问题
I think I must be assuming something from the name boost::interprocess that is not true.
The documents repeat that named_mutex is global here.
I am unable to make it work though. Two copies of the same executable should be run at the same time, and I expect that a named mutex in a library named boost::interprocess might actually BLOCK sometimes. It doesn't. It also doesn't prevent data file corruption in the code below.
Here's some code from the boost docs:
#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;
Here's what I did to it so I could prove to myself the mutex was doing something:
#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;
}
Console Output:
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...
Also, the demo writes to a file, which if you run two copies of the program will be missing some data.
I expect that if I delete file_name and run two copies of the program, I should get interleaved writes to file_name containing 100 rows from each instance.
(Note, that the demo code is clearly not using an ofstream in append mode, instead it simply rewrites the file each time this program runs, so if we wanted a demo to show two processes writing to a file, I'm aware of that reason why it wouldn't work, but what I did expect is for the above code to be a feasible demonstration of mutual exclusion, which it is not. Also calls to a very handy and aptly named ofstream::flush() method could have been included, and weren't.)
Using Boost 1.53 on Visual C++ 2008
回答1:
It turns out that Boost is a wonderful library, and it code examples interspersed in the documentation may sometimes be broken. At least the one for boost::interprocess::named_mutex in the docs is not functional on Windows systems.
*Always deleting a mutex as part of the demo code causes the mutex to not function. *
That should be commented in the demo code at the very least. It fails to pass the "principle of least amazement", although I wondered why it was there, I thought it must be idiomatic and necessary, it's idiotic and unnecessary, in actual fact. Or if it's necessary it's an example of what Joel Spolsky would call a leaky abstraction. If mutexes are really filesystem points under C:\ProgramData in Windows I sure don't want to know about it, or know that turds get left behind that will break the abstraction if I don't detect that case and clean it up. (Sure smells like posix friendly semantics for mutexes in Boost have caused them to use a posix-style implementation instead of going to Win32 API directly and implementing a simple mutex that has no filesystem turds.)
Here's a working demo:
#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;
}
I would love critques and edits on this answer, so that people will have a working demo of using this named mutex .
To use the demo:
- Build it and run two copies of it. Pass a parameter in so you can see which instance wrote which lines (start myexename ABC and start myexename DEF from a command prompt in windows)
- If it's your second run, delete any stray output named "file_name" if you don't want the second run appended to the first.
来源：https://stackoverflow.com/questions/15314726/when-boost-library-interprocess-defines-a-named-mutex-do-those-named-mutexes-w