How come a non-const reference cannot bind to a temporary object?


Question

Why is it not allowed to get non-const reference to a temporary object, which function getx() returns? Clearly, this is prohibited by C++ Standard but I am interested in the purpose of such restriction, not a reference to the standard.

struct X
{
    X& ref() { return *this; }
};

X getx() { return X();}

void g(X & x) {}    

int f()
{
    const X& x = getx(); // OK
    X& x = getx(); // error
    X& x = getx().ref(); // OK
    g(getx()); //error
    g(getx().ref()); //OK
    return 0;
}
  1. It is clear that the lifetime of the object cannot be the cause, because constant reference to an object is not prohibited by C++ Standard.
  2. It is clear that the temporary object is not constant in the sample above, because calls to non-constant functions are permitted. For instance, ref() could modify the temporary object.
  3. In addition, ref() allows you to fool the compiler and get a link to this temporary object and that solves our problem.

In addition:

They say "assigning a temporary object to the const reference extends the lifetime of this object" and " Nothing is said about non-const references though". My additional question. Does following assignment extend the lifetime of temporary object?

X& x = getx().ref(); // OK
1
218
12/6/2011 2:23:14 PM

Accepted Answer

From this Visual C++ blog article about rvalue references:

... C++ doesn't want you to accidentally modify temporaries, but directly calling a non-const member function on a modifiable rvalue is explicit, so it's allowed ...

Basically, you shouldn't try to modify temporaries for the very reason that they are temporary objects and will die any moment now. The reason you are allowed to call non-const methods is that, well, you are welcome to do some "stupid" things as long as you know what you are doing and you are explicit about it (like, using reinterpret_cast). But if you bind a temporary to a non-const reference, you can keep passing it around "forever" just to have your manipulation of the object disappear, because somewhere along the way you completely forgot this was a temporary.

If I were you, I would rethink the design of my functions. Why is g() accepting reference, does it modify the parameter? If no, make it const reference, if yes, why do you try to pass temporary to it, don't you care it's a temporary you are modifying? Why is getx() returning temporary anyway? If you share with us your real scenario and what you are trying to accomplish, you may get some good suggestions on how to do it.

Going against the language and fooling the compiler rarely solves problems - usually it creates problems.


Edit: Addressing questions in comment: 1) X& x = getx().ref(); // OK when will x die? - I don't know and I don't care, because this is exactly what I mean by "going against the language". The language says "temporaries die at the end of the statement, unless they are bound to const reference, in which case they die when the reference goes out of scope". Applying that rule, it seems x is already dead at the beginning of the next statement, since it's not bound to const reference (the compiler doesn't know what ref() returns). This is just a guess however.

2) I stated the purpose clearly: you are not allowed to modify temporaries, because it just does not make sense (ignoring C++0x rvalue references). The question "then why am I allowed to call non-const members?" is a good one, but I don't have better answer than the one I already stated above.

3) Well, if I'm right about x in X& x = getx().ref(); dying at the end of the statement, the problems are obvious.

Anyway, based on your question and comments I don't think even these extra answers will satisfy you. Here is a final attempt/summary: The C++ committee decided it doesn't make sense to modify temporaries, therefore, they disallowed binding to non-const references. May be some compiler implementation or historic issues were also involved, I don't know. Then, some specific case emerged, and it was decided that against all odds, they will still allow direct modification through calling non-const method. But that's an exception - you are generally not allowed to modify temporaries. Yes, C++ is often that weird.

93
10/14/2009 2:42:30 PM

In your code getx() returns a temporary object, a so-called "rvalue". You can copy rvalues into objects (aka. variables) or bind them to to const references (which will extend their life-time until the end of the reference's life). You cannot bind rvalues to non-const references.

This was a deliberate design decision in order to prevent users from accidentally modifying an object that is going to die at the end of the expression:

g(getx()); // g() would modify an object without anyone being able to observe

If you want to do this, you will have to either make a local copy or of the object first or bind it to a const reference:

X x1 = getx();
const X& x2 = getx(); // extend lifetime of temporary to lifetime of const reference

g(x1); // fine
g(x2); // can't bind a const reference to a non-const reference

Note that the next C++ standard will include rvalue references. What you know as references is therefore becoming to be called "lvalue references". You will be allowed to bind rvalues to rvalue references and you can overload functions on "rvalue-ness":

void g(X&);   // #1, takes an ordinary (lvalue) reference
void g(X&&);  // #2, takes an rvalue reference

X x; 
g(x);      // calls #1
g(getx()); // calls #2
g(X());    // calls #2, too

The idea behind rvalue references is that, since these objects are going to die anyway, you can take advantage of that knowledge and implement what's called "move semantics", a certain kind of optimization:

class X {
  X(X&& rhs)
    : pimpl( rhs.pimpl ) // steal rhs' data...
  {
    rhs.pimpl = NULL; // ...and leave it empty, but deconstructible
  }

  data* pimpl; // you would use a smart ptr, of course
};


X x(getx()); // x will steal the rvalue's data, leaving the temporary object empty

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