Is reference binding to a temporary of a temporary undefined behavior?

Question

In the following the snippet, is the lifetime of the sphere extended in such a way that the value of r is not undefined?

 struct Sphere {
   auto& radius() const { return _radius;}
   float _radius{};
 };
 struct Capsule {
   auto sphere() const { return Sphere{12.0}; }
 };
 auto func() {
   auto capsule = Capsule{};
   const auto& r = capsule.sphere().radius();
   std::cout << r;
 }

I know that const-references extend the life time of a temporary, but I'm not sure what happens if a member of a temporary is bound to.

Note: I'm very suspicious that the equivalent of this snippet is causing a bug for me, but neither Clang nor Visual Studio issues a warning.

Same thing as what I observed in [this question of mine](https://stackoverflow.com/q/38339023/1896169). I think there's a dupe somewhere... — Justin, Mar 13 '18 at 20:29
Possible duplicate of [const reference to a temporary object becomes broken after function scope (life time)](https://stackoverflow.com/questions/17236007/const-reference-to-a-temporary-object-becomes-broken-after-function-scope-life) — Justin, Mar 13 '18 at 20:32
In this case, yes `r` is dangling. The lifetime of the object returned from `sphere()` is not extended to the entire lifetime of `r` — Justin, Mar 13 '18 at 20:35
@Justin Yes, technically a duplicate but I think the problem is easier to grasp in this question as it does not involve std::map — Viktor Sehr, Mar 13 '18 at 21:06
This is some of similar with [this question](https://stackoverflow.com/questions/1374468/stringstream-string-and-char-conversion-confusion) — Khurshid Normuradov, Mar 14 '18 at 06:25

score 5 · Accepted Answer · answered Mar 13 '18 at 20:38

In order for lifetime extension to occur, the reference must bind to a prvalue (which is then materialized into a temporary object so that the reference has something to bind to). However, capsule.sphere().radius() is not a prvalue; it is an lvalue that refers to Sphere::_radius. Therefore lifetime extension does not occur, and the Sphere object together with its _radius member will be destroyed at the end of the full-expression.

score 1 · Answer 2 · answered Mar 13 '18 at 20:38

The lifetime of the temporary Sphere-object is extended only until the completion of the full expression containing the call. The full expression is const auto& r = capsule.sphere().radius(); and not longer, such that r lives longer than the full expression and it's temporary. So I'd clearly vote for undefined behaviour.

Is reference binding to a temporary of a temporary undefined behavior?

2 Answers2