Is there another way to achieve the functionality of std::optional without another object being created?

Question

class ScopedClass {
 public:
  ScopedClass() {
    cout << "constructor called\n";
  }

  ~ScopedClass() {
    cout << "destructor called\n";
  }

  ScopedClass(const ScopedClass &t)
  {
      cout << "copy constructor called\n";
  }

  ScopedClass(ScopedClass &&t)
  {
      cout << "move constructor called\n";
  }
}

std::optional<ScopedClass> check_and_return(bool condition) {
  if (condition == true)
    return std::make_optional(ScopedClass());
  return std::nullopt;
}

int main() {
  auto x = check_and_return(true);
}

I want a single ScopedClass object but ScopedClass() is copied/moved when I enclose it within std::optional. This means that the destructor is called when it's copied/moved and it's being called twice. Once inside check_and_return and once when main ends.

I would like the destructor to be called only when main ends.

Is there a way to achieve this? Or some alternate method not using std::optional?

What is your problem with moving? Do you have a proven performance problem with your code when you use std::optional?. I ask because as far as I know std::optional is quite good at avoiding making copies. — Pepijn Kramer
– Pepijn Kramer, Commented Dec 13, 2023 at 16:37
Do you want just std::make_optional<ScopedClass>()? Your code is creating a temporary ScopedClass which is moved into the optional, this defeats the purpose of using make_optional — Alan Birtles
– Alan Birtles, Commented Dec 13, 2023 at 16:38

Caleth · Accepted Answer · 2023-12-13 16:41:13Z

7

You can use the in-place overload of std::make_optional rather than the move overload.

std::optional<ScopedClass> check_and_return(bool condition) {
  if (condition == true)
    return std::make_optional<ScopedClass>(); // constructs ScopedClass in place with the arguments ()
  return std::nullopt;
}

answered Dec 13, 2023 at 16:41

Caleth

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Comments

Pepijn Kramer · Accepted Answer · 2023-12-13 16:49:25Z

3

And yes there is another option, use std::make_unique/std::unique_ptr. This will not move anything and call constructor and destructor only once.

#include <cassert>
#include <memory>
#include <iostream>
#include <string>

class ScopedClass {
public:
    ScopedClass() {
        std::cout << "constructor called\n";
    }

    ~ScopedClass() {
        std::cout << "destructor called\n";
    }

    ScopedClass(const ScopedClass& t)
    {
        std::cout << "copy constructor called\n";
    }

    ScopedClass(const ScopedClass&& t)
    {
        std::cout << "move constructor called\n";
    }

private:
    std::string some_resource;
};

std::unique_ptr<ScopedClass>  check_and_return(bool condition)
{
    if ( condition )
    {
        return std::make_unique<ScopedClass>();
    }

    return nullptr;
}

int main() 
{
    auto x = check_and_return(true);
}

answered Dec 13, 2023 at 16:49

Pepijn Kramer

14k4 gold badges12 silver badges27 bronze badges

3 Comments

Jakkapong Rattananen Over a year ago

Only downside (just lazy reason) for this method. We must check return value with if(x != nullptr) not if(x) like std::optional

Pepijn Kramer Over a year ago

Not if(x) is ok. std::unique_ptr has an implicit bool conversion, that's why I proposed this solution. Also the way I understand it no function calls will be made on ScopedClass (the name seems to imply some RAII resource managment). If call needs to be made the std::optional code would also have to have the if (x) code.

Jakkapong Rattananen Over a year ago

Thank you. I used to use if (x) for check value for std::unique_ptr but it's got unexpected result. I might did something wrong.

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