On 27/04/12 00:45, Scott Aron Bloom wrote:
> Nope. That was not the issue. The issue was that when you append list
> A into list B, you can't get to list A because what's inside list B is a
> copy of A, not A itself. You might want to read again from the first
> post.
>
> _______________________________________________
>
> Nope.. not at all..
>
> The problem is with your testcase Nikos... NOT implicit sharing..
>
> listOfInts.append(10);
>
> Expliticly PREVENTS the modification of the listOfInts variable.
>
> It takes in a const object reference, Guarenteeing the original will NOT
> be modified.. And implying that any modification to the original will
> not effect the copy as well..
I said:
list->append(&someOtherList);
You said:
If someOtherList goes out of scope, you will be pointing to
a deleted list.
Since QList uses implicit sharing, the cost of removing the
inner pointer isn't that much.
QList< QList< int > >
Is much safer.
I said:
Note that with QList< QList<int> > you can't modify the other
lists. You'd only be modifying the copies.
You said:
Not true at all..
I don't see what it is that we're discussing here. Your statement was
simply wrong. My suggestion to be able to modify the other list:
list->append(&someOtherList);
Your suggestion to be able to modify the other list:
list->append(someOtherList);
which doesn't work. You will *not* be able to modify 'someOtherList'
through 'list'. It's basic C++. You *need* to hold a pointer, or else
you're dealing with a copy. Whether that copy uses implicit sharing
doesn't matter at all; implicit sharing is just an optimization.
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