Re: Strange warning from g++ "returning reference to temporary"

From:

Dragan Milenkovic <dragan@plusplus.rs>

Newsgroups:

comp.lang.c++.moderated

Date:

Sat, 6 Dec 2008 08:33:19 CST

Message-ID:

<ghb8qu$tpd$1@aioe.org>

irotas wrote:

Consider the following code:

struct Foo
{

   operator const char* const&() const
   {
     return s; // <-- "warning: returning reference to temporary"
   }

   operator char*&()
   {
     return s; // <-- no warning!
   }

   char* s;
};

Aside from the "why would you want to do that anyway?" (there is a
reason!), could someone please explain the warning, and why there is
no warning on the non-const version?

Should there be a warning on the non-const version, or is it actually
OK?

Just lose the "&":

operator const char * const () const

.... and you're ok.

Compiler is doing something like this:

const char * const & res = s;
return res;

.... and there is no direct binding of T2 to a reference to T1,
where T1 != T2 and T1,T2 are pointers defined in the standard.
I don't know of the _correct_ reason for this, but it's not
a problem since you can return a pointer by its value.

--
Best regards,
Dragan

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