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futex 手册页提供了一个简单的 演示，但我无法得到页面描述的结果，结果似乎是我的机器上的死锁（linux 5.2.1）；父进程不会被其子进程唤醒。手册页错了吗？

我的机器上的输出示例：

[root@archlinux ~]# ./a.out
Child  (12877) 0
Parent (12876) 0
Child  (12877) 1
// block here

我的系统：

[root@archlinux ~]# uname -a
Linux archlinux 5.2.1-arch1-1-ARCH #1 SMP PREEMPT Sun Jul 14 14:52:52 UTC 2019 x86_64 GNU/Linux
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1 回答 1

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实际上，手册页中的示例是错误的，代码在两个地方偏离了相应注释中的正确描述。

       /* Acquire the futex pointed to by 'futexp': wait for its value to
          become 1, and then set the value to 0. */

       static void
       fwait(int *futexp)
       {
           int s;

           /* atomic_compare_exchange_strong(ptr, oldval, newval)
              atomically performs the equivalent of:

                  if (*ptr == *oldval)
                      *ptr = newval;

              It returns true if the test yielded true and *ptr was updated. */

           while (1) {

               /* Is the futex available? */
               const int zero = 0;
               if (atomic_compare_exchange_strong(futexp, &zero, 1))
                   break;      /* Yes */

必须反过来：

               /* Is the futex available? */
               if (atomic_compare_exchange_strong(futexp, &(int){1}, 0))
                   break;      /* Yes */

       /* Release the futex pointed to by 'futexp': if the futex currently
          has the value 0, set its value to 1 and the wake any futex waiters,
          so that if the peer is blocked in fpost(), it can proceed. */

       static void
       fpost(int *futexp)
       {
           int s;

           /* atomic_compare_exchange_strong() was described in comments above */

           const int one = 1;
           if (atomic_compare_exchange_strong(futexp, &one, 0)) {
               …

必须反过来：

           if (atomic_compare_exchange_strong(futexp, &(int){0}, 1)) {
               …
于 2020-01-07T14:50:22.090 回答