Re: A non-const reference may only be bound to an lvalue?

=?Utf-8?B?R2Vvcmdl?= <>
Sat, 15 Dec 2007 06:30:01 -0800
Hi Doug,

To my surprise, I have tested that the following code can compile,

struct X {


void f(X& x) {};

void g()

I think it breaks the rule that non-const reference parameter X& x can not
be binded to a rvalue (return of X(), which is a temporary object, i.e.
rvalue)? Any comments?

I am using Visual Studio 2008.

"Doug Harrison [MVP]" wrote:

On Fri, 14 Dec 2007 08:56:53 -0500, "Igor Tandetnik" <>

To avoid surprises like this:

void f(long& x) { x = 1; }

int y = 0;
f(y); // hypothetical, doesn't compile
assert(y == 1); // fails

If f(y) were allowed to compile, it would generate a temporary of type
long, and the function would modify that temporary, leaving the original
int variable unchanged.

It would also mean that a seemingly benign change in function
signature( from f(int&) to f(long&) ) may silently alter the behavior of
the program. As things stand now, such a change would lead to compiler
errors - much better than silent behavior change.

The thing is, that's got little to do with the typical desired usage, which
is to permit:

void f(X&);

void g()

There are legitimate reasons to want to do that, but I've come to accept it
shouldn't be allowed for this reason. Suppose you have a class:

struct Y
   // Lifetime of x must exceed object's use of it
   explicit Y(X& x)
   : m_x(x)

   X& m_x;

Whether or not you agree with the design of the class, under the existing
rule, you can't get into trouble binding a temporary to m_x, at least not
easily. And when you really need to say f(X()), there's always lvalue_cast.

Doug Harrison
Visual C++ MVP


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