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

14 Nov 2012


      2012/11/14 Adam Badura <abadura@o2.pl>
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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:
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>>>>>
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;
};

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

Andrzej Krzemienski