Declare a variable of a class without creating an instance of it

Question

I'm learning C++.

I want to declare a variable without creating an instance of it.

MyClass variable;

// More code

int main(int argc, char **argv)
{
   // More code
   variable = MyClass(0);

   // More code
}

If I do that, and MyClass only has a constructor declared MyClass::MyClass(int value) it fails.

I need to make it global because I'm going to use it on a CallBack function and I can pass that variable as a parameter.

And also, I don't want to create an instance of the class when I declare the variable and then, another instance when I use the constructor. I think I'm wasting resources and CPU time.

Is it possible to declare a variable without instance it?

Sounds like an XY problem. Why not just declare the object when you are ready to construct it? — NathanOliver
– NathanOliver, Commented Jul 15, 2019 at 19:43
Because I need it on a CallBack function and I can pass that variable as a parameter. — VansFannel
– VansFannel, Commented Jul 15, 2019 at 19:43
If your class is not copyable and not movable, than you can not copy or move it. May be you need two-step initialization with dreaded init function? Or maybe std::optional could be a... hm... option. — SergeyA
– SergeyA, Commented Jul 15, 2019 at 19:46

John Kugelman · Accepted Answer · 2019-07-15 20:15:07Z

6

Use a pointer to delay instantiation. Prefer smart pointers to raw pointers so you don't have to worry about manual memory management.

#include <memory>

std::shared_ptr<MyClass> variable;

// More code

int main(int argc, char **argv)
{
   // More code
   variable = std::make_shared<MyClass>(0);

   // More code
}

Whether you use std::unique_ptr or std::shared_ptr depends on what you plan to do with the pointer.

edited Jul 15, 2019 at 20:15

answered Jul 15, 2019 at 20:13

John Kugelman

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1 Comment

cangozpi Over a year ago

How can we do this with regular pointers ?

Guillaume Racicot · Accepted Answer · 2019-07-15 20:18:06Z

4

You can use a std::optional for this:

#include <optional>

std::optional<MyClass> variable; // not initialized

int main() {
    if (variable) {
        // enters this if only when initialized
    }

    variable = MyClass(0);

    // Access the contained value:
    MyClass value = *variable;
    int member = variable->member;
}

The std::optional type is available in C++17. If your project is constrained to an older version, you can always use a polyfill

edited Jul 15, 2019 at 20:18

answered Jul 15, 2019 at 20:12

Guillaume Racicot

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Comments

Jasmeet Singh · Accepted Answer · 2019-07-15 20:05:40Z

1

I am not sure, but it seems like you have created a constructor like :

MyClass(int v) : value(v) {}

and not written the default constructor, which may look like :

MyClass() {}

Because of this, the compiler cannot find the constructor to instantiate a MyClass object when you write:

MyClass variable;

In short, you have to explicitly write the default constructor when you have written other constructors but also want to use the default one.

And your question is a bit vague, when you declare a variable of a class like above, you are indeed creating an instance of it. Even when doing

variable = MyClass(0);

you are creating a temporary object of MyClass and then using the assignment operator to assign it to the variable object.

answered Jul 15, 2019 at 20:05

Jasmeet Singh

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2 Comments

VansFannel Over a year ago

Yes, I don't want to create an instance when I declare it and then another instance when I use the constructor.

Jasmeet Singh Over a year ago

Yes, I think that a smart pointer is a solution then, as suggested by John.

Einar Pius · Accepted Answer · 2019-07-15 20:41:00Z

1

You could just initialise the variable with some temporary value with the existing constructor when you declare it. Ideally with a value you could easily identify when debugging.

MyClass variable(-1);

Although then it would probably make more sense to just add a default constructor that does that.

answered Jul 15, 2019 at 20:41

Einar Pius

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Comments

Francis Cugler · Accepted Answer · 2019-07-15 23:50:02Z

0

I want to declare a variable without creating an instance of it.

In your code you have:

MyClass variable;

Declared in the global namespace. Here variable is still an instance of MyClass; it just has different storage, linkage and lifetime than if it was within a function.

As to having it declared in a global namespace and initializing it afterwards here is what I have tried this and it seems to work for me. I'm not sure if this is what you are after, but I think this may work for you.

#include <iostream>

class MyClass {
private:
    int value_;

public:
    MyClass() : value_{ 0 } {}
    explicit MyClass(int value) : value_{ value } {}

    MyClass& operator()(int value) {
        value_ = value;
        return *this;
    }

    int value() const { return value_; }
};

MyClass myClass;

int main() {
    std::cout << myClass.value() << '\n';

    myClass = MyClass(2) << '\n';
    std::cout << myClass.value() << '\n';

    myClass(5);
    std::cout << myClass.value() << '\n';

    std::cout << myClass(8).value() << '\n';        

    return 0;
}

Output

Let me know what you think.

answered Jul 15, 2019 at 23:50

Francis Cugler

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1 Comment

VansFannel Over a year ago

Thanks for your answer. I don't want to create an instance of the class when I declare the variable and then, another instance when I use the constructor. I think I'm wasting resources and CPU time.

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