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Home/ Questions/Q 248643
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T21:21:37+00:00 2026-05-11T21:21:37+00:00

I’m reviewing a collegue’s code, and I see he has several constants defined in

  • 0

I’m reviewing a collegue’s code, and I see he has several constants defined in the global scope as:

const string& SomeConstant = "This is some constant text";

Personally, this smells bad to me because the reference is referring to what I’m assuming is an “anonymous” object constructed from the given char array.

Syntactically, it’s legal (at least in VC++ 7), and it seems to run, but really I’d rather have him remove the & so there’s no ambiguity as to what it’s doing.

So, is this TRULY safe and legal and I’m obsessing? Does the temp object being constructed have a guaranteed lifetime? I had always assumed anonymous objects used in this manner were destructed after use…


So my question could also be generalized to anonymous object lifetime. Does the standard dictate the lifetime of an anonymous object? Would it have the same lifetime as any other object in that same scope? Or is it only given the lifetime of the expression?


Also, when doing it as a local, it’s obviously scoped differently:

class A
{
    string _str;

public:
    A(const string& str) :
        _str(str)
    {
        cout << "Constructing A(" << _str << ")" << endl;
    }

    ~A()
    {
        cout << "Destructing A(" << _str << ")" << endl;
    }
};

void TestFun()
{
    A("Outer");
    cout << "Hi" << endl;
}

Shows:

Constructing A(Outer);
Destructing A(Outer);
Hi

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  1. Editorial Team
    Editorial Team
    2026-05-11T21:21:37+00:00Added an answer on May 11, 2026 at 9:21 pm

    It’s completely legal. It will not be destructed until the program ends.

    EDIT: Yes, it’s guaranteed:

    “All objects which do not have dynamic
    storage duration, do not have thread
    storage duration, and are not local
    have static storage duration. The
    storage for these objects shall last
    for the duration of the program
    (3.6.2, 3.6.3).”

    — 2008 Working Draft, Standard for Programming Language C++, § 3.7.1 p. 63

    As Martin noted, this is not the whole answer. The standard draft further notes (§ 12.2, p. 250-1):

    “Temporaries of class type are created
    in various contexts: binding an rvalue
    to a reference (8.5.3) […] Even when
    the creation of the temporary object
    is avoided (12.8), all the semantic
    restrictions shall be respected as if
    the temporary object had been created.
    […] Temporary objects are destroyed
    as the last step in evaluating the
    full-expression (1.9) that (lexically)
    contains the point where they were
    created. […] There are two contexts
    in which temporaries are destroyed at
    a different point than the end of the
    full-expression. […] 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 as specified
    below.”

    I tested in g++ if that makes you feel any better. 😉

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