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Re: debug mode performance patch
On 10 November 2010 21:55, François Dumont wrote:
>
> Well, this is not something my change introduce. For the moment the debug
> mode is using a single mutex for all synchronizations. To keep binary
> compatibility I haven't been able to add a mutex instance to the safe
> sequence so I introduce a small mutex pool to limit contention.
Sorry if my review sounded negative, I think the general idea is very
good. But with a single mutex there is no chance for deadlock, it
just means only one swap can take place at a time.
By introducing multiple mutexes without some checking you make it easy
to deadlock. Slow and correct is better than fast and deadlock ;-)
>> There are two problems with _M_swap in your patch:
>>
>> 1) if two objects are swapped whih share the same mutex then it will
>> try to lock the same mutex twice. ?You should be able to reproduce
>> this like so (untested):
>>
>> __gnu_debug::vector<int> ?v[17];
>> swap(v[0], v[16]);
>>
>> If I'm not mistaken, v[0] and v[16] should get the same mutex from
>> get_mutex and so should lock it twice. That's bad, even in a
>> single-threaded program.
>>
>
> Yes, this is why I asked if mutex are guarantied to be reentrant even if I
> agree that even if it is so it is still bad practice.
When your mail arrived I had just finished some examples to
demonstrate the problems, maybe these should go in the testsuite.
This deadlocks for me:
#include <mutex>
#include <iostream>
std::mutex& get_mutex(void* p)
{
const uintptr_t mask = 0xf;
static std::mutex mutexes[ mask + 1 ];
uintptr_t index = (reinterpret_cast<uintptr_t>(p) / sizeof(void*)) & mask;
std::cerr << index << std::endl;
return mutexes[ index ];
}
struct A { void* p; };
void swap(A& a, A& b)
{
std::lock_guard<std::mutex> l1(get_mutex(&a));
std::lock_guard<std::mutex> l2(get_mutex(&b));
}
int main()
{
A a[17];
swap(a[0], a[16]); // distinct objects, but "share" a mutex
}
>>
>> 2) consider four sequences, A, B, C and D. They happen to have
>> addresses such that:
>>
>> &A.get_mutex() ==&C.get_mutex()
>> and
>> &B.get_mutex() ==&D.get_mutex()
>>
>> now, in my multithreaded program I do this concurrently:
>> swap(A, B); swap(D, C);
>> I'm not accessing any single object in more than one thread, but it
>> will deadlock.
>>
>
> Indeed
e.g this deadlocks for me:
// same definition of A, get_mutex and swap:
#include <thread>
A a[32];
void f(int i, int j) { for (int k=0; k<10; ++k) swap(a[i], a[j]); }
int main()
{
// swap distinct objects in two threads
std::thread t1(f, 0, 1);
std::thread t2(f, 17, 16);
t1.join();
t2.join();
}
>>
>> The solution is to ensure that you don't lock the same mutex twice and
>> you always lock in a consistent order
>>
e.g. this fixes both deadlocks
void swap(A& a, A& b)
{
std::mutex& m1 = get_mutex(&a);
std::mutex& m2 = get_mutex(&b);
if (&m1 < &m2)
std::lock_guard<std::mutex> l(m1);
else
{
std::lock_guard<std::mutex> l1(&m1 < &m2 ? m1 : m2);
std::lock_guard<std::mutex> l2(&m1 < &m2 ? m2 : m1);
}
}
> Ok, so except fixing this lock issue you seem to agree on the modified debug
> mode lock model, great. I will fix this issue and submit a new proposition.
Great - thanks for working on this!
Jonathan