Global static variable initialised with a call to static class function in c++

Question

Not sure how correctly formulate the question but here is the problem.

I have a static lib where I have the following class in a.h:

#pragma once
#include <vector>
class A{
public:
 void Run() {
  data_.push_back(10);
  std::cout << "size: " << data_.size() << std::endl;
 }
private:
 static std::vector<int> data_;
};

a.cpp is as follows:

#include "a.h"
std::vector<int> A::data_;

And I have another class in b.h:

#pragma once
#include <string>
class B
{
  public:
    static std::string Get();
};

And b.cpp:

#include "b.h"
#include "a.h"
std::string B::Get()
{
  static A a;
  a.Run();
  return "foo";
}

Now my main app which is using the above static lib is as follows:

#include <iostream>
#include "a.h"
#include "b.h"

static std::string var1= B::Get();

int main(int argc, char** argv)
{
  A a;
  a.Run();
}

Trying to understand why the output is:

size: 1

There should be a single instance of each static data member for the entire class, so there should be a single call to A::data_ constructor. Am I hitting "static initialization order fiasco"? I.e. data_ is not initialised before I use it, but then I should be getting the crash?

And now lets imagine my data_ holds dynamically initialised items (something none POD). How will it be destructed if at the end data_ holds one item, although I've inserted 2?

And that's what actually is happening in my real life code (it sometimes crashes during destruction of data_).

Getting rid of global static ( static std::string var1= B::Get(); ) solves the problem, but I still want to understand the under the hood problem.

The described case can be reproduced in VS2015 (the real life case is reproducible in gcc 6.2 )

R Sahu · Accepted Answer · 2017-04-02 17:16:25Z

1

Am I hitting "static initialization order fiasco"?

Most likely.

You can remove the problem by making the static data of a class available via a function call. E.g.

class A{
public:
 void Run() {
  getData().push_back(10);
  std::cout << "size: " << getData().size() << std::endl;
 }
private:
 static std::vector<int>& getData();
};

std::vector<int>& A::getData()
{
   static std::vector<int> data;
   return data;
}

When you do that, data will be initialized when A::getData() is called the first time. It removes the static initialization order issue completely.

answered Apr 2, 2017 at 17:16

R Sahu

207k15 gold badges168 silver badges291 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Tadzys Over a year ago

@R Sahu, yes, indeed your suggestion gets rid of the problem. Although there is one more important detail: if in destructor of a class you need to call any static function, make sure to do the same call in costructor, otherwise it might be already destructed.

R Sahu Over a year ago

@Tadzys, I don't recall construction and destruction of static objects inside functions being a problem at all, If you run into any issues, please start another post with the relevant details.

Tadzys Over a year ago

@R Sahu, lets say my getData holds static std::vector<int*>

R Sahu Over a year ago

@Tadzys, that points to poor design. It's not a good idea to store pointers in static objects. You can come up with convoluted logic to deal with them but avoid them if you can. Use std::vector<std::vector<int>> instead.

Tadzys Over a year ago

@R Sahu, lets say getData holds static std::vector<int*> and in destructor I want to clean up the vector, but getData() was called prior to constructor (constructor might not be called at all), so you will get either a leak or a crash. My specific solution was to call getData() in constructor (this way I will guarantee that destructor will be called prior to vector destructor). Of course using smart pointers helps to avoid the problem as well.

|

Collectives™ on Stack Overflow

Global static variable initialised with a call to static class function in c++

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related