blocks|key|1355385|text|回答我的问题，因为我没有在其他问题中找到其他人来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1355386|这可以通过跨平台的方式使用boost/interprocess/sync/named_mutex来实现(我使用了boost+1.63)|offset|length|style|CODE|1355387|我在Linux和Windows上进行了测试，实现防止打开应用程序的第二个实例。|1355388|我发现的问题是，如果我杀死进程(在两个平台上)，互斥对象就不会被删除，因为析构函数~MyApplication没有被调用。因此，只有在系统重新启动之后，我才能再次运行应用程序。|1355389|#include+<boost/interprocess/sync/named_mutex.hpp>
#include+<iostream>

class+MyApplication
{
public:
++++MyApplication()+=+default;

++++~MyApplication()
++++{
++++++++if+(mLockedByThisInstance)
++++++++{
++++++++++++boost::interprocess::named_mutex::remove("myApplicationMutex");
++++++++}
++++}

++++bool+IsAlreadyRunning()
++++{
++++++++mLockedByThisInstance+=+mNamedMutex.try_lock();

++++++++if+(!mLockedByThisInstance)
++++++++{
++++++++++++return+true;
++++++++}

++++++++return+false;
++++}

++++int+Run(int+argc,+char+*argv[])
++++{
++++++++//+Application+main+loop++++++++++++
++++++++return+0;
++++}

private:
++++bool+mLockedByThisInstance+=+false;
++++boost::interprocess::named_mutex+mNamedMutex{+boost::interprocess::open_or_create,
++++++++"myApplicationMutex"+};
};

int+main(int+argc,+char+*argv[])
{
++++MyApplication+myApplication;

++++if+(myApplication.IsAlreadyRunning())
++++{
++++++++std::cout+<<+"MyApplication+is+already+running!\n";
++++++++return+1;
++++}

++++return+myApplication.Run(argc,+argv);
}|code-block|syntax|javascript|1355390|entityMap^0|0|D|Z|1K|A|0|2|5|8|7|0|15|E|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@$D|Z|E|10|F|G]|$D|11|E|12|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|13|8|@$D|14|E|15|F|G]]|9|@]|A|$]]|$1|L|3|M|5|N|7|16|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

Answering my question, as I didn't find someone else addressing this in the other questions.

This can be achieved in a cross platform way using <code>boost/interprocess/sync/named_mutex</code> (I used <code>boost 1.63</code>)

I tested on both <code>Linux</code> and <code>Windows</code> and the implementation prevents opening a second instance of the application. 

The problem that I found is that if I kill the process (on both platforms), the mutex is not removed because the destructor <code>~MyApplication</code> is not called. So only after system restart I will be able to run the application again.

<pre><code>#include &lt;boost/interprocess/sync/named_mutex.hpp&gt;
#include &lt;iostream&gt;

class MyApplication
{
public:
 MyApplication() = default;

 ~MyApplication()
 {
 if (mLockedByThisInstance)
 {
 boost::interprocess::named_mutex::remove("myApplicationMutex");
 }
 }

 bool IsAlreadyRunning()
 {
 mLockedByThisInstance = mNamedMutex.try_lock();

 if (!mLockedByThisInstance)
 {
 return true;
 }

 return false;
 }

 int Run(int argc, char *argv[])
 {
 // Application main loop 
 return 0;
 }

private:
 bool mLockedByThisInstance = false;
 boost::interprocess::named_mutex mNamedMutex{ boost::interprocess::open_or_create,
 "myApplicationMutex" };
};

int main(int argc, char *argv[])
{
 MyApplication myApplication;

 if (myApplication.IsAlreadyRunning())
 {
 std::cout &lt;&lt; "MyApplication is already running!\n";
 return 1;
 }

 return myApplication.Run(argc, argv);
}
</code></pre>

blocks|key|51405|text|这并不是一个完整的答案，但是要查看一些内容，比如保存特定的文件句柄以进行独占访问。您可以像互斥一样使用它，但是操作系统应该自动“清理”进程终止时的句柄。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|51406|我不能告诉您，这在Linux上是否有效，但至少在Windows上，这应该会使您达到同样的效果。希望这能有所帮助。|51407|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

This is not really a complete answer, but look into something like holding a specific file handle for exclusive access. You can use this like a mutex, but with the added benefit that the OS should "clean up" the handle on process termination automatically.

I cannot speak to if this will work on Linux, but at least on Windows, this should let you achieve the same effect. Hope that helps.

These questions <a href="https://stackoverflow.com/questions/171213/how-to-block-running-two-instances-of-the-same-program">How to block running two instances of the same program?</a> , <a href="https://stackoverflow.com/questions/5339200/how-to-create-a-single-instance-application-in-c-or-c">How to create a single instance application in C or C++</a> and <a href="https://stackoverflow.com/questions/8799646/preventing-multiple-instances-of-my-application">Preventing multiple instances of my application</a> address how to prevent opening multiple instances of the same application for a platform specifically (<code>Windows</code>, <code>Linux</code>).

Is there a cross platform way to achieve this?

Cross platform way to prevent opening multiple instances of an application

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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这些问题包括、和如何防止为平台(Windows、Linux)打开同一个应用程序的多个实例。是否有一个跨平台的方法来实现这一点？

问防止打开应用程序的多个实例的跨平台方法
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