最奇怪的优化？ 在`libuv`中。 请解释

Question

The libuv contains next code in core.c:uv_run()

/* The if statement lets the compiler compile it to a conditional store.
 * Avoids dirtying a cache line.
 */
if (loop->stop_flag != 0)
    loop->stop_flag = 0;

What does this mean? Is it some kind of optimization? Why they did not simply assign 0?

Answer 1

Yes, just like the comment says. In case the flag is already 0, there is no need to write any data to the memory, thus avoiding a possible eviction of present data in the cache and replacing it with 0 for the flag. This will provide added value only in extremely time-critical applications.

Answer 2

I would argue this optimization is bad. For example, on gcc with -O3 it gives following code:

foo():
        movl    stop_flag(%rip), %eax
        testl   %eax, %eax
        je      .L3
        movl    $0, stop_flag(%rip)
.L3:
        ret
stop_flag:
        .zero   4

As you see, there is no conditional move, but a branch. And I am sure, branch misprediction is far worse than dirtying the cache line.