Re: Revistiing using return value as reference

James Kanze <>
Sun, 30 Dec 2007 13:52:11 -0800 (PST)
On Dec 27, 12:49 pm, Erik Wikstr=F6m <> wrote:

On 2007-12-27 02:19, johanatan wrote:

On Dec 26, 6:33 am, Erik Wikstr=F6m <> wrote:

On 2007-12-26 06:01, johanatan wrote:

On Dec 25, 3:00 pm, "Jim Langston" <> wrote:

Foo& Bar( int Val )
   return Foo( Val );

Will not work, can not convert Foo to Foo&

Should work, but not a very useful thing to do as local
copy dies immediately.

Should not work, a reference can not bind to a rvalue.

Ok, you're right. I just tried this again in Ch (a C++
interpreter) -- apparently the method wasn't being called
before and so wasn't compiled.

Does not work, same thing, can not convert Foo to Foo&.

Should also work. Something else is going on as that
syntax is compiling fine in my compiler.

Get a new compiler. Once again, you can not bind a
reference to a rvalue.

This is certainly an interesting design decision of C++. I
suspect it was a 'safety' feature built in, but I prefer
languages to be more flexible by inferring as much as
possible from context. What is the danger of binding to an

Consider the following code:

int& r = 5;

r = 6;

If I can bind a reference to a value I can also change that
value. But that is not how a reference works, a reference is a
alias for an object not an object on its own. Therefore a
reference can not bind to a rvalue.

That's an interesting example, and a very good reason for not
allowing using an rvalue to initialize a reference. Still, the
original language (pre-1988, roughly, so we're talking about a
very long time ago) did allow using an rvalue to initialize any
reference, with more or less the same semantics which currently
apply when an rvalue is used to initialize a reference to const.
The rule was changed because it was found, in practice, to be
too great a source of errors. IMHO, the real problem is that
there are too many implicit conversions, and that the result of
an implicit conversion is an rvalue; the errors are typically
cases where an implicit conversion resulted in an rvalue when
the programmer expected an lvalue. Things like:

    increment( int& i )
        ++ i ;

        unsigned x = 0 ;
        increment( x ) ;
        std::cout << x << std::endl ;
        return 0 ;

which, of course, results in 0.

James Kanze (GABI Software)
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