The ACCESS_ONCE() macro (2012)

The ACCESS_ONCE() macro (2012)(lwn.net)

13 points by gaoprea 11 years ago | 11 comments

kazinator 11 years ago |

It has a confusing name, suggesting that it is intended to prevent multiple accesses. (Contrast with `pthread_once` for instance: a mechanism for once-only initialization.)

In fact, the very example given in the article shows that without the ACCESS_ONCE, the compiler moved the access outside of the loop---thereby ensuring that it's accessed just once, ironically!!! Whereas by using ACCESS_ONCE, we ensure that it's accessed as many times as the loop is iterated, not only once.

Basically this just does some kind of volatile access, and so the name should reflect that:

  foo = VOLATILE_READ(bar);

Then the number of times it is accessed is implied from the agreement between abstract and actual semantics. In the abstract semantics, and expression is evaluated once for each invocation of whatever encloses it. If an expression is unconditionally evaluated in a loop that iterates 100 times, then it is evaluated 100 times. Hoisting it out of the loop would be an optimization which is forbidden by agreement with the abstract semantics.

rdc12 11 years ago | |

"In fact, the very example given in the article shows that without the ACCESS_ONCE, the compiler moved the access outside of the loop---thereby ensuring that it's accessed just once, ironically!!! Whereas by using ACCESS_ONCE, we ensure that it's accessed as many times as the loop is iterated, not only once."

But that is just a consequence of putting a call to ACCESS_ONCE inside a loop. But it is retaining the property that it is only read from memory once per iteration.

It would be way more confusing if the programmer only wanted read once for the whole loop and still put the statement inside the loop.

I do think that knowing that it is being called in a loop context, implies that it should only be accessed once per iteration

kazinator 11 years ago | | |

But, say:

   for (;;) {
      /* ... */
      if (pthread_once(&once_var, once_routine))
        /* error */;
      /* ... */
   }

Here, we are ensuring that once_routine is called exactly once.

ACCESS_ONCE to me strongly suggests that the value is accessed once and cached forever. E.g. ACCESS_ONCE(x) could expand to something like:

    ({
      static __flag;
      static typeof(x) __cached;
      if (!__flag) {
        __cached = (x);
        __flag = 1;
      }
      __cached;
    })

ANSI Common Lisp has load-time-value. You can use this anywhere:

   (defun myfunc ()
     (let ((foo (bar))
        ... ;; deeply nested
        (let ((x (load-time-value (whatever))))
          ...))))

The load-time-value form is evaluated when the module is loaded and the value is stashed. Then whenever myfunc is called and that code is evaluated, the previously stashed value is retrieved; the (whatever) is not evaluated any more.

cperciva 11 years ago |

The silly thing here is that this macro is not guaranteed to accomplish anything. The C standard limits how compilers can optimize accesses to volatile objects, but says nothing about accesses to non-volatile objects which are accessed via volatile-qualified pointers.

cnvogel 11 years ago | |

That's certainly true if you mix accesses to the same object via volatile and non-volatile pointers. It's even explicitly undefined in the standard. (C99 §6.7.3/5). [while in practice it will work most of the time]

But if you always access a certain object through casts to ((volatile )&x), I don't see how this should be different than accesses to a globally declared "volatile x" variable, as pointers are guaranteed to not change if casted hence-and-forth (§6.3.2.3/7).

cperciva 11 years ago | | |

The key text is in C99 §6.7.4: "An object that has volatile-qualified type...". Note that this does not say "An object accessed via a pointer to a volatile-qualified type": What matters is the type of the underlying object, i.e., how it is originally defined.

webaholic 11 years ago | |

It is not in the C standard but the way GCC implements it is that it is hands-off with optimizations to any volatile variables and pointers. And since Linux is only compiled with GCC, it works.

nwmcsween 11 years ago | | |

Linux is also compiled with llvm as well as icc