重写新运算符时,std :: mutex锁会挂起

Question

我们有一个内部存储器管理器,我们使用其中一个产品.内存管理器会覆盖new和delete运算符,并且在单线程应用程序中运行良好.但是,我现在的任务是使其适用于多线程应用程序.根据我的理解,下面的伪代码应该可以工作,但它会挂起,即使是try_lock().有任何想法吗？

更新#1

导致"访问冲突":

#include <mutex>

std::mutex g_mutex;

/*!
\brief Overrides the Standard C++ new operator
\param size [in] Number of bytes to allocate
*/
void *operator new(size_t size)
{
   g_mutex.lock(); // Access violation exception
   ...   
}

导致线程在旋转中永远挂起:

#include <mutex>

std::mutex g_mutex;
bool g_systemInitiated = false;


/*!
\brief Overrides the Standard C++ new operator
\param size [in] Number of bytes to allocate
*/
void *operator new(size_t size)
{
   if (g_systemInitiated == false) return malloc(size);
   g_mutex.lock(); // Thread hangs forever here. g_mutex.try_lock() also hangs
   ...   
}

int main(int argc, const char* argv[])
{
   // Tell the new() operator that the system has initiated
   g_systemInitiated = true;
   ...
}

更新#2

递归互斥锁也会导致线程在旋转中永久挂起:

#include <mutex>

std::recursive_mutex g_mutex;
bool g_systemInitiated = false;


/*!
\brief Overrides the Standard C++ new operator
\param size [in] Number of bytes to allocate
*/
void *operator new(size_t size)
{
   if (g_systemInitiated == false) return malloc(size);
   g_mutex.lock(); // Thread hangs forever here. g_mutex.try_lock() also hangs
   ...   
}

int main(int argc, const char* argv[])
{
   // Tell the new() operator that the system has initiated
   g_systemInitiated = true;
   ...
}

更新#3

乔纳森威克利建议我应该尝试unique_lock和/或lock_guard,但锁仍然在旋转.

unique_lock 测试:

#include <mutex>

std::mutex g_mutex;
std::unique_lock<std::mutex> g_lock1(g_mutex, std::defer_lock);
bool g_systemInitiated = false;

/*!
\brief Overrides the Standard C++ new operator
\param size [in] Number of bytes to allocate
*/
void *operator new(size_t size)
{
   if (g_systemInitiated == false) return malloc(size);
   g_lock1.lock(); // Thread hangs forever here the first time it is called
   ...   
}

int main(int argc, const char* argv[])
{
   // Tell the new() operator that the system has initiated
   g_systemInitiated = true;
   ...
}

lock_guard 测试:

#include <mutex>

std::recursive_mutex g_mutex;
bool g_systemInitiated = false;


/*!
\brief Overrides the Standard C++ new operator
\param size [in] Number of bytes to allocate
*/
void *operator new(size_t size)
{
   if (g_systemInitiated == false) return malloc(size);
   std::lock_guard<std::mutex> g_lock_guard1(g_mutex); // Thread hangs forever here the first time it is called
   ...   
}

int main(int argc, const char* argv[])
{
   // Tell the new() operator that the system has initiated
   g_systemInitiated = true;
   ...
}

我认为我的问题是delete在锁定时由C++ 11互斥库调用.delete也像这样被覆盖:

/*!
\brief Overrides the Standard C++ new operator
\param p [in] The pointer to memory to free
*/
void operator delete(void *p)
{
    if (g_systemInitiated == false)
    {
       free(p); 
    }
    else
    {
       std::lock_guard<std::mutex> g_lock_guard1(g_mutex);
       ...
    }
}

这导致死锁情况,我看不到任何好的解决方案除了使我自己的锁定不会产生任何调用new或delete锁定或解锁时.

更新#4

我已经实现了我自己的自定义递归互斥锁,它没有调用,new或者delete它允许同一个线程进入锁定的块.

#include <thread>

std::thread::id g_lockedByThread;
bool g_isLocked = false;
bool g_systemInitiated = false;

/*!
\brief Overrides the Standard C++ new operator
\param size [in] Number of bytes to allocate
*/
void *operator new(size_t size)
{
   if (g_systemInitiated == false) return malloc(size);

   while (g_isLocked && g_lockedByThread != std::this_thread::get_id());
   g_isLocked = true; // Atomic operation
   g_lockedByThread = std::this_thread::get_id();
   ...   
   g_isLocked = false;
}

/*!
\brief Overrides the Standard C++ new operator
\param p [in] The pointer to memory to free
*/
void operator delete(void *p)
{
    if (g_systemInitiated == false)
    {
       free(p); 
    }
    else
    {
       while (g_isLocked && g_lockedByThread != std::this_thread::get_id());
       g_isLocked = true; // Atomic operation
       g_lockedByThread = std::this_thread::get_id();
       ...   
       g_isLocked = false;
    }
}

int main(int argc, const char* argv[])
{
   // Tell the new() operator that the system has initiated
   g_systemInitiated = true;
   ...
}

更新#5

尝试了Jonathan Wakely的建议,并发现微软实施C++ 11 Mutexs似乎有些不对劲; 如果使用/MTd(多线程调试)编译器标志编译,他的示例会挂起,但如果使用/MDd(多线程调试DLL)编译器标志编译,则可以正常工作.正如Jonathan正确指出的那样,std::mutex实现应该是constexpr.这是我用来测试实现问题的VS 2012 C++代码:

#include "stdafx.h"

#include <mutex>
#include <iostream>

bool g_systemInitiated = false;
std::mutex g_mutex;

void *operator new(size_t size)
{
    if (g_systemInitiated == false) return malloc(size);
    std::lock_guard<std::mutex> lock(g_mutex);
    std::cout << "Inside new() critical section" << std::endl;
    // <-- Memory manager would be called here, dummy call to malloc() in stead
    return malloc(size);
}

void operator delete(void *p)
{
    if (g_systemInitiated == false) free(p); 
    else
    {
        std::lock_guard<std::mutex> lock(g_mutex);
        std::cout << "Inside delete() critical section" << std::endl;
        // <-- Memory manager would be called here, dummy call to free() in stead
        free(p);
    }
}

int _tmain(int argc, _TCHAR* argv[])
{
    g_systemInitiated = true;

    char *test = new char[100];
    std::cout << "Allocated" << std::endl;
    delete test;
    std::cout << "Deleted" << std::endl;

    return 0;
}

更新#6

向Microsoft提交了一个错误报告:https: //connect.microsoft.com/VisualStudio/feedback/details/776596/std-mutex-not-a-constexpr-with-mtd-compiler-flag#details

Answer 1

互斥体库使用new, std::mutex 默认情况下不是递归的（即可重入的）。一个先有鸡还是先有蛋的问题。

更新正如下面的评论所指出的，使用 std::recursive_mutex 可能有效。但是，经典的 C++ 全局变量静态初始化顺序定义不明确的问题依然存在，外部访问全局互斥锁的危险也依然存在（最好将其放在匿名命名空间中。）

更新2您可能太早将g_systemInitiated 切换为true，即在互斥体有机会完成其初始化之前，因此“首次通过”调用永远不会发生malloc()。要强制执行此操作，您可以尝试通过调用分配器模块中的初始化函数来替换 main() 中的分配：

namespace {
    std::recursive_mutex    g_mutex;
    bool                    g_initialized = false;
}
void initialize()
{
    g_mutex.lock();
    g_initialized = true;
    g_mutex.unlock();
}