如何使用condition_variable真正wait_for不超过一定的持续时间

Question

正如事实证明,condition_variable::wait_for的确可以称为condition_variable::wait_for_or_possibly_indefinitely_longer_than,因为它需要之前确实超时并返回到重新获取锁.

请参阅此程序以进行演示.

有没有办法表达,"看,我真的只有两秒钟.如果myPredicate()当时仍然是假的和/或锁仍然锁定,我不在乎,只是随身携带,并给我一个方法来检测那."

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.
};

Answer 1

我认为你误用了condition_variable锁.它仅用于保护条件,而不是用于保护耗时的工作.

您的示例可以通过将其拆分mutex为两个来轻松修复- 一个用于临界区,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();
}

它的输出:

@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...