Lifetime of temporary passed through function by reference

Question

I can do something like this:

const int &i = 5;

and have the lifetime of the temporary extended to the lifetime of i.

But how about

const int &fun (const int &i){
    return i;
}

int main () {
    const int &r = fun(5);
    // Can I use r here?
}

Is the lifetime of the proxy-5 still extended? Or is r a dangling reference?

5 is an rvalue. The function will create a temporary for you, but then delete it after the function is over. So yes, this is a dangling reference. — Trevor Hickey, May 29 '15 at 00:17
Possible almost-duplicate of http://stackoverflow.com/q/14735630/560648 — Lightness Races in Orbit, May 29 '15 at 00:20
@Barry: I think it's one of those "touches the same subject and basically asks the same question but isn't a 'duplicate'" situations :P Since that original question was mine I didn't want to seem improper. — Lightness Races in Orbit, May 29 '15 at 00:22
Well the "dupe" has answers that answer my question. But I would not have come up with the wording "only works once". — Baum mit Augen, May 29 '15 at 00:25

score 6 · Accepted Answer · answered May 29 '15 at 00:17

It's a dangling reference. From [class.temporary]/4-5:

There are two contexts in which temporaries are destroyed at a different point than the end of the fullexpression. The first context is when a default constructor is called [ ... ]

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:

A temporary bound to a reference member in a constructor’s ctor-initializer [ ...]

A temporary bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.

[...]

The 5 persists until the completion of the full-expression containing the call, which is to say:

const int &r = fun(5);
// <== no more 5

score 4 · Answer 2 · edited May 23 '17 at 10:31

4

No, I don't believe so. You bound 5 to the reference that's the argument to fun, so it lasts for as long as that argument lasts. The argument only lasts for the duration of the call to fun.

The relevant standard text was explored in this previous question.

edited May 23 '17 at 10:31

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answered May 29 '15 at 00:16

Lightness Races in Orbit

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Lifetime of temporary passed through function by reference

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