一、造成死锁的原因

一般性原则，每个临界资源都需要一个线程锁来进行保护，假设有多个线程锁该怎么设计程序呢？先来看以下的一个示例（代码很简单就不加注释了）：

#include <QCoreApplication>
#include <QThread>
#include <QMutex>
#include <QDebug>int g_i_current_data = 0;
QMutex g_mutex_1;
QMutex g_mutex_2;class ThreadMutexA : public QThread
{
protected:void run(){qDebug() << objectName() << ": run begin...";while( true ){g_mutex_1.lock();qDebug() << objectName() << "get mutex_1...";g_mutex_2.lock();qDebug() << objectName() << "get mutex_2...";g_i_current_data++;qDebug() << objectName() << ": " << g_i_current_data;g_mutex_2.unlock();g_mutex_1.unlock();msleep(1);}qDebug() << objectName() << ": run end...";}
};class ThreadMutexB : public QThread
{
protected:void run(){qDebug() << objectName() << ": run begin...";while( true ){g_mutex_2.lock();qDebug() << objectName() << "get mutex_2...";g_mutex_1.lock();qDebug() << objectName() << "get mutex_1...";g_i_current_data--;qDebug() << objectName() << ": " << g_i_current_data;g_mutex_1.unlock();g_mutex_2.unlock();msleep(1);}qDebug() << objectName() << ": run end...";}
};int main(int argc, char *argv[])
{QCoreApplication a(argc, argv);qDebug() << "main begin...";ThreadMutexA thread_a;ThreadMutexA thread_a1;ThreadMutexA thread_a2;ThreadMutexB thread_b;ThreadMutexB thread_b1;thread_a.setObjectName("ThreadMutexA");thread_a1.setObjectName("ThreadMutexA_1");thread_a2.setObjectName("ThreadMutexA_2");thread_b.setObjectName("ThreadMutexB");thread_b1.setObjectName("ThreadMutexB_1");thread_a.start();thread_a1.start();thread_a2.start();thread_b.start();thread_b1.start();qDebug() << "main end...";return a.exec();
}

现象分析

1、现象：五个线程分别都进入了run()函数中，线程锁1和线程锁2分别被线程A和线程B获取（这两个线程动作比较快优先获取到线程数）后，程序就阻塞在这里了，不继续往下执行了。

2、分析：

（1）当线程A获取线程锁1时，就要继续获取线程锁2，但是此时线程锁2被线程B获取，线程A就只好等待线程锁2被释放；

（2）线程B获取线程锁2后，需要继续获取线程锁1，而此时线程锁1被线程A锁获取并没有释放，线程B也就只好等待线程A释放线程锁1了；

（3）两个线程一直等待彼此释放相应线程锁，等到花都谢了也等不到，程序无法继续执行。

3、原因：造成了线程的死锁

二、死锁的概念和发生条件

1、死锁概念：多线程之间相互等待临界资源而造成彼此无法继续执行。

2、死锁发生条件：（1）系统中存在多个临界资源且临界资源不可抢占；（2）每个线程需要多个临界资源才能继续执行。

三、死锁的避免

1、对所有临界资源都分配唯一的序号（R1, R2, R3, ..., Rn)；

2、对应的线程锁也分配同样的序号（M1, M2, M3, ..., Mn)；

3、系统中的每个线程必须按照严格的升序的次序去请求资源。

    ThreadMutexA A;mutex_1.lock();mutex_2.lock();mutex_3.lock();mutex_4.lock();

四、避免死锁的示例代码

#include <QCoreApplication>
#include <QThread>
#include <QMutex>
#include <QDebug>int g_i_current_data = 0;
QMutex g_mutex_1;
QMutex g_mutex_2;class ThreadMutexA : public QThread
{
protected:void run(){qDebug() << objectName() << ": run begin...";while( true ){g_mutex_1.lock();qDebug() << objectName() << "get mutex_1...";g_mutex_2.lock();qDebug() << objectName() << "get mutex_2...";g_i_current_data++;qDebug() << objectName() << ": " << g_i_current_data;g_mutex_2.unlock();g_mutex_1.unlock();msleep(1);}qDebug() << objectName() << ": run end...";}
};class ThreadMutexB : public QThread
{
protected:void run(){qDebug() << objectName() << ": run begin...";while( true ){g_mutex_1.lock();qDebug() << objectName() << "get mutex_1...";g_mutex_2.lock();qDebug() << objectName() << "get mutex_2...";g_i_current_data--;qDebug() << objectName() << ": " << g_i_current_data;g_mutex_2.unlock();g_mutex_1.unlock();msleep(1);}qDebug() << objectName() << ": run end...";}
};int main(int argc, char *argv[])
{QCoreApplication a(argc, argv);qDebug() << "main begin...";ThreadMutexA thread_a;ThreadMutexA thread_a1;ThreadMutexA thread_a2;ThreadMutexB thread_b;ThreadMutexB thread_b1;thread_a.setObjectName("ThreadMutexA");thread_a1.setObjectName("ThreadMutexA_1");thread_a2.setObjectName("ThreadMutexA_2");thread_b.setObjectName("ThreadMutexB");thread_b1.setObjectName("ThreadMutexB_1");thread_a.start();thread_a1.start();thread_a2.start();thread_b.start();thread_b1.start();qDebug() << "main end...";return a.exec();
}