What does `auto && e` do in range-based for-loops?
Solution 1:
When and if you should use auto&&
in for loops has been explained very nicely by Howard Hinnant here.
This leaves the question what x
in
auto &&x = ...expr...
actually is. And it is handled as if there there were a function template definition
template <class U> void f(const U& u);
and the type of x
is deduced by the same rules as u
[§7.1.6.4.(7)].
This means it is not handled as a RValue Reference, but as a "Universal/Forwarding Reference" -- the "Reference Collapsing Rules" apply.
This also holds for
const auto &&x = ...expr...
as the example in §7.1.6.4.(7) states, at least for const auto &x
.
But, as PiotrS says in the questions comments, any qualifiers nullifies the URef-ness:
no, because neither
T
intemplate<class T> void f(const T&&)
is a forwarding reference, norconst auto&&
is. The fact thatT&&
occurs in parameter declaration does not imply it is forwarding reference. Only pureT&&
with no qualifiers likeconst
orvolatile
is forwarding reference, meaning it has to betemplate<class T> void f(T&&)
orauto&&
, and neverconst T&&
orconst auto&&