多次运行的持久存储

发布于 2024-11-06 21:50:48 字数 828 浏览 4 评论 0原文

我想知道在不使用文件系统或外部数据库的输入输出的情况下，拥有一个存储容器在多次执行（运行）后不会丢失其内容的最佳解决方案是什么。

假设我有一个存储整数的 foo() 类。我想从 main() 调用一个添加整数的方法，并且该类不会忘记其以前的内容。

//
// Data storage accross different runs
// This should go into the daemon process
//

#include<iostream>
#include<list>

using namespace std;

class foo {
public:
  foo(int add): add(add) {}
  void store(int i) {
    vec.push_back( i + add);
  }
private:
  list<int> vec;
  int       add;
};

main 函数应该检查是否有已经运行的守护进程 - 如果没有启动它。

//
// Main program. Should check whether daemon runs already, if not starts it.
//

void main(int argc, char *argv[]) {

  // if (daemon is not running)
  //  start daemon( some_number )

  // call daemon::add( atoi(argv[1]) );
}

如何使用共享库或守护进程来最好地做到这一点？存储和调用程序位于同一台Linux主机上。

原文

I was wondering what would be the best solution for having a storage container that does not loose its contents over several execution times (runs) without using input-output to the filesystem or external database.

Say I have a class foo() which stores integers. From main() I want to call a method that adds an integer and the class does not forget about its former contents.

//
// Data storage accross different runs
// This should go into the daemon process
//

#include<iostream>
#include<list>

using namespace std;

class foo {
public:
  foo(int add): add(add) {}
  void store(int i) {
    vec.push_back( i + add);
  }
private:
  list<int> vec;
  int       add;
};

The main function should check for an already running daemon - if not starts it.

//
// Main program. Should check whether daemon runs already, if not starts it.
//

void main(int argc, char *argv[]) {

  // if (daemon is not running)
  //  start daemon( some_number )

  // call daemon::add( atoi(argv[1]) );
}

How would one do this best with shared libraries or with a daemon process? Storage and caller program are on the same Linux host.

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