A friend recently asked me whether Example 1 below is legal, and if so what it means. It led to a nice discussion I thought I’d post here. Since it was in close to GotW style already, I thought I’d do another honorary one after all these years… no, I have not made a New Year’s Resolution to resume writing regular GotWs. :-)

JG Questions

Q1: Is the following code legal C++?

// Example 1 string f() { return "abc"; } void g() {

const string& s = f();

cout << s << endl; // can we still use the "temporary" object?

}

A1: Yes.

This is a C++ feature… the code is valid and does exactly what it appears to do.

Normally, a temporary object lasts only until the end of the full expression in which it appears. However, C++ deliberately specifies that binding a temporary object to a reference to const on the stack lengthens the lifetime of the temporary to the lifetime of the reference itself, and thus avoids what would otherwise be a common dangling-reference error. In the example above, the temporary returned by f() lives until the closing curly brace. (Note this only applies to stack-based references. It doesn’t work for references that are members of objects.)

Does this work in practice? Yes, it works on all compilers I tried (except Digital Mars 8.50, so I sent a bug report to Walter to rattle his cage :-) and he quickly fixed it for the Digital Mars 8.51.0 beta).

Q2: What if we take out the const… is Example 2 still legal C++?

// Example 2 string f() { return "abc"; } void g() {

string& s = f(); // still legal?

cout << s << endl;

}

A2: No.

The "const" is important. The first line is an error and the code won’t compile portably with this reference to non-const, because f() returns a temporary object (i.e., rvalue) and only lvalues can be bound to references to non-const.

Note: Visual C++ does allow Example 2 but emits a "nonstandard extension used" warning by default. A conforming C++ compiler can always allow otherwise-illegal C++ code to compile and give it some meaning — hey, it could choose to allow inline COBOL if some kooky compiler writer was willing to implement that extension, maybe after a few too many Tequilas. For some kinds of extensions the C++ standard requires that the compiler at least emit some diagnostic to say that the code isn’t valid ISO C++, as this compiler does.

I once heard Andrei Alexandrescu give a talk on ScopeGuard (invented by Petru Marginean) where he used this C++ feature and called it "the most important const I ever wrote." And this brings us to the Guru Question, which highlights the additional subtlety that Andrei’s code deftly leveraged…

Guru Question

Q3: When the reference goes out of scope, which destructor gets called?

A3: The same destructor that would be called for the temporary object. It’s just being delayed.

Corollary: You can take a const Base& to a Derived temporary and it will be destroyed without virtual dispatch on the destructor call.

This is nifty. Consider the following code:

// Example 3 Derived factory(); // construct a Derived object void g() {

const Base& b = factory(); // calls Derived::Derived here

// … use b …

} // calls Derived::~Derived directly here — not Base::~Base + virtual dispatch!

Does this work in practice on real compilers? Yes: Every compiler I have access to calls the correct Derived destructor, including even ancient Borland 5.5 and Visual C++ 6.0 (and Digital Mars, though DM calls the destructor at the wrong time, as noted above).

Andrei leverages this subtlety (of course) in his ScopeGuard implementation to avoid making the implementation classes’ destructors virtual at all, which is okay in that case because those classes otherwise have no need for one.

Updates: