How to return a reference of a QPointer ?

[jezz]

Hi, I need to do something like this:

class Apple {
private:
  QPointer<Tree> m_tree;
public:
  Tree &tree();
}
 
Tree &Apple::tree()
{
  return m_tree;
}

class Apple {
private:
  QPointer<Tree> m_tree;
public:
  Tree &tree();
}

Tree &Apple::tree()
{
  return m_tree;
}

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But the compiler says this for the line 'return m_tree':
error C2440: 'return' : cannot convert from 'QPointer<T>' to 'Tree &'

How to change the line to make it work?
Thanks.

[Zlatomir]

What "work" means, what are you trying to achieve.
Do you want to return a QPointer<Tree> or a dereferenced QPointer?

[jezz]

Thanks, just want to return a reference to m_tree.

[Zlatomir]

Why not returning the m_tree pointer? Since it's just a pointer not the whole object?
Always try to keep the code simple.

QPointer<Tree>Apple::tree()
{
  return m_tree;
}

QPointer<Tree>Apple::tree()
{
  return m_tree;
}

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[jezz]

What about something like :

return *m_tree;

According to the documentation here http://doc.qt.nokia.com/4.7/qpointer.html , operator* should return a reference.

Thanks, this works.

Why not returning the m_tree pointer? Since it's just a pointer not the whole object?
Always try to keep the code simple.

QPointer<Tree>Apple::tree()
{
  return m_tree;
}

QPointer<Tree>Apple::tree()
{
  return m_tree;
}

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Thanks, but the usage would look like this

QPointer<Tree> blah = apple->tree();

QPointer<Tree> blah = apple->tree();

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It is cleaner like this:

Tree blah = apple->tree();

Tree blah = apple->tree();

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[Zlatomir]

Well, you said you need the pointer when i asked the first question (what you are trying to achieve), i asked that because you were somewhere in the "middle" between the two solutions.

LE: for that you didn't really needed the reference return. Since you are coping the object you can return the value.

[stampede]

It is cleaner like this:

Tree blah = apple->tree();

Tree blah = apple->tree();

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You can always use typedef:

typedef QPointer<Tree> TreePtr;
TreePtr p = apple->tree();

typedef QPointer<Tree> TreePtr;
TreePtr p = apple->tree();

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Looks clean enough to me And its just a pointer, so it does not require to copy the whole tree

[jezz]

Well, you said you need the pointer when i asked the first question (what you are trying to achieve), i asked that because you were somewhere in the "middle" between the two solutions.

LE: for that you didn't really needed the reference return. Since you are coping the object you can return the value.

I mean &m_tree (reference to m_tree), not pointer.

You can always use typedef:

typedef QPointer<Tree> TreePtr;
TreePtr p = apple->tree();

typedef QPointer<Tree> TreePtr;
TreePtr p = apple->tree();

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Looks clean enough to me And its just a pointer, so it does not require to copy the whole tree

It would need lots of typedef if there are lots of pointers to different classes.

[Zlatomir]

I mean &m_tree (reference to m_tree), not pointer.

Q: Now you have a reference to m_tree? A: NO
You have a copy of a Tree object.

You use pointers or references to prevent the copy of whole objects when you pass them from one place to another.

[stampede]

Tree blah = apple->tree();

Tree blah = apple->tree();

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There is no difference here if tree() returns a reference or value, each time Tree::Tree(const Tree& t) copy constructor is invoked.
If you want to prevent copying, return a pointer or:

Tree& blah = apple->tree();

Tree& blah = apple->tree();

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It would need lots of typedef if there are lots of pointers to different classes.

One per class, not that much

[jezz]

My bad, I should have put it this way

Tree& blah = apple->tree();

Tree& blah = apple->tree();

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[Zlatomir]

Just be careful so that you don't end-up with "dangling" references.

The references can't be NULL (that is why some assume they are safer than pointers) but with the help of pointers they can get invalid...

[mpalomas]

What about something like :

return *m_tree;

According to the documentation here http://doc.qt.nokia.com/4.7/qpointer.html , operator* should return a reference.