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c++11多线程学习笔记之二 mutex使用

时间:2015-06-07 22:57:58      阅读:277      评论:0      收藏:0      [点我收藏+]
// 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

c++11多线程学习笔记之二 mutex使用

原文:http://www.cnblogs.com/itdef/p/4559146.html

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