103 lines
2.5 KiB
C++
103 lines
2.5 KiB
C++
#include <thread>
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#include <mutex>
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#include <queue>
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#include <iostream>
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using namespace std;
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#define LOOP_TIMES 100
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void test1() {
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mutex mtx;
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int res = 0;
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thread t1([&]() {
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for (int i = 0; i < LOOP_TIMES;i++) {
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this_thread::sleep_for(chrono::milliseconds(1));
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res = res + 1;
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}
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});
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thread t2([&]() {
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for (int i = 0; i < LOOP_TIMES; i++) {
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this_thread::sleep_for(chrono::milliseconds(1));
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mtx.lock();
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res = res - 1;
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mtx.unlock();
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}
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});
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t1.join();
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t2.join();
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std::cout << "test1:: res:: " << res << " mutexsize::" << sizeof(mutex) << endl;
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}
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void test2() {
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recursive_mutex mtx;
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int res = 0;
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thread t1([&]() {
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for (int i = 0; i < LOOP_TIMES / 10; i++) {
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this_thread::sleep_for(chrono::milliseconds(i));
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lock_guard<recursive_mutex> lock(mtx);
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res = res + 1;
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}
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});
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thread t2([&]() {
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for (int i = 0; i < LOOP_TIMES / 10; i++) {
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this_thread::sleep_for(chrono::milliseconds(i));
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unique_lock<recursive_mutex> lock(mtx);
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res = res - 1;
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}
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});
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t1.join();
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t2.join();
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std::cout << "test2:: res:: " << res << " lock_guard::" << sizeof(lock_guard<mutex>)
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<< " unique_lock::" << sizeof(unique_lock<mutex>) << endl;
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}
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void test3() {
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mutex mtx;
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int res = 0;
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int error_count = 0;
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queue<int> mq;
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thread t1([&]() {
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for (int i = 0; i < LOOP_TIMES; i++) {
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this_thread::sleep_for(chrono::milliseconds(i));
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lock_guard<mutex> lock(mtx);
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mq.push(i / 2);
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res = res + 1;
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}
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});
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thread t2([&]() {
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int a = 0;
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for (int i = 0; i < LOOP_TIMES;) {
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this_thread::sleep_for(chrono::milliseconds(a));
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unique_lock<mutex> lock(mtx);
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if (mq.empty()) {
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error_count++;
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continue;
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}
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i++;
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a = mq.front();
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mq.pop();
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res = res - 1;
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}
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});
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t1.join();
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t2.join();
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std::cout << "test3:: res:: " << res << " error_count::" << error_count << endl;
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}
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int main() {
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/*
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进程是房子，线程是住在房子里的人，厕所、厨房...是房子里的共享资源。
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*/
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test1();
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test2();
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/*
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mutex 原子性 排队性
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//保证资源的正确使用，如一个厕所只能一个人用
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//递归锁可以让前面的人重复上锁，避免函数递归时解锁出错
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//排队等待前一个人离开，才能使用
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//超时锁，可以先进入自旋，超时后阻塞
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*/
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test3();
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/*
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锁 如果存在单向依赖，无法保证资源是否存在，会空转CPU
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生产者知道什么时候有资源，消费者知道什么时候没有资源
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消费者上锁等待，生产者解锁通知， 锁只能自己锁自己解锁
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*/
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return 0;
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} |