Will range based for loop in c++ preserve the index order

Question

In c++11, if I use a range based for loop on vector, will it guarantee the iteration order?

i.e. are the following code blocks guaranteed to produce the same output?

vector<T> output;
vector<U> V;
for( auto v: V) output.push_back(f(v));

vs

for(int i =0; i < V.size(); ++i) output.push_back(f(V[i])); 

what if it is not vector but map, etc?

Answer 1

Yes the two codes are guaranteed to do the same. Though I don't have a link to the standard you can have a look here. I quote: You can read that as "for all x in v" going through starting with v.begin() and iterating to v.end().

Answer 2

Yes, they are equivalent. The standard guarantees in 6.5.4:

For a range-based for statement of the form

for ( for-range-declaration : expression ) statement

let range-init be equivalent to the expression surrounded by parentheses ( expression )

and for a range-based for statement of the form

for ( for-range-declaration : braced-init-list ) statement

let range-init be equivalent to the braced-init-list. In each case, a range-based for statement is equivalent to

{
  auto && __range = range-init;
  for ( auto __begin = begin-expr,
      __end = end-expr;
      __begin != __end;
      ++__begin ) {
    for-range-declaration = *__begin;
    statement
  }
}

where __range, __begin, and __end are variables defined for exposition only, and _RangeT is the type of the expression, and begin-expr and end-expr are determined as follows:

— if _RangeT is an array type, begin-expr and end-expr are __range and __range + __bound, respectively, where __bound is the array bound. If _RangeT is an array of unknown size or an array of incomplete type, the program is ill-formed;

— if _RangeT is a class type, the unqualified-ids begin and end are looked up in the scope of class _RangeT as if by class member access lookup (3.4.5), and if either (or both) finds at least one declaration, begin-expr and end-expr are __range.begin() and __range.end(), respectively;

— otherwise, begin-expr and end-expr are begin(__range) and end(__range), respectively, where begin and end are looked up with argument-dependent lookup (3.4.2). For the purposes of this name lookup, namespace std is an associated namespace.

Though your question about map is a bit nonsensical. If it's an ordered map and you iterate through the map properly, then they're equivalent. If it's an unordered map then your question doesn't really make much sense.