value-initialization of a const member reference

Question

I am looking at code of the following form:

class foo
{
  public:
    foo() {}

  //...
};

class bar
{
  public:
    bar() : ref() {}

  private:
    const foo &ref;
};

Is initializing a reference using a temporary in this way correct? I know that it is possible to initialize a const reference that's a local variable with a temporary, and that doing so extends the lifetime of the temporary, e.g.

const foo &tmp = funcThatReturnsByValue(); //OK

However, one of the answers to the related initialize reference in initialization list suggests that there is a difference between "short-lived" and "long-lived" references, and that initializing ref as above is undefined behavior (even though ref is a const reference).

12.2.5 in the standard says, in part, "A temporary bound to a reference member in a constructor's ctor-initializer persists until the constructor exits." Is that describing this situation?

Could you fix your code so that it actually involves a temporary? — Cheers and hth. - Alf, Nov 10 '10 at 19:34
ref() doesn't bind ref to a temporary instance of foo? Sorry if my terminology is imprecise... I'm doing my best to understand this code snippet. — user168715, Nov 10 '10 at 19:42
Maybe this comment is worthless (I don't know the context you're looking at this code in), but the problem disappears if `ref` is a pointer. The point is that references cannot be value-initialized, but pointers can (they get zero-initialized). Moreover, most compilers (at least MSVC) issue a warning for being unable to generate a default assignment operator. As a guideline, whenever you want a reference member, you want actually a pointer... — Alexandre C., Nov 10 '10 at 21:03
The extension of the temporary's lifetime doesn't seem to be so dependable, after all: http://stackoverflow.com/questions/2604206/c-constant-reference-lifetime. It seems to be concocted in order to allow temporaries created during type conversions to work correctly, but isn't intended for other uses. — rwong, Nov 10 '10 at 21:13
@rwong: What do you mean it's not "so dependable?" It is dependable, but only under two circumstances, both of which are described in the question to which you've linked. — James McNellis, Nov 10 '10 at 21:38
@James McNellis: the rule only applies when the result is a true temporary *of the caller* (hence the reason the asker names the function `funcThatReturnsByValue()` ) If the function returns by reference (or other stuff) instead of by value, then it is a temporary whose lifetime is controlled by something else (such as the called function) and would have been destroyed before it is returned. So, *temporary* has a narrow meaning, and is not apparent to people outside the C++ compiler circle. It also would not be apparent to any code reviewers. — rwong, Nov 11 '10 at 07:35

score 4 · Accepted Answer · answered Nov 10 '10 at 19:34

4

This code is ill-formed. You can't default initialize or value initialize a reference.

If you actually had an expression inside of ref(), then yes, 12.2.5 would apply and the temporary would be destroyed when the constructor exits.

answered Nov 10 '10 at 19:34

James McNellis

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You'd think, but I can't get GCC 4.3.4 to actually do that. – Steve Jessop Nov 10 '10 at 19:47

score 2 · Answer 2 · answered Nov 10 '10 at 21:06

Your example isn't creating a temporary - to do that you need to change to:

    bar() : ref(foo()) {}

Now you're binding the reference to a temporary, and that temporary object will be destroyed at the end of the constructor. Your reference will be invalid, and that is Not A Good Thing.

score 1 · Answer 3 · answered Nov 10 '10 at 20:40

I guess what you want to do is:

bar() : ref(foo()) {}

but don't naively think that the lifetime of a temporary is extended until there's a reference to it. No, it actually is not. So, whether const or not, you sould initialize the reference with a normal object.

value-initialization of a const member reference

3 Answers3