问题描述
根据,有一个在Qt容器上使用C ++ 11范围库 for
时的某些缺陷。考虑:
QList< MyStruct>列表;
pre>
for(const MyStruct& item:list)
{
// ...
}
根据谈话,陷阱来自隐式共享。在引擎盖下,基于范围的获取从容器的迭代器。但是因为容器不是const,所以interator将是非常量,这显然足以使容器分离。
当你控制容器的生命周期时,这是容易修复,一个只是通过const引用容器,强制它使用const_iterator和不分离。
QList< MyStruct>列表;
pre>
const Qlist< MyStruct> & constList = list;
for(const MyStruct& item:constList)
{
// ...
}
但是例如容器作为返回值。
QList< MyStruct> foo(){// ...}
void main()
{
for(const MyStruct& item:foo())
{
}
}
容器是否仍被复制?直观地,我会说这是为了避免这可能需要做。
QList< MyStruct> foo(){// ...}
main()
{
for(const MyStruct& item:const_cast< const QList< MyStruct> )))
{
}
}
当然。我知道这是一个有点更冗长,但我需要这是因为我使用范围基于对巨大的容器上的循环很多,所以这种谈话类型与我正确的字符串。
到目前为止,我使用一个辅助函数将容器转换为const引用,但如果有一个更简单/更短的方法来实现相同的,我想听到它。
解决方案模板< class T&
std :: remove_reference_t< T>常数& as_const(T& t){return t;}
隐式共享对象返回的值可以隐式地检测由于非const迭代造成的写剔除(和detatch)。
这会给你:
for(auto&& item:as_const(foo()))
{
}
它可以让你以一种const方式(非常清楚)进行迭代。
如果您需要参考有效期延长工作,有2个重载:
模板< class T&
T const as_const(T& t){return std :: forward< T>(t);}
template< class T>
T const& as_const(T& t){return t;}
但是遍历const值和关心它经常一个设计错误:他们被丢弃的副本,为什么它重要,如果你编辑他们?如果你的行为方式不同,基于const限定,那将在其他地方咬你。
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 interator 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 easier/shorter way to achieve the same I would like to hear it.
解决方案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.
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