Question

The following minimum code examples runs fine when it is terminated normally (by pressing enter):

#include <stdio.h>
#include <iostream>
#include <string>

using namespace std;

class SingletonLogClass
{

private:

    SingletonLogClass() {}

public:

    ~SingletonLogClass()
    {
        LogMessage("~SingletonLogClass()");
    }

    static SingletonLogClass& getInstance(void)
    {
        static SingletonLogClass inst;
        return inst;
    }

    void LogMessage(string msg)
    {
        //printf("%s\n", msg.c_str());
        cout << msg << endl;
    }

};

int main(int argc, char* argv[])
{
    SingletonLogClass::getInstance().LogMessage("This is a message");
    getchar();
    return 0;
}

When I terminate the program by closing the console window it depends on the implementation of the LogMessage function. If it’s implemented using printf everything is fine. But when it’s implemented using cout I get an access violation.

Can someone explain this?
What exactly happens when the program is terminated by closing the console window?
Why does it work with printf but not with cout?

I’m working with VC++ 2010.

Answer 1

cout is a global object. Your singleton instance (defined as static in the getInstance static member function) is also a global object.

In C++, you cannot have control over the order of construction, neither destruction of global objects. As such, it is possible that the cout object is destructed before your SingletonLogClass is. As the destructor of your SingletonLogClass logs something, it cannot use anymore cout (whereas printf is fine).

The difference in behaviour when the program is terminated normally (pressing enter) and thus exiting the main function, and when the program is terminated abruptly by closing the shell comes from the order of destruction of globals. You cannot have control over the order of destruction of global objects.