std::unique_lock::lock
void lock(); | ? | (since C++11) |
|---|
锁定相关的互斥对象。有效呼叫mutex()->lock()...
参数
%280%29
返回值
%280%29
例外
- 引发的任何异常
mutex()->lock()
- 如果没有关联的互斥，
std::system_error错误代码为std::errc::operation_not_permitted
- 如果互斥锁已被
unique_lock%28，换言之，拥有[医]锁是%29，std::system_error错误代码为std::errc::resource_deadlock_would_occur
例
下面的示例使用lock重新获得一个未锁定的互斥体。
二次
#include <mutex>
#include <thread>
#include <iostream>
#include <vector>
#include <chrono>
int main()
{
int counter = 0;
std::mutex counter_mutex;
std::vector<std::thread> threads;
auto worker_task = [&](int id) {
std::unique_lock<std::mutex> lock(counter_mutex);
++counter;
std::cout << id << ", initial counter: " << counter << '\n';
lock.unlock();
// don't hold the lock while we simulate an expensive operation
std::this_thread::sleep_for(std::chrono::seconds(1));
lock.lock();
++counter;
std::cout << id << ", final counter: " << counter << '\n';
};
for (int i = 0; i < 10; ++i) threads.emplace_back(worker_task, i);
for (auto &thread : threads) thread.join();
}二次
可能的产出：
二次
0, initial counter: 1
1, initial counter: 2
2, initial counter: 3
3, initial counter: 4
4, initial counter: 5
5, initial counter: 6
6, initial counter: 7
7, initial counter: 8
8, initial counter: 9
9, initial counter: 10
6, final counter: 11
3, final counter: 12
4, final counter: 13
2, final counter: 14
5, final counter: 15
0, final counter: 16
1, final counter: 17
7, final counter: 18
9, final counter: 19
8, final counter: 20二次
另见
try_lock | tries to lock the associated mutex, returns if the mutex is not available (public member function) |
|---|---|
unlock | unlocks the associated mutex (public member function) |
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