用boost :: lock_guard进行线程饥饿

Question

我正在尝试学习Boost库，并正在研究boost :: thread的示例。

下面的示例说明了boost :: lock_guard用于线程同步的用法，确保对std :: cout的访问不是并发的：

#include <boost/thread.hpp>
#include <boost/format.hpp>
#include <iostream>

void wait(const int secs) {
    boost::this_thread::sleep(boost::posix_time::seconds(secs));
}

boost::mutex mutex;

void thread1() {
    for (int i = 0; i < 10; ++i) {
        wait(1); // <-- all works fine if wait is placed here
        boost::lock_guard<boost::mutex> lock(mutex);
        std::cout << boost::format("thread A here %d\n") % i ;
    }
}

void thread2() {
    for (int i = 0; i < 10; ++i) {
        wait(1); //  <-- all works fine if wait is placed here
        boost::lock_guard<boost::mutex> lock(mutex);
        std::cout << boost::format("thread B here %d\n") % i;
    }

}

int main() {
    boost::thread t1(thread1);
    boost::thread t2(thread2);
    t1.join();
    t2.join();
}

结果几乎是人们所期望的，即通过打印的两个线程交替显示消息：

thread A here 0
thread B here 0
thread A here 1
thread B here 1
thread A here 2
thread B here 2
thread A here 3
thread B here 3
thread A here 4
thread B here 4
...

然而，一个小的修改-移动锁定挡板的范围内等待呼叫-导致了一个惊喜：

void thread1() {
    for (int i = 0; i < 10; ++i) {
        boost::lock_guard<boost::mutex> lock(mutex);
        wait(1); // <== !
        std::cout << boost::format("thread A here %d\n") % i ;
    }
}

void thread2() {
    for (int i = 0; i < 10; ++i) {
        boost::lock_guard<boost::mutex> lock(mutex);
        wait(1);  // <== !
        std::cout << boost::format("thread B here %d\n") % i;
    }

现在，thead1或thread2赢得互斥量的初始“竞争”，然后在每次循环迭代中一次又一次获胜，从而使另一个线程饿死！

输出示例：

thread B here 0
thread B here 1
thread B here 2
thread B here 3
thread B here 4
thread B here 5
thread B here 6
thread B here 7
thread B here 8
thread B here 9
thread A here 0
thread A here 1
thread A here 2
thread A here 3
thread A here 4
thread A here 5
thread A here 6
thread A here 7
thread A here 8
thread A here 9

谁能解释为什么会这样吗？

Answer 1

这是因为在获取锁定后，等待调用导致第二个线程开始执行。 由于第二个线程无法获取锁定，因此它将进入等待状态，直到锁定可用为止。 在您的情况下，只有在第一个线程完成循环之前，锁才变得可用。