Question

I have a base class(Base) whose constructor takes a reference as argument. In my derived class its constructor, I call the superclass-constructor and of course I need to pass a reference as argument. But I have to obtain that argument from a method of which the return type is by value…

I will give a short example:

class Base
{
public:
    Base(MyType &obj) { /* do something with the obj */}
};

class Derived : public Base
{
public:
    Derived(MyOtherType *otherType) :
         Base(otherType->getMyTypeObj()) // <--- Here is the error because (see *)
    {
         // * 
         // getMyTypeObj() returns a value and
         // the Base constructor wants a reference...
    }
};

class MyOtherType
{
public:
    MyType getMyTypeObj()
    {
         MyType obj;
         obj.setData( /* blah, blah, blah... Some data */);
         return obj; // Return by value to avoid the returned reference goes out of scope.
    }
};

How can I solve this problem?

Answer 1

Change the Base class to:
class Base
{
public:
Base(const MyType &obj) { /* do something with the obj */}
};


Update: If you want to modify obj you cannot obviously have a const reference. In that case you can either:

1)Pass the parameter by value. That will have the overhead for the copy but avoid having to free it explicitly later.

2) Change MyOtherType::getMyTypeObj() to

MyType& MyOtherType::getMyTypeObj()
{
    MyType* obj = new MyType();
    obj->setData( /* blah, blah, blah... Some data */);
    return *obj;

}

In this case, remember to delete the object after you are done with it.