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单一进程中的僵局?
原标题:deadlock in a single process -?

单一程序/校对试图获得同一顶顶垫时的状态是否被称作“lock锁”?

mutex_lock(&mutex) - success
...
mutex_lock(&mutex) - blocked, mutex not released

Seems as condition of "nonshearable resource" doesn t apply: here is only single process/thread so that no case of nonshearable resource hold and wait: here is only single process so that no wait for another process to release the requested resource

最佳回答

是的,这是考虑的,或者有可能导致僵局。

该词不一定要经过深层或过程——但requests

浏览你的24小时(mutex>)为not,在变换时将你的资源用于锁定,从多种来源转移资源是一种错误。 当/如果您的单一阅读过程试图在突变和要求锁定的不宽容锁锁内进行数据的新变换时,会发生什么情况? 僵局

可能的话,因为你没有确切说明你处理哪类mut子。

问题回答

无,它不会死锁。

<><><><><><><>>><>>><>可同时出现四种条件。

  • Mutual exclusion: The resources involved must be unshareable; otherwise, the processes would not be prevented from using the resource when necessary. Only one process can use the resource at any given instant of time.
  • Hold and wait or resource holding: a process is currently holding at least one resource and requesting additional resources which are being held by other processes.
  • No preemption: a resource can be released only voluntarily by the process holding it.
  • Circular wait: each process must be waiting for a resource which is being held by another process, which in turn is waiting for the first process to release the resource. In general, there is a set of waiting processes, P = {P1, P2, …, PN}, such that P1 is waiting for a resource held by P2, P2 is waiting for a resource held by P3 and so on until PN is waiting for a resource held by P1.

https://en.wikipedia.org/wiki/Deadlock”rel=“nofollow noreferer”>more.

With a single process, the Circular Wait condition can never be met, thus deadlock can never arise with a single process.





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