std::try_lock
出自cppreference.com
| 在標頭 <mutex> 定義
|
||
| template< class Lockable1, class Lockable2, class... LockableN > int try_lock( Lockable1& lock1, Lockable2& lock2, LockableN&... lockn ); |
(C++11 起) | |
嘗試鎖定每個給定的可鎖定 (Lockable) 對象 lock1、lock2、...、lockn，從頭開始依次調用 try_lock。
若調用 try_lock 失敗，則不再繼續調用 try_lock，並對任何已鎖對象調用 unlock，返回鎖定失敗對象的基於 0 的索引。
若調用 try_lock 拋出異常，則在重拋前對任何已鎖對象調用 unlock。
目錄 |
[編輯] 參數
| lock1, lock2, ... , lockn | - | 要鎖定的可鎖定 (Lockable) 對象 |
[編輯] 返回值
成功時為 -1，否則為鎖定失敗對象的基於 0 的索引值。
[編輯] 示例
下列示例用 std::try_lock 周期地記錄並重置運行於分離線程的計數器。
運行此代碼
#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'; }
可能的輸出：
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
[編輯] 參閱
| (C++11) |
鎖定指定的互斥體，若任何一個不可用則阻塞 (函數模板) |
