模拟实现一个死锁


/**
 * Created by fangruibin
 * 测试死锁产生的场景
 */
 
#include <iostream>
#include <pthread.h>
#include <unistd.h>
 
//定义两把锁
pthread_mutex_t m_mutex1;
pthread_mutex_t m_mutex2;
int A = 0, B = 0;
 
//线程1
void* threadFunc1(void* p)
{
    //printf("thread 1 running..\n");
    pthread_mutex_lock(&m_mutex1);
    A = 1;
    printf("thread 1 write source A\n");
    usleep(100);
    
    pthread_mutex_lock(&m_mutex2);
    B = 1;
    printf("thread 1 write source B\n");
    
    //解锁，实际上是跑不到这里的，因为前面已经死锁了
    pthread_mutex_unlock(&m_mutex2);
    pthread_mutex_unlock(&m_mutex1);
    
    return NULL;
}
 
//线程2
void* threadFunc2(void* p)
{
    //printf("thread 2 running..\n");
    pthread_mutex_lock(&m_mutex2);
    B = 1;
    printf("thread 2 write source B\n");
    usleep(100);
    
    pthread_mutex_lock(&m_mutex1);
    A = 1;
    printf("thread 2 write source A\n");
    
    //解锁，实际上是跑不到这里的，因为前面已经死锁了
    pthread_mutex_unlock(&m_mutex1);
    pthread_mutex_unlock(&m_mutex2);
    
    return NULL;
}
 
int main()
{
    //初始化锁
    if (pthread_mutex_init(&m_mutex1, 0) != 0)
    {
        printf("init mutex 1 failed\n");
        return -1;
    }
    if (pthread_mutex_init(&m_mutex2, 0) != 0)
    {
        printf("init mutex 2 failed\n");
        return -1;
    }
    
    //初始化线程
    pthread_t hThread1;
    pthread_t hThread2;
    if (pthread_create(&hThread1, NULL, &threadFunc1, NULL) != 0)
    {
        printf("create thread 1 failed\n");
        return -1;
    }
    if (pthread_create(&hThread2, NULL, &threadFunc2, NULL) != 0)
    {
        printf("create thread 2 failed\n");
        return -1;
    }
    
    while (1)
    {
        sleep(1);
    }
    
    pthread_mutex_destroy(&m_mutex1);
    pthread_mutex_destroy(&m_mutex2);
    return 0;
}

面试官要求用c++写一个死锁的程序。
目前想到两种简单的写法，一种是单线程对一个资源重复申请上锁；第二种是两个线程对两个资源申请上锁，形成环路。

实现一：单线程对一个资源重复申请上锁

#include <iostream>
#include <thread>
#include <mutex>
#include <unistd.h>

using namespace std;

int data = 1;
mutex mt1,mt2;

void a2() {
    data = data * data;
    mt1.lock();  //第二次申请对mt1上锁，但是上不上去
    cout<<data<<endl;
    mt1.unlock();
}
void a1() {
    mt1.lock();  //第一次对mt1上锁
    data = data+1;
    a2();
    cout<<data<<endl;
    mt1.unlock();
}

int main() {
    thread t1(a1);
    t1.join();
    cout<<"main here"<<endl;
    return 0;
}

实现二、两个线程对两个资源申请上锁，形成环路

#include <iostream>
#include <thread>
#include <mutex>
#include <unistd.h>

using namespace std;

int data = 1;
mutex mt1,mt2;

void a2() {
    mt2.lock();
    sleep(1);
    data = data * data;
    mt1.lock();  //此时a1已经对mt1上锁，所以要等待
    cout<<data<<endl;
    mt1.unlock();
    mt2.unlock();
}
void a1() {
    mt1.lock();
    sleep(1);
    data = data+1;
    mt2.lock();  //此时a2已经对mt2上锁，所以要等待
    cout<<data<<endl;
    mt2.unlock();
    mt1.unlock();
}

int main() {
    thread t2(a2);
    thread t1(a1);
    
    t1.join();
    t2.join();
    cout<<"main here"<<endl;  //要t1线程、t2线程都执行完毕后才会执行
    return 0;
}

最后编辑于：2020.09.25 15:47:59