什么情况造成互斥锁死锁-编程知识

什么情况造成互斥锁死锁

由于互斥锁的使用不当，导致多个线程无法进行下一步的代码运行，也就是说竞争锁的两个线程互相锁住，导致整个进程无法往下运行。

举个例子：

两个锁，两个线程，两个线程运行的条件都是需要同时获得这两把锁，但是这两个线程一人获得了一把锁，又想去获得对方手里的那把锁，但是互相都不让出手里的锁，就会出现死锁的情况。

#include <stdio.h>
#include <pthread.h>pthread_mutex_t mutex1;
pthread_mutex_t mutex2;void *func1(void *arg)
{int i;pthread_mutex_lock(&mutex1);sleep(1);pthread_mutex_lock(&mutex2);for(i=0;i<3;i++){printf("t1:%ld thread is create\n",(unsigned long)pthread_self());printf("t1:param is %d\n",*((int *)arg));sleep(1);}pthread_mutex_unlock(&mutex1);pthread_mutex_unlock(&mutex2);}void *func2(void *arg)
{pthread_mutex_lock(&mutex2);sleep(1);pthread_mutex_lock(&mutex1);printf("t2:%ld thread is create\n",(unsigned long)pthread_self());printf("t2:param is %d\n",*((int *)arg));pthread_mutex_unlock(&mutex1);pthread_mutex_unlock(&mutex2);
}int main()
{int ret;int param = 100;pthread_t t1;pthread_t t2;pthread_mutex_init(&mutex1,NULL);pthread_mutex_init(&mutex2,NULL);ret = pthread_create(&t1, NULL, func1,(void *)&param);if(ret == 0){printf("main:create t1 success\n");}ret = pthread_create(&t2, NULL, func2,(void *)&param);if(ret == 0){printf("main:create t2 success\n");}printf("main:%ld\n",(unsigned long)pthread_self());pthread_join(t1,NULL);pthread_join(t2,NULL);pthread_mutex_destroy(&mutex1);pthread_mutex_destroy(&mutex2);return 0;
}