CODEWITHSUNDEEP
c++variablesstaticdestructor
Tojmak
Tojmak

Reputation: 155

Static variable destructor

i am wondering why if i have code like this:

class Test2 {
 public:
 Test2() { std::cout << "TEST2 Constructor\n"; } 
 ~Test2() { std::cout << "TEST2 Destructor\n"; }  
};

class Test {
 public:
 static Test2& get() { static Test2 test{}; return test; }
};

int main()
{
    auto test = Test::get();
    std::cout << "Created\n";
    
    auto test1 = Test::get();
    std::cout << "Created\n";
    
    auto test2 = Test::get();
    std::cout << "Created\n";    
}

i've got output like this:

> TEST2 Constructor 
> Created 
> Created 
> Created 
> TEST2 Destructor 
> TEST2 Destructor 
> TEST2 Destructor 
> TEST2 Destructor

Why destructor is beeing called four times? Isn't there should be only one instance of Test?

Upvotes: 0

Views: 347

Answers (2)

Wander3r
Wander3r

Reputation: 1881

Destructor is being called 4 times as there are 4 objects that got created. This statement

auto test1 = Test::get();

calls the copy-constructor of Test class. You can verify by having a copy constructor with cout statement. auto resolves to Test not Test&. If you want to get the reference of the object, it has to be said explicitly

auto& test1 = Test::get();

Always follow Rule of 3/5/0 if you are definining your own implementations for destructor/copy/move constructor.

Upvotes: 1

Quentin
Quentin

Reputation: 63124

test, test1 and test2 are all instances of Test, and they get destructed at the end of main as expected. You do not see them get constructed because you didn't instrument the move-constructor with which they are initialized.

Upvotes: 0

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