模拟死锁产生代码cpp

以下是一个简单的模拟死锁产生的代码示例：

#include <iostream>
#include <thread>
#include <mutex>

std::mutex mutex1, mutex2;

void thread1() {
    std::unique_lock<std::mutex> lock1(mutex1);
    std::cout << "Thread 1 has acquired mutex 1" << std::endl;
    std::this_thread::sleep_for(std::chrono::milliseconds(1000));

    std::unique_lock<std::mutex> lock2(mutex2);
    std::cout << "Thread 1 has acquired mutex 2" << std::endl;

    lock2.unlock();
    lock1.unlock();
}

void thread2() {
    std::unique_lock<std::mutex> lock2(mutex2);
    std::cout << "Thread 2 has acquired mutex 2" << std::endl;
    std::this_thread::sleep_for(std::chrono::milliseconds(1000));

    std::unique_lock<std::mutex> lock1(mutex1);
    std::cout << "Thread 2 has acquired mutex 1" << std::endl;

    lock1.unlock();
    lock2.unlock();
}

int main() {
    std::thread t1(thread1);
    std::thread t2(thread2);

    t1.join();
    t2.join();

    return 0;
}

这段代码中，两个线程分别尝试获取两个互斥锁 mutex1 和 mutex2。线程1首先获取了 mutex1，然后尝试获取 mutex2；而线程2则是先获取了 mutex2，再尝试获取 mutex1。由于两个线程的互斥锁获取顺序相反，所以在特定条件下，可能会发生死锁。