为什么嵌套锁不会导致死锁？

发布于 2024-10-17 16:47:25 字数 203 浏览 4 评论 0原文

为什么这段代码不会导致死锁？

   private static readonly object a = new object();

...

   lock(a)
   {
      lock(a)
      {
         ....
      }
   }

原文

Why does this code not cause a deadlock?

   private static readonly object a = new object();

...

   lock(a)
   {
      lock(a)
      {
         ....
      }
   }

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浅沫记忆 2024-10-24 16:47:25

如果线程已经持有锁，那么它可以毫无问题地再次“获取该锁”。

至于为什么（以及为什么这是一个好主意），请考虑以下情况，其中我们在 a -> 程序的其他地方定义了锁定顺序。 b：

void f()
{
    lock(a)
    { /* do stuff inside a */ }
}

void doStuff()
{
    lock(b)
    {
        //do stuff inside b, that involves leaving b in an inconsistent state
        f();
        //do more stuff inside b so that its consistent again
    }
}

哎呀，我们刚刚违反了我们的锁定顺序，并且我们手上有一个潜在的死锁。

我们确实需要能够执行以下操作：

function doStuff()
{
    lock(a)
    lock(b)
    {
        //do stuff inside b, that involves leaving b in an inconsistent state
        f();
        //do more stuff inside b so that its consistent again
    }
}

以便在调用f()时保持我们的锁顺序，而不会出现自死锁。

If a thread already holds a lock, then it can "take that lock" again without issue.

As to why that is, (and why it's a good idea), consider the following situation, where we have a defined lock ordering elsewhere in the program of a -> b:

void f()
{
    lock(a)
    { /* do stuff inside a */ }
}

void doStuff()
{
    lock(b)
    {
        //do stuff inside b, that involves leaving b in an inconsistent state
        f();
        //do more stuff inside b so that its consistent again
    }
}

Whoops, we just violated our lock ordering and have a potential deadlock on our hands.

We really need to be able to do the following:

function doStuff()
{
    lock(a)
    lock(b)
    {
        //do stuff inside b, that involves leaving b in an inconsistent state
        f();
        //do more stuff inside b so that its consistent again
    }
}

So that our lock ordering is maintained, without self-deadlocking when we call f().

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