问题
A range-based for statement is defined in §6.5.4 to be equivalent to:
{
auto && __range = range-init;
for ( auto __begin = begin-expr,
__end = end-expr;
__begin != __end;
++__begin ) {
for-range-declaration = *__begin;
statement
}
}
where range-init is defined for the two forms of range-based for as:
for ( for-range-declaration : expression ) => ( expression )
for ( for-range-declaration : braced-init-list ) => braced-init-list
(the clause further specifies the meaning of the other sub-expressions)
Why is __range given the deduced type auto&&? My understanding of auto&& is that it's useful for preserving the original valueness (lvalue/rvalue) of an expression by passing it through std::forward. However, __range isn't passed anywhere through std::forward. It's only used when getting the range iterators, as one of __range, __range.begin(), or begin(__range).
What's the benefit here of using the "universal reference" auto&&? Wouldn't auto& suffice?
Note: As far as I can tell, the proposal doesn't say anything about the choice of auto&&.
回答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<int>{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.
来源：https://stackoverflow.com/questions/13241108/why-does-a-range-based-for-statement-take-the-range-by-auto