std::try_lock
定义于头文件 <mutex>
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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
,通过以从头开始的顺序调用 try_lock
。
若调用 try_lock
失败,则不再进一步调用 try_lock
,并对任何已锁对象调用 unlock
,返回锁定失败对象的 0
底下标。
若调用 try_lock
抛出异常,则在重抛前对任何已锁对象调用 unlock
。
参数
lock1, lock2, ... , lockn | - | 要锁定的可锁定 (Lockable) 对象 |
返回值
成功时为 -1
,否则为锁定失败对象的 0
底下标值。
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 <mutex> #include <vector> #include <thread> #include <iostream> #include <functional> #include <chrono> 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, const char *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 << "Done processing\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 Done processing foo: 0 bar: 0 overall: 20
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) |