std::try_lock
来自cppreference.com
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| 在标头 <mutex> 定义
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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，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。
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基于 0 的索引值。
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 <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, 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 << "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
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) |