C++11: The range-based for statement: “range-init” lifetime?

Question

In the latest C++ standard it implies that:

for (foo : bar)
    baz;

is equivilant to:

{
    auto && r = bar;
    f         


        

Answer 1

The reasoning is not correct because boo returns a temporary object by value. Binding this temporary object to a reference implies that the lifetime of the temporary is extended. Standard quote (§ 12.2/5):

[…] The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference […]

The reasoning would be correct if boo returned a reference. An example for an expression returning a reference to a temporary is string("a") += string("b"); using this value in a range-based for loop gives rise to undefined behavior.

Answer 2

Is this reasoning correct? If not, why not?

It is correct up until this point:

And so the temporary return value of boo() is destroyed at the end of the statement "auto&&r=boo()" [...]

Binding a temporary to a reference extends its lifetime to be that of the reference. So the temporary lasts for the whole loop (that's also why there is an extra set of {} around the whole construct: to correctly limit the lifetime of that temporary).

This is according to paragraph 5 of §12.2 of the C++ standard:

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:

[various exceptions that don't apply here]

This is an interesting property that allows abusing the ranged-for loop for non-rangey things: http://ideone.com/QAXNf