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Access the owners counter used by std::recursive_mutex

I have a case where my algorithm's decisions are based on the depth of a shared std::recursive_mutex.

#include <iostream>
#include <mutex>
#include <thread>

int g_i = 0;
std::recursive_mutex g_i_mutex;

void bar() {
  std::lock_guard<std::recursive_mutex> lock(g_i_mutex);
  switch (get_counter(g_i_mutex)) { // some way to find the number of owners
    case 1: std::cout << "depth 1\n"; break;
    case 2: std::cout << "depth 2\n"; break;
    default:;
  }
}

void foo() {
   std::lock_guard<std::recursive_mutex> lock(g_i_mutex);
   std::cout << "hello\n";
   bar();
}

int main() {
  foo(); //print hello then depth 2
  bar(); //print depth 1
}

I've read that recursive mutexes hold a use count of some sort, and they increase and decrease it with each call to lock/unlock, is there a way to access that information?

like image 523
DsCpp Avatar asked Sep 26 '17 11:09

DsCpp


1 Answers

No you cannot.

This is impossible because the counter you mentioned is an implementation solution, it may or may not exist. If you knew a specific implementation of the Standard Library used a counter, you could use some magic (pointer arithmetic and casts) to get it, but that would be undefined behavior.

That being said, nothing forbids you to define your own recursive_mutex:

#include <iostream>
#include <mutex>
#include <atomic>

class recursive_mutex
{
    std::recursive_mutex _mutex;
    std::atomic<unsigned> _counter;
public:
    recursive_mutex() : _mutex(), _counter(0) {}
    recursive_mutex(recursive_mutex&) = delete;
    void operator=(recursive_mutex&) = delete;
    void lock() { _mutex.lock(); ++_counter; }
    bool try_lock() { bool result = _mutex.try_lock(); _counter += result; return result; }
    void unlock() { --_counter; _mutex.unlock(); }
    unsigned counter() { return _counter; }
};

int main() {
  recursive_mutex m;
  m.lock();
  m.lock();
  std::cout << m.counter() << "\n";
  m.unlock();
  std::cout << m.counter() << "\n";
  m.unlock();
}

2
1

demo

like image 174
YSC Avatar answered Sep 23 '22 06:09

YSC