CP.43: Minimize time spent in a critical section
CP.43:尽量减少花费在临界区中的时间
Reason（原因）
The less time is spent with a mutex taken, the less chance that another thread has to wait, and thread suspension and resumption are expensive.
获得mutex锁之后花费的时间越短，其他线程需要等待的机会就越小。线程阻塞和唤醒的代价太高了。
Example（示例）
代码语言：javascript复制void do_something() // bad
{
unique_lock<mutex> lck(my_lock);
do0(); // preparation: does not need lock
do1(); // transaction: needs locking
do2(); // cleanup: does not need locking
}
Here, we are holding the lock for longer than necessary: We should not have taken the lock before we needed it and should have released it again before starting the cleanup. We could rewrite this to
这里，我们保持锁定的时间超出必要的限度了：我们不应该在不需要的时候获取锁，另一方面，应该在开始清理之前就释放锁。我们可以这样重写代码：
代码语言：javascript复制void do_something() // bad
{
do0(); // preparation: does not need lock
my_lock.lock();
do1(); // transaction: needs locking
my_lock.unlock();
do2(); // cleanup: does not need locking
}
But that compromises safety and violates the use RAII rule. Instead, add a block for the critical section:
但是这种做法在安全方面进行了妥协，还违反了RAII准则。作为改善，可以为临界区增加一个代码块：
代码语言：javascript复制void do_something() // OK
{
do0(); // preparation: does not need lock
{
unique_lock<mutex> lck(my_lock);
do1(); // transaction: needs locking
}
do2(); // cleanup: does not need locking
}Enforcement（实施建议）
Impossible in general. Flag "naked" lock() and unlock().
一般情况下不可能。标记暴露的lock和unlock操作。
原文链接
https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGuidelines.md#cp43-minimize-time-spent-in-a-critical-section
