我想我一定是从名称 boost::interprocess
中假设了一些不正确的东西。
文档重复 named_mutex
是全局的 here .
虽然我无法让它工作。同一个可执行文件的两个拷贝应该同时运行,我希望名为 boost::interprocess
的库中的命名互斥体有时实际上可能会阻塞。它没有。它也不能防止以下代码中的数据文件损坏。
这是 boost 文档中的一些代码:
#include <boost/interprocess/sync/scoped_lock.hpp>
#include <boost/interprocess/sync/named_mutex.hpp>
#include <fstream>
#include <iostream>
#include <cstdio>
int main ()
{
using namespace boost::interprocess;
try{
struct file_remove
{
file_remove() { std::remove("file_name"); }
~file_remove(){ std::remove("file_name"); }
} file_remover;
struct mutex_remove
{
mutex_remove() { named_mutex::remove("fstream_named_mutex"); }
~mutex_remove(){ named_mutex::remove("fstream_named_mutex"); }
} remover;
//Open or create the named mutex
named_mutex mutex(open_or_create, "fstream_named_mutex");
std::ofstream file("file_name");
for(int i = 0; i < 10; ++i){
//Do some operations...
//Write to file atomically
scoped_lock<named_mutex> lock(mutex);
file << "Process name, ";
file << "This is iteration #" << i;
file << std::endl;
}
}
catch(interprocess_exception &ex){
std::cout << ex.what() << std::endl;
return 1;
}
return 0;
这是我对它所做的,这样我就可以向自己证明互斥体正在做某事:
#include <windows.h>
#include <boost/interprocess/sync/interprocess_mutex.hpp>
#include <boost/lambda/lambda.hpp>
#include <boost/interprocess/sync/scoped_lock.hpp>
#include <boost/interprocess/sync/named_mutex.hpp>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <cstdio>
int main (int argc, char *argv[])
{
srand((unsigned) time(NULL));
using namespace boost::interprocess;
try{
/*
struct file_remove
{
file_remove() { std::remove("file_name"); }
~file_remove(){ std::remove("file_name"); }
} file_remover;
*/
struct mutex_remove
{
mutex_remove() { named_mutex::remove("fstream_named_mutex"); }
~mutex_remove(){ named_mutex::remove("fstream_named_mutex"); }
} remover;
//Open or create the named mutex
named_mutex mutex(open_or_create, "fstream_named_mutex");
std::ofstream file("file_name");
for(int i = 0; i < 100; ++i){
//Do some operations...
//Write to file atomically
DWORD n1,n2;
n1 = GetTickCount();
scoped_lock<named_mutex> lock(mutex);
n2 = GetTickCount();
std::cout << "took " << (n2-n1) << " msec to acquire mutex";
int randomtime = rand()%10;
if (randomtime<1)
randomtime = 1;
Sleep(randomtime*100);
std::cout << " ... writing...\n";
if (argc>1)
file << argv[1];
else
file << "SOMETHING";
file << " This is iteration #" << i;
file << std::endl;
file.flush(); // added in case this explains the corruption, it does not.
}
}
catch(interprocess_exception &ex){
std::cout << "ERROR " << ex.what() << std::endl;
return 1;
}
return 0;
}
控制台输出:
took 0 msec to acquire mutex ... writing...
took 0 msec to acquire mutex ... writing...
took 0 msec to acquire mutex ... writing...
took 0 msec to acquire mutex ... writing...
此外,该演示会写入一个文件,如果您运行该程序的两个拷贝,该文件将丢失一些数据。
我希望如果我删除 file_name
并运行该程序的两个拷贝,我应该对 file_name
进行交错写入,每个实例包含 100 行。
(请注意,演示代码显然没有在附加模式下使用 ofstream
,而是在每次程序运行时简单地重写文件,所以如果我们想要一个演示来显示两个进程写入到一个文件,我知道它不起作用的原因,但我确实期望上面的代码是互斥的可行演示,但事实并非如此。还调用了一个非常方便和恰当的方法名为 ofstream::flush()
的方法本可以包含在内,但没有包含。)
在 Visual C++ 2008 上使用 Boost 1.53
最佳答案
事实证明,Boost 是一个很棒的库,它穿插在文档中的代码示例有时可能会被破坏。至少文档中用于 boost::interprocess::named_mutex
的那个在 Windows 系统上不起作用。
*在演示代码中始终删除互斥锁会导致互斥锁不起作用。 *
至少应该在演示代码中对此进行注释。它没有通过“最小惊奇原则”,虽然我想知道它为什么在那里,我认为它必须是惯用的和必要的,它是愚蠢的和不必要的,实际上。或者,如果有必要,它是 Joel Spolsky 称之为泄漏抽象的示例。如果互斥量确实是 Windows 中 C:\ProgramData
下的文件系统点,我肯定不想知道它,或者知道如果我没有检测到它会留下大便会破坏抽象箱并清理干净。 (确实,Boost 中互斥量的 posix 友好语义使他们使用 posix 风格的实现,而不是直接转到 Win32 API 并实现一个没有文件系统垃圾的简单互斥量。)
这是一个工作演示:
#include <windows.h>
#include <boost/interprocess/sync/interprocess_mutex.hpp>
#include <boost/lambda/lambda.hpp>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <boost/interprocess/sync/scoped_lock.hpp>
#include <boost/interprocess/sync/named_mutex.hpp>
#include <fstream>
#include <iostream>
#include <cstdio>
#include <windows.h>
int main (int argc, char *argv[])
{
srand((unsigned) time(NULL));
using namespace boost::interprocess;
try{
/*
// UNCOMMENT THIS IF YOU WANT TO MAKE THIS DEMO IMPOSSIBLE TO USE TO DEMO ANYTHING
struct file_remove
{
file_remove() { std::remove("file_name"); }
~file_remove(){ std::remove("file_name"); }
} file_remover;
// UNCOMMENT THIS IF YOU WANT TO BREAK THIS DEMO HORRIBLY:
struct mutex_remove
{
mutex_remove() { named_mutex::remove("fstream_named_mutex"); }
~mutex_remove(){ named_mutex::remove("fstream_named_mutex"); }
} remover;
*/
//Open or create the named mutex
named_mutex mutex(open_or_create, "fstream_named_mutex");
std::ofstream file("file_name", std::ios_base::app );
int randomtime = 0;
for(int i = 0; i < 100; ++i){
//Do some operations...
//Write to file atomically
DWORD n1,n2;
n1 = GetTickCount();
{
scoped_lock<named_mutex> lock(mutex);
n2 = GetTickCount();
std::cout << "took " << (n2-n1) << " msec to acquire mutex";
randomtime = rand()%10;
if (randomtime<1)
randomtime = 1;
std::cout << " ... writing...\n";
if (argc>1)
file << argv[1];
else
file << "SOMETHING";
file << "...";
Sleep(randomtime*100);
file << " This is iteration #" << i;
file << std::endl;
file.flush();
}
Sleep(randomtime*100); // let the other guy in.
}
}
catch(interprocess_exception &ex){
std::cout << "ERROR " << ex.what() << std::endl;
return 1;
}
return 0;
}
我很乐意对这个答案进行评论和编辑,这样人们就会有一个使用这个命名互斥体的工作演示。
要使用演示:
- 构建它并运行它的两个拷贝。传入一个参数,以便您可以查看哪个实例写入了哪些行(start myexename ABC
和 start myexename DEF
从 Windows 中的命令提示符)
- 如果这是您的第二次运行,如果您不想将第二次运行附加到第一次运行,请删除任何名为“file_name”的杂散输出。
关于c++ - 当 boost 库 "interprocess"定义一个 named_mutex 时,这些 named_mutex 是否在不同进程之间正常工作,或者只与线程一起工作?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15314726/