linux多线程编程--使用互斥锁的简单程序

来源：互联网发布：金融大数据案例分析编辑：程序博客网时间：2024/04/29 18:04

银行应用中，一个帐号给另一个帐号汇款时，如果另一个帐号又有读取存款的操作，就会用到互斥锁。

使用互斥锁之前的简单程序如下：

// 这个程序使用锁来模拟银行的存取款#include <iostream>#include <pthread.h>using namespace std;struct data {    int m;    int n;};void* changeValue(void*);void* print(void*);int main() {   struct data data0;   data0.m = 5;   data0.n = 3;      pthread_t thread1;   pthread_create(&thread1, NULL, changeValue, &data0);      pthread_t thread2;   pthread_create(&thread2, NULL, print, &data0);      pthread_join(thread1, NULL);   pthread_join(thread2, NULL);      return 0;}void* changeValue(void* data_in) {    sleep(1.5);    struct data* data = (struct data*)data_in;    data->m = data->m - 2;    sleep(1.5);    data->n = data->n + 2;    sleep(1.5);    pthread_exit(NULL);}void* print(void* data_in) {    struct data* data0 = (struct data*)data_in;    double sleep_time = 1.0, sum_time(0);    while (sum_time<5.0) {        cout << "i= " << data0->m << "j= " << data0->n << endl;        cout << "i+j= " << data0->m + data0->n << endl;        sleep(sleep_time);        sum_time += sleep_time;    }    pthread_exit(NULL);}

加互斥锁的程序

// 这个程序使用锁来模拟银行的存取款#include <iostream>#include <pthread.h>using namespace std;struct data {    int m;    int n;};void* changeValue(void*); // 函数最后都定义为这样的格式void* print(void*);pthread_mutex_t lock; // 定义全局互斥锁，否则不能在函数中使用int main(){   struct data data0;   data0.m = 5;   data0.n = 3;      pthread_mutex_init(&lock, NULL); // 初始化互斥锁      pthread_t thread1;   pthread_create(&thread1, NULL, changeValue, &data0);      pthread_t thread2;   pthread_create(&thread2, NULL, print, &data0);      pthread_join(thread1, NULL);   pthread_join(thread2, NULL);      pthread_mutex_destroy(&lock); // 最后要销毁锁      return 0;}void* changeValue(void* data_in) {    pthread_mutex_lock(&lock); // 占用锁资源    sleep(1.5);    struct data* data = (struct data*)data_in;    data->m = data->m - 2;    sleep(1.5);    data->n = data->n + 2;    sleep(1.5);    pthread_mutex_unlock(&lock); // 解除锁资源    pthread_exit(NULL);}void* print(void* data_in) {    sleep(0.1);    pthread_mutex_lock(&lock);    struct data* data0 = (struct data*)data_in;    double sleep_time = 1.0, sum_time(0);    while (sum_time<5.0) {        cout << "i= " << data0->m << "j= " << data0->n << endl;        cout << "i+j= " << data0->m + data0->n << endl;        sleep(sleep_time);        sum_time += sleep_time;    }    pthread_mutex_unlock(&lock);    pthread_exit(NULL);}

0 0