How to return a reference of a QPointer ?

Hi, I need to do something like this:

Code:

---

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

---

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.

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

Thanks, just want to return a reference to m_tree.

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.

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

Code:

---

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

---

Quote:

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Originally Posted by 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.

---

Thanks, this works.

Quote:

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Originally Posted by 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.

Code:

---

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

---

---

Thanks, but the usage would look like this

Code:

---

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

---

It is cleaner like this:

Code:

---

Tree blah = apple->tree();

---

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.

Quote:

---

It is cleaner like this:

Code:

---

Tree blah = apple->tree();

---

---

You can always use typedef:

Code:

---

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

---

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

Quote:

---

Originally Posted by 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.

---

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

Quote:

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Originally Posted by stampede

You can always use typedef:

Code:

---

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

---

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

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

Quote:

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Originally Posted by jezz

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

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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.

Code:

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Tree blah = apple->tree();

---

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:

Code:

---

Tree& blah = apple->tree();

---

Quote:

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

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One per class, not that much :)

My bad, I should have put it this way :)

Code:

---

Tree& blah = apple->tree();

---

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...

Quote:

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It is cleaner like this:
Qt Code:
Switch view

Tree blah = apple->tree();

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Even if it is Tree& blah = apple->tree();
I would disagree it is cleaner code.
If you use pointers, you know you are modifying something which you haven't created. But in case of reference it is hard to know.

Also if references are returned, I would prefer to use it on left hand side. For eg if you have QList<int> numbers;
Then having a reference with [] operator,,,, I can assign value like - numbers[index] = 10;
Lets say I wrote code like -

Code:

---

int & temp = numbers[index];
...
...  // assume you have many lines of code here...
temp = 10;  // Can you still know you are actually modifying something ? oops ,, did you forget ?
// numbers[index] = 10;  // with this line you know you are modifying something..

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Rest depends on personal choice :)

Uh - what's the point of this exercise? You declare m_tree private, but then you hand over a direct reference to it whenever the user asks. There's no point in making m_tree private in such a case, since you throw away privacy by passing back the object.

Here's a more sane solution that accomplishes exactly the same thing with far less code: make m_tree public and access it directly via apple.m_tree. There's no difference between that and the far more convoluted constructs you're wrestling with above.