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

From:

=?Utf-8?B?R2Vvcmdl?= <George@discussions.microsoft.com>

Newsgroups:

microsoft.public.vc.language

Date:

Sat, 15 Dec 2007 06:43:02 -0800

Message-ID:

<B58BC723-F8FC-4EB4-99EA-74294E4C3272@microsoft.com>

Hi Doug,

I do not quite understand why the following code can convert rvalue to
lvalue, could you provide more description please?

template<typename T>
inline T&
lvalue_cast(const T& rvalue)
{
return const_cast<T&>(rvalue);
}

"Doug Harrison [MVP]" wrote:

On Fri, 14 Dec 2007 13:02:20 -0500, David Wilkinson <no-reply@effisols.com>
wrote:

Of course it is easily fixed by doing

void g()
{
X x;
f(x);
}

Alternatively, you can use the lvalue_cast I mentioned and keep the
original syntax:

template<typename T>
inline T&
lvalue_cast(const T& rvalue)
{
   return const_cast<T&>(rvalue);
}

void g()
{
   f(lvalue_cast(X()));
}

Or for some X, you can use the "joke" I posted in the thread Carl linked
to:

   f(X() = X()); // Not serious

Only in C++ could assigning one rvalue to another rvalue produce a
modifiable lvalue. (This is actually the default for class types.)

--
Doug Harrison
Visual C++ MVP

regards,
George