How to implement a Singleton in an application wit

2020-04-17 06:24发布

问题:

I have an application (in MS Visual Studio) that contains 3 projects:

  • main (the one that contains the main function)
  • device (models some hardware device)
  • config (contains some configuration for both other projects)

So the dependency graph is:

  • main depends on device, which depends on config
  • main depends on config

The config project contains a Singleton, which holds some configuration parameters.

I decided to turn the device project into a DLL. When i did this, it seems that i got two instances of the Singleton in the config project! I guess this is a classic problem, which might have a good solution. So how can i fix this?

I reproduced the problem with the following (relatively small) code. Of course, in my case there are some 30 projects, not just 3. And i would like to make just 1 DLL (if possible).


// config.h
#pragma once
#include <string>
#include <map>
class Config
{
public:
    static void Initialize();
    static int GetConfig(const std::string& name);

private:
    std::map<std::string, int> data;
};

// config.cpp
#include "config.h"

static Config g_instance;

void Config::Initialize()
{
    g_instance.data["one"] = 1;
    g_instance.data["two"] = 2;
}

int Config::GetConfig(const std::string& name)
{
    return g_instance.data[name];
}

// device.h
#pragma once

#ifdef _DLL
#define dll_cruft __declspec( dllexport )
#else
#define dll_cruft __declspec( dllimport )
#endif

class dll_cruft Device
{
public:
    void Work();
};

// device.cpp
#include "device.h"
#include <iostream>
#include "config.h"

void Device::Work()
{
    std::cout << "Device is working: two = " << Config::GetConfig("two") << '\n';
}

// main.cpp
#include <iostream>
#include "config.h"
#include "device.h"

int main()
{
    std::cout << "Before initialization in application: one = " << Config::GetConfig("one") << '\n';
    Config::Initialize();
    std::cout << "After initialization in application: one = " << Config::GetConfig("one") << '\n';
    Device().Work();
    std::cout << "After working in application: two = " << Config::GetConfig("two") << '\n';
}

Output:

Before initialization in application: one = 0

After initialization in application: one = 1

Device is working: two = 0

After working in application: two = 2

Some explanations on what the code does and why:

  1. Main application starts
  2. The first print is just to show that the singleton is not initialized yet
  3. Main application initializes the singleton
  4. The first print shows that the initialization worked
  5. Main application starts the "hardware device"
  6. Inside the DLL, the singleton is not initialized! I expect it to output two = 2
  7. The last print shows that the singleton is still initialized in main application

回答1:

You can decide where singleton should reside and then expose it to other consumers.


Edited by OP:

For example, i want that the config instance appear only in the EXE (not DLL).

  1. Turn the instance into a pointer

    static Config* g_instance;
    
  2. Add a separate initializing function to device's exported functions:

    void InitializeWithExisting(Config* instance) {g_instance=instance;}
    
  3. After initializing the singleton normally, use the second initialization:

    Config::Initialize();
    Config::InitializeWithExisting();
    


回答2:

When I ran into this same problem I solved it by creating another DLL whose sole purpose is to manage the singleton instance. All attempts to get a pointer to the singleton call the function inside this new DLL.



回答3:

I believe that defining and accessing singleton instance this way might solve your problem:

Config& getInstance()
{
  static Config config;
  return config;
}

This way you also don't need to have (and call) the Initialize method, you can use constructor for initializing, that will be called automatically when you call getInstance for the first time.