> Most of our implicitly shared types already had a move-assignment operator,
> because it's just
>
> QFoo &operator=(QForr &&other) { swap(other); return *this; }
And that's indeed how QVector's move assignment is implemented.
It's also broken.
Just because other is a rvalue reference does not mean that it is a temporary
and thus the is no guarantee that other's dtor gets called.
And thus, this line:
QVector<Klass> vector = std::move(something);
does not gurantees that the dtors for the old contents of vector are called.
For standard containers, this is specified in *container.requirements*
To quote, with: a being a container and rv a non-const rvalue of the same
type:
a = rv
"All existing elements of a are either move assigned to or destroyed,"
I do not see any reason why our containers should not give the same guarantee
for move assignment and consider this a bug.
daniel
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