guaranteed when taking into account the different memory cache levels (or other optimizations)?

  2. On the read macros, the volatile type is used in a casting trick. Anyone has a clue how this affects the atomicity here? (Note that it is not used in the write operation)

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  1. Editorial Team
    Editorial Team
    2026-05-29T22:41:33+00:00Added an answer on May 29, 2026 at 10:41 pm

    I think you are misunderstanding the (very much vague) usage of the word “atomic” and “volatile” here. Atomic only really means that the words will be read or written atomically (in one step, and guaranteeing that the contents of this memory position will always be one write or the other, and not something in between). And the volatile keyword tells the compiler to never assume the data in that location due to an earlier read/write (basically, never optimize away the read).

    What the words “atomic” and “volatile” do NOT mean here is that there’s any form of memory synchronization. Neither implies ANY read/write barriers or fences. Nothing is guaranteed with regards to memory and cache coherence. These functions are basically atomic only at the software level, and the hardware can optimize/lie however it deems fit.

    Now as to why simply reading is enough: the memory models for each architecture are different. Many architectures can guarantee atomic reads or writes for data aligned to a certain byte offset, or x words in length, etc. and vary from CPU to CPU. The Linux kernel contains many defines for the different architectures that let it do without any atomic calls (CMPXCHG, basically) on platforms that guarantee (sometimes even only in practice even if in reality their spec says the don’t actually guarantee) atomic reads/writes.

    As for the volatile, while there is no need for it in general unless you’re accessing memory-mapped IO, it all depends on when/where/why the atomic_read and atomic_write macros are being called. Many compilers will (though it is not set in the C spec) generate memory barriers/fences for volatile variables (GCC, off the top of my head, is one. MSVC does for sure.). While this would normally mean that all reads/writes to this variable are now officially exempt from just about any compiler optimizations, in this case by creating a “virtual” volatile variable only this particular instance of a read/write is off-limits for optimization and re-ordering.

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