Question

所以我有一些shared_mutex并完成了这个：

        boost::upgrade_lock<boost::shared_mutex> lock(f->mutex);
        boost::upgrade_to_unique_lock<boost::shared_mutex> uniqueLock(lock);

现在我想“解锁”或者至少将其降级为：

boost::shared_lock<boost::shared_mutex> lock_r(f->mutex);

怎么做这样的事情？有可能吗？

Answer 1

如果您让upgrade_to_unique_lock超出范围，它将自动降级回升级所有权。

例如

void foo() {
   boost::upgrade_lock<boost::shared_mutex> lock(f->mutex);

   // Do shared operations, as mutex is held upgradeable
   // ...

   if(need_to_get_unique)
   {
      boost::upgrade_to_unique_lock<boost::shared_mutex> uniqueLock(lock); 

      // Do exclusive operations, as mutex is held uniquely
      // ... 
      // At end of scope unique is released back to upgradeable
   }
   // Only shared operations here, as it's only held upgradeable
   // ...

   // At end of scope mutex is completely released
}

编辑：另一件事。如果给定函数仅需要排他锁，则可以使用boost::unique_lock并唯一锁定，而无需同时执行upgrade和upgrade_to_unique锁定。

如何解锁boost :: upgrade_to_unique_lock（由boost :: shared_mutex制作）？

1 个答案: