我想我必须从名字中假设一些boost::interprocess不正确的东西。文件在这里重复说named_mutex是全球性的。
我无法让它工作。同一个可执行文件的两个副本应该同时运行，我希望一个名为的库中的一个命名互斥锁boost::interprocess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。
这是来自 boost 文档的一些代码：
#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，所以我可以向自己证明互斥锁正在做某事：
#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副本，该文件将丢失一些数据。
我希望如果我删除file_name并运行程序的两个副本，我应该得到交错写入，file_name每个实例包含 100 行。
（注意，演示代码显然没有使用ofstream附加模式，而是在每次程序运行时简单地重写文件，所以如果我们想要演示显示两个进程写入文件，我知道这个原因为什么它不起作用,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.还可以ofstream::flush()包括对一个非常方便且恰当命名的方法的调用,但没有.)
在 Visual C++ 2008 上使用 Boost 1.53
