Re: [boost] [optional] Types without assignment operator

Subject: Re: [boost] [optional] Types without assignment operator
From: Andrzej Krzemienski (akrzemi1_at_[hidden])
Date: 2012-11-14 07:33:28

2012/11/14 Adam Badura <abadura_at_[hidden]>

> >
>> > So instead of accessing the object directly I always access it by a
>> > function. That function first checks if the object is already >
>> constructed
>> > (boost::optional is initialized) and if so just returns it. Otherwise it
>> > first constructs it.
>> >
>>
>> I am having problems imagining the situation. Can't you change your
>> function so that returns real object rather than optional object if it is
>> initialized? Perhaps using optional reference to an object would solve
>> your
>> problem?
>>
>
> Example follows:
>
>
>>>>>>>>>>>
> class Owner {
> public:
>    Owner()
>        : m_value()
>    {
>    }
>
>    SomeType const& getValue() const {
>        if ( !m_value )
>            m_value = constructValue();
>        return *m_value;
>    }
>
> protected:
>    virtual int virtualFunction() const = 0;
>
> private:
>    SomeType constructValue() const {
>        SomeType object( /*...*/ );
>        object.callThis();
>        object.assignHere = virtualFunction();
>        return object;
>    }
>
>    mutable boost::optional< SomeType > m_value;
> };
>
> <<<<<<<<<<
>

Would the following work for you?

class Owner {
public:
   Owner() : m_value() {}

   SomeType const& getValue() const {
       if ( !m_value )
           constructValue(m_value);
       return *m_value;
   }

protected:
   virtual int virtualFunction() const = 0;

private:
   static void constructValue(SomeType & target) const {
       SomeType object( /*...*/ );
       object.callThis();
       object.assignHere = virtualFunction();
       target = boost::in_place(object);
   }

   mutable boost::optional< SomeType > m_value;
};