Issue
According to this talk there is a certain pitfall when using C++11 range base for
on Qt containers. Consider:
QList<MyStruct> list;
for(const MyStruct &item : list)
{
//...
}
The pitfall, according to the talk, comes from the implicit sharing. Under the hood the ranged-based for
gets the iterator from the container. But because the container is not const
the iterator will be non-const
and that is apparently enough for the container to detach.
When you control the lifetime of a container this is easy to fix, one just passes the const
reference to the container to force it to use const_iterator
and not to detach.
QList<MyStruct> list;
const Qlist<MyStruct> &constList = list;
for(const MyStruct &item : constList)
{
//...
}
However what about for
example containers as return values.
QList<MyStruct> foo() { //... }
void main()
{
for(const MyStruct &item : foo())
{
}
}
What does happen here? Is the container still copied? Intuitively I would say it is so to avoid that this might need to be done?
QList<MyStruct> foo() { //... }
main()
{
for(const MyStruct &item : const_cast<const QList<MyStruct>>(foo()))
{
}
}
I am not sure. I know it is a bit more verbose but I need this because I use ranged based for loops heavily on huge containers a lot so the talk kind of struck the right string with me.
So far I use a helper function to convert the container to the const
reference but if there is a shorter/easier way to achieve the same I would like to hear it.
Solution
template<class T>
std::remove_reference_t<T> const& as_const(T&&t){return t;}
might help. An implicitly shared object returned an rvalue can implicitly detect write-shraring (and detatch) due to non-const iteration.
This gives you:
for(auto&&item : as_const(foo()))
{
}
which lets you iterate in a const way (and pretty clearly).
If you need reference lifetime extension to work, have 2 overloads:
template<class T>
T const as_const(T&&t){return std::forward<T>(t);}
template<class T>
T const& as_const(T&t){return t;}
But iterating over const rvalues and caring about it is often a design error: they are throw away copies, why does it matter if you edit them? And if you behave very differently based off const qualification, that will bite you elsewhere.
Answered By – Yakk – Adam Nevraumont
Answer Checked By – Clifford M. (BugsFixing Volunteer)