I'm trying to create a simple pool object, which I would like to more-or-less fairly allocate access to a set of shared resources to any threads that ask for it. In windows, I would typically have an array of Mutexes and do a WaitForMultipleObjects, with bWaitAll=FALSE (see windows_pool_of_n_t below). But I'm hoping to someday be able to port this to other OSes, so I'd like to stick with the standard. A deque of resources, with a condition_variable on size()!=0 seemed like the obvious solution (see pool_of_n_t below).
But for reasons I don't understand, that code serializes thread access. I'm not expecting strict fairness, but this is pretty much the worst possible case - the thread that had the lock last time always gets the lock the next time. It's not that std::mutex doesn't conform to Windows more-or-less fair scheduling, since using just a mutex without the condition variable works as expected, although only for a pool of one, of course (see pool_of_one_t below).
Can anyone explain this? Is there a way around this?
the results:
C:\temp\stdpool>bin\stdpool.exe
pool:pool_of_one_t
thread 0:19826 ms
thread 1:19846 ms
thread 2:19866 ms
thread 3:19886 ms
thread 4:19906 ms
thread 5:19926 ms
thread 6:19946 ms
thread 7:19965 ms
thread 8:19985 ms
thread 9:20004 ms
pool:windows_pool_of_n_t(1)
thread 0:19819 ms
thread 1:19838 ms
thread 2:19858 ms
thread 3:19878 ms
thread 4:19898 ms
thread 5:19918 ms
thread 6:19938 ms
thread 7:19958 ms
thread 8:19978 ms
thread 9:19997 ms
pool:pool_of_n_t(1)
thread 9:3637 ms
thread 0:4538 ms
thread 6:7558 ms
thread 4:9779 ms
thread 8:9997 ms
thread 2:13058 ms
thread 1:13997 ms
thread 3:17076 ms
thread 5:17995 ms
thread 7:19994 ms
pool:windows_pool_of_n_t(2)
thread 1:9919 ms
thread 0:9919 ms
thread 2:9939 ms
thread 3:9939 ms
thread 5:9958 ms
thread 4:9959 ms
thread 6:9978 ms
thread 7:9978 ms
thread 9:9997 ms
thread 8:9997 ms
pool:pool_of_n_t(2)
thread 2:6019 ms
thread 0:7882 ms
thread 4:8102 ms
thread 5:8182 ms
thread 1:8382 ms
thread 8:8742 ms
thread 7:9162 ms
thread 9:9641 ms
thread 3:9802 ms
thread 6:10201 ms
pool:windows_pool_of_n_t(5)
thread 4:3978 ms
thread 3:3978 ms
thread 2:3979 ms
thread 0:3980 ms
thread 1:3980 ms
thread 9:3997 ms
thread 7:3999 ms
thread 6:3999 ms
thread 5:4000 ms
thread 8:4001 ms
pool:pool_of_n_t(5)
thread 2:3080 ms
thread 0:3498 ms
thread 8:3697 ms
thread 3:3699 ms
thread 6:3797 ms
thread 7:3857 ms
thread 1:3978 ms
thread 4:4039 ms
thread 9:4057 ms
thread 5:4059 ms
the code:
#include <iostream>
#include <deque>
#include <vector>
#include <mutex>
#include <thread>
#include <sstream>
#include <chrono>
#include <iomanip>
#include <cassert>
#include <condition_variable>
#include <windows.h>
using namespace std;
class pool_t {
public:
virtual void check_in(int size) = 0;
virtual int check_out() = 0;
virtual string pool_name() = 0;
};
class pool_of_one_t : public pool_t {
mutex lock;
public:
virtual void check_in(int resource) {
lock.unlock();
}
virtual int check_out() {
lock.lock();
return 0;
}
virtual string pool_name() {
return "pool_of_one_t";
}
};
class windows_pool_of_n_t : public pool_t {
vector<HANDLE> resources;
public:
windows_pool_of_n_t(int size) {
for (int i=0; i < size; ++i)
resources.push_back(CreateMutex(NULL, FALSE, NULL));
}
~windows_pool_of_n_t() {
for (auto resource : resources)
CloseHandle(resource);
}
virtual void check_in(int resource) {
ReleaseMutex(resources[resource]);
}
virtual int check_out() {
DWORD result = WaitForMultipleObjects(resources.size(),
resources.data(), FALSE, INFINITE);
assert(result >= WAIT_OBJECT_0
&& result < WAIT_OBJECT_0+resources.size());
return result - WAIT_OBJECT_0;
}
virtual string pool_name() {
ostringstream name;
name << "windows_pool_of_n_t(" << resources.size() << ")";
return name.str();
}
};
class pool_of_n_t : public pool_t {
deque<int> resources;
mutex lock;
condition_variable not_empty;
public:
pool_of_n_t(int size) {
for (int i=0; i < size; ++i)
check_in(i);
}
virtual void check_in(int resource) {
unique_lock<mutex> resources_guard(lock);
resources.push_back(resource);
resources_guard.unlock();
not_empty.notify_one();
}
virtual int check_out() {
unique_lock<mutex> resources_guard(lock);
not_empty.wait(resources_guard,
[this](){return resources.size() > 0;});
auto resource = resources.front();
resources.pop_front();
bool notify_others = resources.size() > 0;
resources_guard.unlock();
if (notify_others)
not_empty.notify_one();
return resource;
}
virtual string pool_name() {
ostringstream name;
name << "pool_of_n_t(" << resources.size() << ")";
return name.str();
}
};
void worker_thread(int id, pool_t& resource_pool)
{
auto start_time = chrono::system_clock::now();
for (int i=0; i < 100; ++i) {
auto resource = resource_pool.check_out();
this_thread::sleep_for(chrono::milliseconds(20));
resource_pool.check_in(resource);
this_thread::yield();
}
static mutex cout_lock;
{
unique_lock<mutex> cout_guard(cout_lock);
cout << "thread " << id << ":"
<< chrono::duration_cast<chrono::milliseconds>(
chrono::system_clock::now() - start_time).count()
<< " ms" << endl;
}
}
void test_it(pool_t& resource_pool)
{
cout << "pool:" << resource_pool.pool_name() << endl;
vector<thread> threads;
for (int i=0; i < 10; ++i)
threads.push_back(thread(worker_thread, i, ref(resource_pool)));
for (auto& thread : threads)
thread.join();
}
int main(int argc, char* argv[])
{
test_it(pool_of_one_t());
test_it(windows_pool_of_n_t(1));
test_it(pool_of_n_t(1));
test_it(windows_pool_of_n_t(2));
test_it(pool_of_n_t(2));
test_it(windows_pool_of_n_t(5));
test_it(pool_of_n_t(5));
return 0;
}
答案 0 :(得分：7)
我在Linux上测试pool:pool_of_n_t(2)并在
this_thread::yield();
在我的comp上查看测试池的结果：pool_of_n_t（2）：
1）this_thread :: yield（）：
$./a.out
pool:pool_of_n_t(2)
thread 0, run for:2053 ms
thread 9, run for:3721 ms
thread 5, run for:4830 ms
thread 6, run for:6854 ms
thread 3, run for:8229 ms
thread 4, run for:8353 ms
thread 7, run for:9441 ms
thread 2, run for:9482 ms
thread 1, run for:10127 ms
thread 8, run for:10426 ms
它们与你的相似。
2）当我用this_thread::yield()替换pthread_yield()时进行同样的测试：
$ ./a.out
pool:pool_of_n_t(2)
thread 0, run for:7922 ms
thread 3, run for:8853 ms
thread 4, run for:8854 ms
thread 1, run for:9077 ms
thread 5, run for:9364 ms
thread 9, run for:9446 ms
thread 7, run for:9594 ms
thread 2, run for:9615 ms
thread 8, run for:10170 ms
thread 6, run for:10416 ms
更公平。您假设this_thread :: yield（）确实将CPU提供给另一个线程，但它没有给出它。
这是gcc 4.8的this_thread :: yield的disas：
(gdb) disassembl this_thread::yield
Dump of assembler code for function std::this_thread::yield():
0x0000000000401fb2 <+0>: push %rbp
0x0000000000401fb3 <+1>: mov %rsp,%rbp
0x0000000000401fb6 <+4>: pop %rbp
0x0000000000401fb7 <+5>: retq
End of assembler dump.
我没有看到任何重新安排
这是pthread_yield的disas：
(gdb) disassemble pthread_yield
Dump of assembler code for function pthread_yield:
0x0000003149c084c0 <+0>: jmpq 0x3149c05448 <sched_yield@plt>
End of assembler dump.
(gdb) disassemble sched_yield
Dump of assembler code for function sched_yield:
0x00000031498cf520 <+0>: mov $0x18,%eax
0x00000031498cf525 <+5>: syscall
0x00000031498cf527 <+7>: cmp $0xfffffffffffff001,%rax
0x00000031498cf52d <+13>: jae 0x31498cf530 <sched_yield+16>
0x00000031498cf52f <+15>: retq
0x00000031498cf530 <+16>: mov 0x2bea71(%rip),%rcx # 0x3149b8dfa8
0x00000031498cf537 <+23>: xor %edx,%edx
0x00000031498cf539 <+25>: sub %rax,%rdx
0x00000031498cf53c <+28>: mov %edx,%fs:(%rcx)
0x00000031498cf53f <+31>: or $0xffffffffffffffff,%rax
0x00000031498cf543 <+35>: jmp 0x31498cf52f <sched_yield+15>
End of assembler dump.
答案 1 :(得分：2)
我不认为条件变量是罪魁祸首。
Linux "Completely Fair Queue"和Windows线程调度程序都假设理想的目标是为每个线程提供整个时间片（即公平。）他们认为，如果一个线程产生在它消耗整个时间片之前它接近队列的前端[这是一个粗略的简化]，因为这是“公平”的事情。
我发现这非常不幸。如果你有三个线程，其中一个可以工作而另外两个被阻塞等待那个线程，Windows和Linux调度程序将在阻塞的线程之间来回反复多次，然后给“正确的”线程一个机会