Range based for loop on temporary

Question

Consider

std::vector<abc> fun() { return a;}
..
..

for( auto itr = fun().begin(); itr< fun().end(); ++itr) //A
 {
  ..
 }
for( auto & itr : fun()) //B
 {
 }

Are both loops here are unsafe? (iterators not being compatible?)

The 2nd version can't work. Both calls to `fun()` returns a different `std::vector`. You cannot compare iterators from different containers. — François Andrieux, Jun 02 '17 at 17:59
the ranged for you wrote is using vs's extention, it should be const reference — Sopel, Jun 02 '17 at 18:00
(I read this questions as you being aware the first variant is wrong.) — Baum mit Augen, Jun 02 '17 at 18:00
@DanielSchepler It is 100% safe. You should reread the reference: http://en.cppreference.com/w/cpp/language/range-for — NathanOliver, Jun 02 '17 at 18:07
@NathanOliver Oh right, I was confusing it with the case `boost::optional f(); ... for (auto c : *(f())) { ... }` which isn't safe because `boost::optional::operator*() &&` returns a `T&&` but the `f()` temporary then isn't lifetime-extended. Sorry for the confusion. — Daniel Schepler, Jun 02 '17 at 18:13

Cory Kramer · Accepted Answer · 2017-06-02T18:07:58.007

This is undefined behavior.

for( auto itr = fun().begin(); itr< fun().end(); ++itr) //A
{
    ..
}

The std::vector from fun().begin() will be a completely different std::vector than that returned from fun().end().

The second version (B) will work fine as described in this post.

for (auto & itr : fun()) //B
{
}

1 Answers1