std::try_lock
来自cppreference.com
在标头 <mutex> 定义 | ||
template<class Lockable1, class Lockable2, class... LockableN> int try_lock( Lockable1& lock1, Lockable2& lock2, LockableN&... lockn); | (C++11 起) | |
尝试锁定每个给定的可锁定(Lockable) 对象 lock1、lock2、...、lockn，从头开始依次调用 try_lock。
若调用 try_lock 失败，则不再继续调用 try_lock，并对任何已锁对象调用 unlock，返回锁定失败对象的基于 0 的索引。
若调用 try_lock 抛出异常，则在重抛前对任何已锁对象调用 unlock。
目录 |
[编辑]参数
| lock1, lock2, ... , lockn | - | 要锁定的可锁定(Lockable) 对象 |
[编辑]返回值
成功时为 -1，否则为锁定失败对象的基于 0 的索引值。
[编辑]示例
下列示例用 std::try_lock 周期地记录并重置运行于分离线程的计数器。
运行此代码
#include <chrono>#include <functional>#include <iostream>#include <mutex>#include <thread>#include <vector> int main(){int foo_count =0;std::mutex foo_count_mutex;int bar_count =0;std::mutex bar_count_mutex;int overall_count =0;bool done =false;std::mutex done_mutex; auto increment =[](int& counter, std::mutex& m, constchar* desc){for(int i =0; i <10;++i){std::unique_lock<std::mutex> lock(m);++counter;std::cout<< desc <<": "<< counter <<'\n'; lock.unlock();std::this_thread::sleep_for(std::chrono::seconds(1));}}; std::thread increment_foo(increment, std::ref(foo_count), std::ref(foo_count_mutex), "foo");std::thread increment_bar(increment, std::ref(bar_count), std::ref(bar_count_mutex), "bar"); std::thread update_overall([&](){ done_mutex.lock();while(!done){ done_mutex.unlock();int result = std::try_lock(foo_count_mutex, bar_count_mutex);if(result ==-1){ overall_count += foo_count + bar_count; foo_count =0; bar_count =0;std::cout<<"overall: "<< overall_count <<'\n'; foo_count_mutex.unlock(); bar_count_mutex.unlock();}std::this_thread::sleep_for(std::chrono::seconds(2)); done_mutex.lock();} done_mutex.unlock();}); increment_foo.join(); increment_bar.join(); done_mutex.lock(); done =true; done_mutex.unlock(); update_overall.join(); std::cout<<"处理完成\n"<<"foo: "<< foo_count <<'\n'<<"bar: "<< bar_count <<'\n'<<"overall: "<< overall_count <<'\n';}
可能的输出：
bar: 1 foo: 1 foo: 2 bar: 2 foo: 3 overall: 5 bar: 1 foo: 1 bar: 2 foo: 2 bar: 3 overall: 10 bar: 1 foo: 1 bar: 2 foo: 2 overall: 14 bar: 1 foo: 1 bar: 2 overall: 17 foo: 1 bar: 1 foo: 2 overall: 20 处理完成 foo: 0 bar: 0 overall: 20
[编辑]参阅
(C++11) | 锁定指定的互斥体，若任何一个不可用则阻塞 (函数模板) |
