Why does a range-based for statement take the range by auto&&?

Question

A range-based for statement is defined in §6.5.4 to be equivalent to:

{
  auto && __range = range-init;
  for ( auto __begin =          


        

Answer 1

Wouldn't auto& suffice?

No, it wouldn't. It wouldn't allow the use of an r-value expression that computes a range. auto&& is used because it can bind to an l-value expression or an r-value expression. So you don't need to stick the range into a variable to make it work.

Or, to put it another way, this wouldn't be possible:

for(const auto &v : std::vector{1, 43, 5, 2, 4})
{
}

Wouldn't const auto& suffice?

No, it wouldn't. A const std::vector will only ever return const_iterators to its contents. If you want to do a non-const traversal over the contents, that won't help.