为什么std :: condition_variable不能通过锁类型模板化？ [英] Why isn't std::condition_variable templated by lock type?
问题描述
在调试模式下具有断言能力是非常有用的，因为开销相当小，互斥锁是否被锁定。查看已知选项，我选择了实现此选项由于低开销，使用 std :: mutex 子类。
It's useful to have the ability to assert in debug mode, with reasonably small overhead, whether a mutex is locked. Viewing the known options, I've chosen to implement this using an std::mutex subclass due to the low overheads.
子类的接口是 std :: mutex 的超集，因此大多数事情都很好用它。例如， std :: unique_lock 模板化以利用具有特定接口的任何锁类型。
The interface of the subclass is a superset of that of std::mutex, and so most things work well with it. E.g., std::unique_lock is templated to utilize any lock type that has a specific interface.
问题出在 std :: condition_variable ，特别是等待成员，例如：
The problem is with std::condition_variable, in particular the wait members, e.g.:
template<class Predicate>
void wait(std::unique_lock<std::mutex> &lock, Predicate pred);
可以看出，该方法需要非常具体的 unique_lock / mutex 组合。不幸的是， Liskov原则不适用于 container< derived> ; $
As can be seen, the method requires a very specific unique_lock/mutex combination. Unfortunately, also, the Liskov principle doesn't extend for container<derived> being converted into container<base>.
我不明白 / p>
I don't understand
- 为什么会这样？
如果意图是强制使用 std :: unique_lock ，那么为什么不能使用以下内容：
Even if the intent was to enforce the use of std::unique_lock, then why couldn't the following be used:
template<class Predicate, class Lock=std::mutex>
void wait(std::unique_lock<Lock> &lock, Predicate pred);
- li>
- how to reasonably get around this?
编辑
并进一步指出@TC，绝对正确和非常简单的解决方案是使用 <$
As explained by @Lingxi, and further pointed out by @T.C, the absolutely correct and very simple solution here is to use condition_variable_any, which was designed for stuff like this.
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