c++11多线程学习笔记之二 mutex使用
// 1111111.cpp : 定义控制台应用程序的入口点。 // #include "stdafx.h" #include <iostream> #include <thread> #include <mutex> int gcounter = 0; std::mutex gmtx; std::mutex gmtxOutput; void Increases() { for (int i = 0; i<10000; ++i) { if (gmtx.try_lock()) { // only increase if currently not locked: ++gcounter; gmtx.unlock(); } else{ gmtxOutput.lock(); std::cout << "try lock failed" << std::endl; gmtxOutput.unlock(); } } } int _tmain(int argc, _TCHAR* argv[]) { std::thread threads[10]; for (int i = 0; i<10; ++i) threads[i] = std::thread(Increases); for (auto& th : threads) th.join(); std::cout << "counter is " << gcounter << std::endl; return 0; }
输出:
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
try lock failed
counter is 99983
请按任意键继续. . .
这个例子说明了 try_lock() 与 lock()的区别
try_lock()会对能否上锁进行测试 并返回布尔值
而lock()则直接进行锁定 不能锁定则阻塞直到锁定
作 者: itdef
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欢迎转帖 请保持文本完整并注明出处
技术博客 http://www.cnblogs.com/itdef/
B站算法视频题解
https://space.bilibili.com/18508846
qq 151435887
gitee https://gitee.com/def/
欢迎c c++ 算法爱好者 windows驱动爱好者 服务器程序员沟通交流
如果觉得不错,欢迎点赞,你的鼓励就是我的动力

