scoped_lock的析构函数怎么理解?cppreference会不会出错?
How to understand destructor of scoped_lock?Does cppreference make a mistake?
~scoped_lock()
{ std::apply([](auto&... __m) { (__m.unlock(), ...); }, _M_devices); }
如何理解[](auto&... __m) { (__m.unlock(), ...);
?我不明白 lambda 中的 ...
,我不知道这个实现如何以相反的顺序释放互斥量。
正如@HolyBlackCat所说,
(__m.unlock(), ...)
表示(__m1.unlock(),(__m2.unlock(), (__m3.unlock(), (...))))
,但没有实现逆序解锁
在cppreference.com中:
When control leaves the scope in which the scoped_lock object was created, the scoped_lock is destructed and the mutexes are released, in reverse order.
我做了如下实验来证实这一点:
#include <chrono>
#include <iostream>
#include <mutex>
#include <thread>
class mymutex : public std::mutex {
public:
void lock() {
std::mutex::lock();
std::cout << "mutex " << _i << " locked" << std::endl;
}
mymutex(int i): _i(i){}
bool try_lock() {
bool res = std::mutex::try_lock();
if (res) {
std::cout << "mutex " << _i << " try locked" << std::endl;
}
return res;
}
void unlock() {
std::mutex::unlock();
std::cout << "mutex " << _i << " unlocked" << std::endl;
}
private:
int _i;
};
class Speaking {
private:
int a;
mymutex my1;
mymutex my2;
mymutex my3;
public:
Speaking() : a(0), my1(1), my2(2), my3(3){};
~Speaking() = default;
void speak_without_lock();
void speak_with_three_lock();
};
void Speaking::speak_without_lock() {
std::cout << std::this_thread::get_id() << ": " << a << std::endl;
a++;
}
void Speaking::speak_with_three_lock()
{
std::scoped_lock<mymutex, mymutex, mymutex> scoped(my1, my2, my3);
speak_without_lock();
}
int main() {
Speaking s;
s.speak_with_three_lock();
return 0;
}
mutex 1 locked
mutex 2 try locked
mutex 3 try locked
1: 0
mutex 1 unlocked
mutex 2 unlocked
mutex 3 unlocked
那么cppreference是不是出错了?
我认为 cppreference.com 在这个细节上是不正确的。 C++17 说:
~scoped_lock();
Effects: For all i in [0, sizeof...(MutexTypes)), get(pm).unlock()
这意味着释放锁的顺序与获取锁的顺序相同。
请注意,为了防止死锁,没有必要以与获取锁相反的顺序释放锁 - 只需要始终以相同的顺序获取锁。
~scoped_lock()
{ std::apply([](auto&... __m) { (__m.unlock(), ...); }, _M_devices); }
如何理解[](auto&... __m) { (__m.unlock(), ...);
?我不明白 lambda 中的 ...
,我不知道这个实现如何以相反的顺序释放互斥量。
正如@HolyBlackCat所说,
(__m.unlock(), ...)
表示(__m1.unlock(),(__m2.unlock(), (__m3.unlock(), (...))))
,但没有实现逆序解锁
在cppreference.com中:
When control leaves the scope in which the scoped_lock object was created, the scoped_lock is destructed and the mutexes are released, in reverse order.
我做了如下实验来证实这一点:
#include <chrono>
#include <iostream>
#include <mutex>
#include <thread>
class mymutex : public std::mutex {
public:
void lock() {
std::mutex::lock();
std::cout << "mutex " << _i << " locked" << std::endl;
}
mymutex(int i): _i(i){}
bool try_lock() {
bool res = std::mutex::try_lock();
if (res) {
std::cout << "mutex " << _i << " try locked" << std::endl;
}
return res;
}
void unlock() {
std::mutex::unlock();
std::cout << "mutex " << _i << " unlocked" << std::endl;
}
private:
int _i;
};
class Speaking {
private:
int a;
mymutex my1;
mymutex my2;
mymutex my3;
public:
Speaking() : a(0), my1(1), my2(2), my3(3){};
~Speaking() = default;
void speak_without_lock();
void speak_with_three_lock();
};
void Speaking::speak_without_lock() {
std::cout << std::this_thread::get_id() << ": " << a << std::endl;
a++;
}
void Speaking::speak_with_three_lock()
{
std::scoped_lock<mymutex, mymutex, mymutex> scoped(my1, my2, my3);
speak_without_lock();
}
int main() {
Speaking s;
s.speak_with_three_lock();
return 0;
}
mutex 1 locked
mutex 2 try locked
mutex 3 try locked
1: 0
mutex 1 unlocked
mutex 2 unlocked
mutex 3 unlocked
那么cppreference是不是出错了?
我认为 cppreference.com 在这个细节上是不正确的。 C++17 说:
~scoped_lock();
Effects: For all i in [0, sizeof...(MutexTypes)), get(pm).unlock()
这意味着释放锁的顺序与获取锁的顺序相同。
请注意,为了防止死锁,没有必要以与获取锁相反的顺序释放锁 - 只需要始终以相同的顺序获取锁。