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网站建设方案和报价,家装设计师工资高吗,四川seo整站优化费用,网站体验方案对于多线程来说导致死锁的原因较多，最简单的一个原因是锁定一个互斥量后，没有释放这个互斥量，导致其它的线程获取这个互斥量时，会永远进入阻塞状态，往往发生在程序分支较多或者抛出异常时忘记主动释放，解决…

对于多线程来说导致死锁的原因较多，最简单的一个原因是锁定一个互斥量后，没有释放这个互斥量，导致其它的线程获取这个互斥量时，会永远进入阻塞状态，往往发生在程序分支较多或者抛出异常时忘记主动释放，解决方法：lock_guard或者unique_lock这样的自动对象，对互斥量进行管理，这样离开作用域后，在析构函数中自动释放互斥量。

另一个原因，同时对多个互斥量进行锁定和解锁时，由于获得和释放互斥量顺序不同容易造成死锁。

例如

#include<array>
#include<thread>
#include<iostream>
#include<mutex>
using namespace std;class DemoClass
{
public:void func1(){{lock_guard<mutex> lock1(mtx1);lock_guard<mutex>lock2(count_mtx);count++;cout << "[func1]count=" << count << "\n";}this_thread::sleep_for(1ms);}void func2(){{lock_guard<mutex>lock1(count_mtx);lock_guard<mutex>lock2(mtx1);count--;cout << "[fuc2]Count=" << count << "\n";}this_thread::sleep_for(1ms);}void calc(int n){for (int i = 0; i < n; i++){if (n % 2){func2();}else{func1();}}	}
private:mutex mtx1;mutex count_mtx;int count = 0;
};
void main()
{DemoClass demo;const int n1 = 10000;const int n2 = 10001;thread th1(&DemoClass::calc,&demo,n1);thread th2(&DemoClass::calc,&demo, n2);th1.join();th2.join();
}

运行结果

程序运行一段时间后发生死锁。

线程1获取mtx1等待获取count_mtx，而线程2获取count_mtx等待获取mtx这样程序永远进入阻塞状态。

改进方法1代码如下

#include<array>
#include<thread>
#include<iostream>
#include<mutex>
using namespace std;class DemoClass
{
public:void func1(){while (true){unique_lock<timed_mutex> lock1(mtx1,defer_lock);unique_lock<timed_mutex>lock2(count_mtx,defer_lock);if (!lock1.try_lock_for(100ms))continue;if (!lock2.try_lock_for(100ms))continue;count++;cout << "[func1]count=" << count << "\n";}this_thread::sleep_for(1ms);}void func2(){while(true){unique_lock<timed_mutex>lock1(count_mtx,defer_lock);unique_lock<timed_mutex>lock2(mtx1,defer_lock);if (!lock1.try_lock_for(100ms))continue;if (!lock2.try_lock_for(100ms))continue;count--;cout << "[fuc2]Count=" << count << "\n";}this_thread::sleep_for(1ms);}void calc(int n){for (int i = 0; i < n; i++){if (n % 2){func2();}else{func1();}}	}
private:timed_mutex mtx1;timed_mutex count_mtx;int count = 0;
};
void main()
{DemoClass demo;const int n1 = 10000;const int n2 = 10001;thread th1(&DemoClass::calc,&demo,n1);thread th2(&DemoClass::calc,&demo, n2);th1.join();th2.join();
}

使用计时锁如果在100ms内为获取到想获取得锁，释放当前拥有得锁，进入下一次循环，这样程序进入到活锁得状态，两个线程不断得释放自己当前拥有得锁，而又得不到想获取得锁。

解决活锁得方法：保证顺序获得互斥量

#include<array>
#include<thread>
#include<iostream>
#include<mutex>
using namespace std;class DemoClass
{
public:void func1(){lock(mtx1,count_mtx);lock_guard<mutex> lock1(mtx1,adopt_lock);lock_guard<mutex> lock2(count_mtx, adopt_lock);count++;cout << "[func1]count=" << count << "\n";this_thread::sleep_for(1ms);}void func2(){lock(mtx1, count_mtx);lock_guard<mutex> lock1(mtx1, adopt_lock);lock_guard<mutex> lock2(count_mtx, adopt_lock);count--;cout << "[fuc2]Count=" << count << "\n";this_thread::sleep_for(1ms);}void calc(int n){for (int i = 0; i < n; i++){if (n % 2){func2();}else{func1();}}	}
private:mutex mtx1;mutex count_mtx;int count = 0;
};
void main()
{DemoClass demo;const int n1 = 10000;const int n2 = 10001;thread th1(&DemoClass::calc,&demo,n1);thread th2(&DemoClass::calc,&demo, n2);th1.join();th2.join();
}

算法大致原理