Many times I saw code like this:
template<typename Collection>
void Foo(Collection&& c)
{
for (auto&& i : std::forward<Collection>(c))
// do something with i
}
For all STL containers (except vector<bool>) i has type of lvalue reference. Is any practical sense to type auto&& in this case?
As you said, the perfect example of where you don't have an lvalue is std::vector<bool>. Using auto& will not compile, as a prvalue is returned from the iterator.
Also, it happened to me some times to make ranges that did not returned an lvalue from its iterator.
Also, the upside of using auto&& is that there is no cases where it won't work. Even of you have a bizarre case where your iterator yield a const rvalue reference, auto&& will bind to it.
For teaching, it's also easier to tell "use auto&& in your for loops." because it will not cause copy and work everywhere.
There where also a proposal to allow implicit auto&& and enable the syntax for (x : range) (I cannot remember which one is it. If you know it, please tell me in the comments)
Related
const auto& would suffice if I want to perform read-only operations. However, I have bumped into
for (auto&& e : v) // v is non-const
a couple of times recently. This makes me wonder:
Is it possible that in some obscure corner cases there is some performance benefit in using forwarding references, compared to auto& or const auto&?
(shared_ptr is a suspect for obscure corner cases)
Update
Two examples that I found in my favorites:
Any disadvantage of using const reference when iterating over basic types?
Can I easily iterate over the values of a map using a range-based for loop?
Please concentrate on the question: why would I want to use auto&& in range-based for loops?
The only advantage I can see is when the sequence iterator returns a proxy reference and you need to operate on that reference in a non-const way. For example consider:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto& e : v)
e = true;
}
This doesn't compile because rvalue vector<bool>::reference returned from the iterator won't bind to a non-const lvalue reference. But this will work:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto&& e : v)
e = true;
}
All that being said, I wouldn't code this way unless you knew you needed to satisfy such a use case. I.e. I wouldn't do this gratuitously because it does cause people to wonder what you're up to. And if I did do it, it wouldn't hurt to include a comment as to why:
#include <vector>
int main()
{
std::vector<bool> v(10);
// using auto&& so that I can handle the rvalue reference
// returned for the vector<bool> case
for (auto&& e : v)
e = true;
}
Edit
This last case of mine should really be a template to make sense. If you know the loop is always handling a proxy reference, then auto would work as well as auto&&. But when the loop was sometimes handling non-proxy references and sometimes proxy-references, then I think auto&& would become the solution of choice.
Using auto&& or universal references with a range-based for-loop has the advantage that you captures what you get. For most kinds of iterators you'll probably get either a T& or a T const& for some type T. The interesting case is where dereferencing an iterator yields a temporary: C++ 2011 got relaxed requirements and iterators aren't necessarily required to yield an lvalue. The use of universal references matches the argument forwarding in std::for_each():
template <typename InIt, typename F>
F std::for_each(InIt it, InIt end, F f) {
for (; it != end; ++it) {
f(*it); // <---------------------- here
}
return f;
}
The function object f can treat T&, T const&, and T differently. Why should the body of a range-based for-loop be different? Of course, to actually take advantage of having deduced the type using universal references you'd need to pass them on correspondingly:
for (auto&& x: range) {
f(std::forward<decltype(x)>(x));
}
Of course, using std::forward() means that you accept any returned values to be moved from. Whether objects like this makes much sense in non-template code I don't know (yet?). I can imagine that using universal references can offer more information to the compiler to do the Right Thing. In templated code it stays out of making any decision on what should happen with the objects.
I virtually always use auto&&. Why get bitten by an edge case when you don't have to? It's shorter to type too, and I simply find it more... transparent. When you use auto&& x, then you know that x is exactly *it, every time.
const auto& would suffice if I want to perform read-only operations. However, I have bumped into
for (auto&& e : v) // v is non-const
a couple of times recently. This makes me wonder:
Is it possible that in some obscure corner cases there is some performance benefit in using forwarding references, compared to auto& or const auto&?
(shared_ptr is a suspect for obscure corner cases)
Update
Two examples that I found in my favorites:
Any disadvantage of using const reference when iterating over basic types?
Can I easily iterate over the values of a map using a range-based for loop?
Please concentrate on the question: why would I want to use auto&& in range-based for loops?
The only advantage I can see is when the sequence iterator returns a proxy reference and you need to operate on that reference in a non-const way. For example consider:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto& e : v)
e = true;
}
This doesn't compile because rvalue vector<bool>::reference returned from the iterator won't bind to a non-const lvalue reference. But this will work:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto&& e : v)
e = true;
}
All that being said, I wouldn't code this way unless you knew you needed to satisfy such a use case. I.e. I wouldn't do this gratuitously because it does cause people to wonder what you're up to. And if I did do it, it wouldn't hurt to include a comment as to why:
#include <vector>
int main()
{
std::vector<bool> v(10);
// using auto&& so that I can handle the rvalue reference
// returned for the vector<bool> case
for (auto&& e : v)
e = true;
}
Edit
This last case of mine should really be a template to make sense. If you know the loop is always handling a proxy reference, then auto would work as well as auto&&. But when the loop was sometimes handling non-proxy references and sometimes proxy-references, then I think auto&& would become the solution of choice.
Using auto&& or universal references with a range-based for-loop has the advantage that you captures what you get. For most kinds of iterators you'll probably get either a T& or a T const& for some type T. The interesting case is where dereferencing an iterator yields a temporary: C++ 2011 got relaxed requirements and iterators aren't necessarily required to yield an lvalue. The use of universal references matches the argument forwarding in std::for_each():
template <typename InIt, typename F>
F std::for_each(InIt it, InIt end, F f) {
for (; it != end; ++it) {
f(*it); // <---------------------- here
}
return f;
}
The function object f can treat T&, T const&, and T differently. Why should the body of a range-based for-loop be different? Of course, to actually take advantage of having deduced the type using universal references you'd need to pass them on correspondingly:
for (auto&& x: range) {
f(std::forward<decltype(x)>(x));
}
Of course, using std::forward() means that you accept any returned values to be moved from. Whether objects like this makes much sense in non-template code I don't know (yet?). I can imagine that using universal references can offer more information to the compiler to do the Right Thing. In templated code it stays out of making any decision on what should happen with the objects.
I virtually always use auto&&. Why get bitten by an edge case when you don't have to? It's shorter to type too, and I simply find it more... transparent. When you use auto&& x, then you know that x is exactly *it, every time.
const auto& would suffice if I want to perform read-only operations. However, I have bumped into
for (auto&& e : v) // v is non-const
a couple of times recently. This makes me wonder:
Is it possible that in some obscure corner cases there is some performance benefit in using forwarding references, compared to auto& or const auto&?
(shared_ptr is a suspect for obscure corner cases)
Update
Two examples that I found in my favorites:
Any disadvantage of using const reference when iterating over basic types?
Can I easily iterate over the values of a map using a range-based for loop?
Please concentrate on the question: why would I want to use auto&& in range-based for loops?
The only advantage I can see is when the sequence iterator returns a proxy reference and you need to operate on that reference in a non-const way. For example consider:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto& e : v)
e = true;
}
This doesn't compile because rvalue vector<bool>::reference returned from the iterator won't bind to a non-const lvalue reference. But this will work:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto&& e : v)
e = true;
}
All that being said, I wouldn't code this way unless you knew you needed to satisfy such a use case. I.e. I wouldn't do this gratuitously because it does cause people to wonder what you're up to. And if I did do it, it wouldn't hurt to include a comment as to why:
#include <vector>
int main()
{
std::vector<bool> v(10);
// using auto&& so that I can handle the rvalue reference
// returned for the vector<bool> case
for (auto&& e : v)
e = true;
}
Edit
This last case of mine should really be a template to make sense. If you know the loop is always handling a proxy reference, then auto would work as well as auto&&. But when the loop was sometimes handling non-proxy references and sometimes proxy-references, then I think auto&& would become the solution of choice.
Using auto&& or universal references with a range-based for-loop has the advantage that you captures what you get. For most kinds of iterators you'll probably get either a T& or a T const& for some type T. The interesting case is where dereferencing an iterator yields a temporary: C++ 2011 got relaxed requirements and iterators aren't necessarily required to yield an lvalue. The use of universal references matches the argument forwarding in std::for_each():
template <typename InIt, typename F>
F std::for_each(InIt it, InIt end, F f) {
for (; it != end; ++it) {
f(*it); // <---------------------- here
}
return f;
}
The function object f can treat T&, T const&, and T differently. Why should the body of a range-based for-loop be different? Of course, to actually take advantage of having deduced the type using universal references you'd need to pass them on correspondingly:
for (auto&& x: range) {
f(std::forward<decltype(x)>(x));
}
Of course, using std::forward() means that you accept any returned values to be moved from. Whether objects like this makes much sense in non-template code I don't know (yet?). I can imagine that using universal references can offer more information to the compiler to do the Right Thing. In templated code it stays out of making any decision on what should happen with the objects.
I virtually always use auto&&. Why get bitten by an edge case when you don't have to? It's shorter to type too, and I simply find it more... transparent. When you use auto&& x, then you know that x is exactly *it, every time.
const auto& would suffice if I want to perform read-only operations. However, I have bumped into
for (auto&& e : v) // v is non-const
a couple of times recently. This makes me wonder:
Is it possible that in some obscure corner cases there is some performance benefit in using forwarding references, compared to auto& or const auto&?
(shared_ptr is a suspect for obscure corner cases)
Update
Two examples that I found in my favorites:
Any disadvantage of using const reference when iterating over basic types?
Can I easily iterate over the values of a map using a range-based for loop?
Please concentrate on the question: why would I want to use auto&& in range-based for loops?
The only advantage I can see is when the sequence iterator returns a proxy reference and you need to operate on that reference in a non-const way. For example consider:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto& e : v)
e = true;
}
This doesn't compile because rvalue vector<bool>::reference returned from the iterator won't bind to a non-const lvalue reference. But this will work:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto&& e : v)
e = true;
}
All that being said, I wouldn't code this way unless you knew you needed to satisfy such a use case. I.e. I wouldn't do this gratuitously because it does cause people to wonder what you're up to. And if I did do it, it wouldn't hurt to include a comment as to why:
#include <vector>
int main()
{
std::vector<bool> v(10);
// using auto&& so that I can handle the rvalue reference
// returned for the vector<bool> case
for (auto&& e : v)
e = true;
}
Edit
This last case of mine should really be a template to make sense. If you know the loop is always handling a proxy reference, then auto would work as well as auto&&. But when the loop was sometimes handling non-proxy references and sometimes proxy-references, then I think auto&& would become the solution of choice.
Using auto&& or universal references with a range-based for-loop has the advantage that you captures what you get. For most kinds of iterators you'll probably get either a T& or a T const& for some type T. The interesting case is where dereferencing an iterator yields a temporary: C++ 2011 got relaxed requirements and iterators aren't necessarily required to yield an lvalue. The use of universal references matches the argument forwarding in std::for_each():
template <typename InIt, typename F>
F std::for_each(InIt it, InIt end, F f) {
for (; it != end; ++it) {
f(*it); // <---------------------- here
}
return f;
}
The function object f can treat T&, T const&, and T differently. Why should the body of a range-based for-loop be different? Of course, to actually take advantage of having deduced the type using universal references you'd need to pass them on correspondingly:
for (auto&& x: range) {
f(std::forward<decltype(x)>(x));
}
Of course, using std::forward() means that you accept any returned values to be moved from. Whether objects like this makes much sense in non-template code I don't know (yet?). I can imagine that using universal references can offer more information to the compiler to do the Right Thing. In templated code it stays out of making any decision on what should happen with the objects.
I virtually always use auto&&. Why get bitten by an edge case when you don't have to? It's shorter to type too, and I simply find it more... transparent. When you use auto&& x, then you know that x is exactly *it, every time.
const auto& would suffice if I want to perform read-only operations. However, I have bumped into
for (auto&& e : v) // v is non-const
a couple of times recently. This makes me wonder:
Is it possible that in some obscure corner cases there is some performance benefit in using forwarding references, compared to auto& or const auto&?
(shared_ptr is a suspect for obscure corner cases)
Update
Two examples that I found in my favorites:
Any disadvantage of using const reference when iterating over basic types?
Can I easily iterate over the values of a map using a range-based for loop?
Please concentrate on the question: why would I want to use auto&& in range-based for loops?
The only advantage I can see is when the sequence iterator returns a proxy reference and you need to operate on that reference in a non-const way. For example consider:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto& e : v)
e = true;
}
This doesn't compile because rvalue vector<bool>::reference returned from the iterator won't bind to a non-const lvalue reference. But this will work:
#include <vector>
int main()
{
std::vector<bool> v(10);
for (auto&& e : v)
e = true;
}
All that being said, I wouldn't code this way unless you knew you needed to satisfy such a use case. I.e. I wouldn't do this gratuitously because it does cause people to wonder what you're up to. And if I did do it, it wouldn't hurt to include a comment as to why:
#include <vector>
int main()
{
std::vector<bool> v(10);
// using auto&& so that I can handle the rvalue reference
// returned for the vector<bool> case
for (auto&& e : v)
e = true;
}
Edit
This last case of mine should really be a template to make sense. If you know the loop is always handling a proxy reference, then auto would work as well as auto&&. But when the loop was sometimes handling non-proxy references and sometimes proxy-references, then I think auto&& would become the solution of choice.
Using auto&& or universal references with a range-based for-loop has the advantage that you captures what you get. For most kinds of iterators you'll probably get either a T& or a T const& for some type T. The interesting case is where dereferencing an iterator yields a temporary: C++ 2011 got relaxed requirements and iterators aren't necessarily required to yield an lvalue. The use of universal references matches the argument forwarding in std::for_each():
template <typename InIt, typename F>
F std::for_each(InIt it, InIt end, F f) {
for (; it != end; ++it) {
f(*it); // <---------------------- here
}
return f;
}
The function object f can treat T&, T const&, and T differently. Why should the body of a range-based for-loop be different? Of course, to actually take advantage of having deduced the type using universal references you'd need to pass them on correspondingly:
for (auto&& x: range) {
f(std::forward<decltype(x)>(x));
}
Of course, using std::forward() means that you accept any returned values to be moved from. Whether objects like this makes much sense in non-template code I don't know (yet?). I can imagine that using universal references can offer more information to the compiler to do the Right Thing. In templated code it stays out of making any decision on what should happen with the objects.
I virtually always use auto&&. Why get bitten by an edge case when you don't have to? It's shorter to type too, and I simply find it more... transparent. When you use auto&& x, then you know that x is exactly *it, every time.