c++ - 如何使用condition_variable真正地等待_不超过特定持续时间

作为it turns out，condition_variable::wait_for应该真正称为condition_variable::wait_for_or_possibly_indefinitely_longer_than，因为它需要在真正超时和返回之前重新获取锁。

请参见this program进行演示。

有没有一种表达方式，“看，我真的只有两个秒。如果当时myPredicate()仍然为false和/或锁仍处于锁定状态，我不在乎，只需继续进行，然后给我一个的方法。”

就像是:

bool myPredicate();
auto sec = std::chrono::seconds(1);

bool pred;
std::condition_variable::cv_status timedOut;

std::tie( pred, timedOut ) =
    cv.really_wait_for_no_longer_than( lck, 2*sec, myPredicate );

if( lck.owns_lock() ) {
    // Can use mutexed resource.
    // ...
    lck.unlock();
} else {
    // Cannot use mutexed resource. Deal with it.
};

最佳答案

我认为您滥用了condition_variable的锁。它仅用于保护状态，而不是用于保护耗时的工作。

通过将mutex分为两部分，可以轻松解决您的示例-一个用于关键部分，另一个用于保护ready条件的修改。这是修改后的片段:

typedef std::unique_lock<std::mutex> lock_type;
auto sec = std::chrono::seconds(1);
std::mutex mtx_work;
std::mutex mtx_ready;
std::condition_variable cv;
bool ready = false;

void task1() {
    log("Starting task 1. Waiting on cv for 2 secs.");
    lock_type lck(mtx_ready);
    bool done = cv.wait_for(lck, 2*sec, []{log("Checking condition..."); return ready;});
    std::stringstream ss;
    ss << "Task 1 finished, done==" << (done?"true":"false") << ", " << (lck.owns_lock()?"lock owned":"lock not owned");
    log(ss.str());
}

void task2() {
    // Allow task1 to go first
    std::this_thread::sleep_for(1*sec);
    log("Starting task 2. Locking and sleeping 2 secs.");
    lock_type lck1(mtx_work);
    std::this_thread::sleep_for(2*sec);
    lock_type lck2(mtx_ready);
    ready = true; // This happens around 3s into the program
    log("OK, task 2 unlocking...");
    lck2.unlock();
    cv.notify_one();
}

输出:

@2 ms: Starting task 1. Waiting on cv for 2 secs.
@2 ms: Checking condition...
@1002 ms: Starting task 2. Locking and sleeping 2 secs.
@2002 ms: Checking condition...
@2002 ms: Task 1 finished, done==false, lock owned
@3002 ms: OK, task 2 unlocking...